A 280-m-wide river flows due east at a uniform speed of 4.7m/s. A boat with a speed of 7.1m/s relative to the water leaves the south bank pointed in a direction 26o west of north. What is the (a) magnitude and (b) direction of the boat's velocity relative to the ground

Answers

Answer 1

Answer:

(a) The speed is 7.96 m/s

(b) The direction is 76 degree from positive X axis in counter clockwise direction.  

Explanation:

Width of river = 280 m

speed of river, vR = 4.7 m/s towards east

speed of boat with respect to water, v(B,R) = 7.1 m/s at 26 degree west of north

[tex]vR = 4.7 i \\\\v(B,R) = 7.1 (- sin 26 i + cos 26 j) = - 3.1 i + 6.4 j[/tex]

(a) The velocity of boat with respect to ground is

[tex]\overrightarrow{v}_{(B,R)}=\overrightarrow{v}_{(B,G)}-\overrightarrow{v}_{(R,G)}\\\\- 3.1 \widehat{i} +6.4 \widehat{j}=\overrightarrow{v}_{(B,G)} - 4.7 \widehat{i}\\\\\overrightarrow{v}_{(B,G)} = 1.6 \widehat{i} + 6.4 \widehat{j}\\\\{v}_{(B,G)} = \sqrt{1.6^2 + 6.4^2}=6.96 m/s[/tex]

(b) The direction is given  by

[tex]tan\theta = \frac{6.4}{1.6} =4\\\\\theta = 76^o[/tex]


Related Questions


If you pitch a baseball with twice the kinetic energy you gave it in the
previous pitch, the magnitude of its momentum is

Answers

Answer:

the magnitude of momentum is √2≈ b

Explanation:

hope that helped

A car starts from rest and accelerates uniformly in a straight line in the positive x direction. After 25 seconds, its speed is 90 km/h.
a) Determine the acceleration of the object. [5]
b) How far does the object travel during the first 25 seconds? [3]
c) What is the average velocity of the object during the first 25 seconds?

Answers

Answer:

A. 1 m/s²

B. 312.5 m

C. 12.5 m/s

Explanation:

We'll begin by converting the velocity i.e 90 Km/h to m/s. This can be obtained as follow:

Velocity (Km/h) = 90 Km/h

Velocity (m/s) =?

Velocity (m/s) = Velocity (Km/h) × 1000 / 3600

Velocity (m/s) = 90 × 1000 / 3600

Velocity (m/s) = 90000 / 3600

Velocity (m/s) = 25 m/s

A. Determination of the acceleration.

Initial velocity (u) = 0 m/s

Final velocity (v) = 25 m/s

Time (t) = 25 s

Acceleration (a) =?

v = u + at

25 = 0 + (a × 25)

25 = 0 + 25a

25 = 25a

Divide both side by 25

a = 25/25

a = 1 m/s²

B. Determination of the distance travelled.

Initial velocity (u) = 0 m/s

Final velocity (v) = 25 m/s

Acceleration (a) = 1 m/s²

Distance travelled (s) =?

v² = u² + 2as

25² = 0 + (2 × 1 × s)

625 = 0 + 2s

625 = 2s

Divide both side by 2

s = 625 / 2

s = 312.5 m

C. Determination of the average velocity.

Total distance travelled = 312.5 m

Total time = 25 s

Average velocity =?

Average velocity = Total distance / total time

Average velocity = 312.5 / 25

Average velocity = 12.5 m/s

which object has potential energy but not kinetic energy

Answers

Answer:

A ball resting on the edge of a cliff

Explanation:

An object must be in motion or be moving to have kinetic energy. Since the ball is resting on the edge of a cliff, it is not actually moving so it does not have kinetic energy but the resting place of the ball is potential energy.

Answer:
A ball at the top of a ramp.
Explanation:
The higher the object is raised above the ground, the more energy it possesses.
(There are many other alternative answers like: fossil fuels, food, batteries but the explanation is different from the ball at the top of a ramp)

A 50mf capacitor, a 0.3H inductor and an 80 ohm resistor is connected in series with a 120v, 60Hz power source
What is the impedance of a circuit? ​

Answers

Answer:

Z = 138.5 Ω

Explanation:

In a series RLC circuit the impedance is

          Z = [tex]\sqrt{R^2 + ( X_L - X_C)^2 }[/tex]

the capacitive impedance is

         X_C = 1 / wC

the inductive impedance is

         X_L = wL

in this exercise indicate that C = 50 10⁻³ F, L = 0.3 H and the frequency is f=60 Hz

angular velocity and frequency are related

         w = 2π f

         w = 2π 60

         w = 376.99 rad / s

let's calculate

        Z = [tex]\sqrt{80^2 + ( 376.99 \ 0.3 - \frac{1}{376.99 \ 50 \ 10^{-3}} )^2 }[/tex]

        Z = [tex]\sqrt{6400 + ( 113.1 - 0.053)^2}[/tex]

        Z = √19179.6

        Z = 138.5 Ω

The engine of a locomotive exerts a constant force of 6.8 105 N to accelerate a train to 80 km/h. Determine the time (in min) taken for the train of mass 1.1 107 kg to reach this speed from rest.

Answers

Answer:

t = 6 minutes

Explanation:

Given that,

Force,[tex]F=6.8\times 10^5\ N[/tex]

Initial speed of the train, u = 0

Final speed of the train, v = 80 km/h = 22.22 m/s

The mass of the train, [tex]m=1.1\times 10^7\ kg[/tex]

We need to find the time taken by the train to come to rest. We know that,

F = ma

[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.1\times10^7\times (22.22-0)}{6.8\times 10^5}\\\\t=359.44\ s[/tex]

or

t = 6 minutes (approx)

So, the required time is equal to 6 minutes.

Unpolarized light of intensity 0.0288 W/m2 is incident on a single polarizing sheet. What is the rms value of the electric field component transmitted

Answers

Answer:

the rms value of the electric field component transmitted is 3.295 V/m

Explanation:

Given;

intensity of the unpolarized light, I = 0.0288 W/m²

For unpolarized light, the relationship between the amplitude electric field and intensity is given as;

[tex]E_{max} = \sqrt{2\mu_0cI} \\\\E_{max} = \sqrt{2(4\pi \times 10^{-7})(3\times 10^8)(0.0288)} \\\\E_{max} = 4.66 \ V/m[/tex]

The relationship between the rms value of the electric field and the amplitude electric field is given as;

[tex]E_{rms} = \frac{E_0}{\sqrt{2} } =\frac{E_{max}}{\sqrt{2} } \\\\E_{rms} = \frac{4.66}{\sqrt{2} }\\\\E_{rms} = 3.295 \ V/m[/tex]

Therefore, the rms value of the electric field component transmitted is 3.295 V/m

how many protons does a neutral atom of oxygen-16 have

Answers

Answer:

eight

Explanation:

it's atomic number is 8 which mean that an oxygen atom has eight protons in it's nucleus

A force of 1.35 newtons is required to accelerate a book by 1.5 meters/second2 along a frictionless surface. What is the mass of the book?

Answers

Answer:

0.9 kg

Explanation:

We would use the equation F=m*a to solve this equation. First, we would need to get mass by itself therefore we divide out acceleration from both sides ( F/a=m*a / a ) acceleration would cancel out and the end equation should look like this ( F/a = m or m = F/a)  After we do that we plug in the numbers 1.35 N / 1.5 m/s^2 we get 0.9 kg, assuming you are using kg.

Answer: 0.90 kilograms

Explanation:

A soccer player kicks a ball. Why does the action force exerted by the player's foot cause a different motion than the reaction force?

The reaction force is greater than the action force.

The action force and the reaction force act on different objects.

The action force is greater than the reaction force.

The action force and the reaction force act in opposite directions.

Answers

Answer:

D because of POE

Explanation:

the reaction force is the ball exerting the same newtons of force back to your leg opposite of the ball, but the reaction force and the action force is never stronger than each-other. and action is only being done on the soccer ball, so process of elimination, the answer is D

Answer:

B

Explanation:

it can only have a different motion if it is acted upon a different ball

A train is moving with a speed of 100 m/s. If the train is traveling south, at what position will it be 3 minutes after passing the +1,000-meter position marker ?
Remember, south is the negative direction and when you use the time, it must be in units of seconds. You will be applying one of the equations from above to solve this problem. And you must include a + sign if the final position is positive or a - sign if the final position is negative.

Answers

Answer:

The position after 3 minutes is - 800 m.

Explanation:

speed, v = 100 m/s

time, t = 3min = 180 s

initial position, x = 1000 m

let the distance traveled in 3 minutes is d

d = 100 x 180 = - 18000 m

So, the position is

=  - 18000 + 1000 = - 800 m

Three ice skaters, numbered 1, 2, and 3, stand in a line, each with her hands on the shoulders of the skater in front. Skater 3, at the rear, pushes forward on skater 2. Assume the ice is frictionless.
Identify all the action/reaction pairs of forces between the three skaters. Draw a free-body diagram for skater 2, in the middle.

Answers

Answer:

Following are the responses to the given question:

Explanation:

The mass (mg) is downwards, the typical [tex]N_2[/tex] pressure is upward. This pushing force of [tex]\vec{F}_{3 \to 2}[/tex] is pushed forward by skater 3 on skater 2. (considered as rightward direction). The strength of the slater 2 in reply is the slater [tex]\vec{F}_{2\to 3}[/tex] Skater three moves to the left.

Its push force [tex]\vec{F}_{2\to1}[/tex] imparted to the skateboarder 2 in relation those above forces The force[tex]\vec{F}_ {1\to 2}[/tex] on skater 2 by skater 1 is as a response, to a left. Skater 2's free body system is as follows:

During one trial, the acceleration is 2m/s^2 to the right. What calculation will give the tensions in actin filaments during this trial

Answers

Answer: hello your question is poorly written attached below is the complete question

answer :

TA = 1.6*10^-24 * 60 * 2,  TB = 1.6*10^-24 * ( 60 + 30 ) * 2  -- ( option 1 )

Explanation:

a = 2m/s^2

Ta = m₁ a = 60 * 1.6 * 10^-24 * 2 ц

Tb - Ta = m₂ a

∴ Tb = m₂ a  + Ta

       = ( 30 * 1.6 * 10^-24 * 2 ) +  ( 60 * 1.6 * 10^-24 * 2 )

= ( 30 + 60 ) * 1.6 * 10^-24 * 2 ц

In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer spce. From stationary positions, they push against each other. Bonzo flies off with a velocity of 1.1 m/s, while Ender recoils with a velocity of -4.3 m/s. Determine the ratio Bonzo/mEnder of the masses of these two enemies.

Answers

Answer:

the ratio Bonzo/mEnder of the masses of these two enemies is 3.91

Explanation:

Given the data in the question;

Velocity of Bonzo [tex]V_{Bonzo[/tex] = 1.1 m/s

Velocity of Ender [tex]V_{Ender[/tex] = -4.3 m/s

the ratio Bonzo/mEnder of the masses of these two enemies = ?

Now, using the law of conservation of momentum.

momentum of both Bonzo and Ender are conserved

so

Initial momentum = final momentum

we have

0 = [tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] + [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex]

[tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] = -[ [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex]  ]

[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex]  = -[ [tex]V_{Ender[/tex] / [tex]V_{Bonzo[/tex] ]

we substitute

[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex]  = -[ -4.3 m/s / 1.1 m/s ]

[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex]  = -[ -3.9090 ]

[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex]  = 3.91

Therefore, the ratio Bonzo/mEnder of the masses of these two enemies is 3.91

A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal energy of the box, determine the magnitude of the increase.

Answers

Answer:

Explanation:

From the given information:

The initial PE [tex](PE)_i[/tex] = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = [tex]P.E_f -P.E_i[/tex]

ΔP.E = 0 - [tex](PE)_i[/tex]

ΔP.E = [tex]-P.E_i[/tex]

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

[tex]\Delta P.E + \Delta K.E + \Delta U = 0[/tex]

[tex]\Delta U = -\Delta P.E - \Delta K.E[/tex]

this can be re-written as:

[tex]\Delta U =- (-\Delta P.E_i) - \Delta K.E[/tex]

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

[tex]\Delta U =(70\%) \Delta P.E_i-0[/tex]

[tex]\Delta U =(0.70) (490.5)[/tex]

ΔU = 343.35  J

Thus, the magnitude of the increase is = 343.35 J

jonatha want to put ketchup o​​​​​n his hamburger.he truns the ketchup bottle at an angle toward his plate and smacks the bottom of the bottle until the ketchup comes.what is unblanced force​​​​​​​​​​​​​​​​​​​​​​​​​​​​​

Answers

Answer:

The force is Inertia

Explanation:

This is an example: He turns the ketchup bottle at an angle toward his plate and smacks the bottom of the bottle until the ketchup comes out. How does inertia affect the ketchup in the bottle? The Inertia keeps the ketchup in the bottle. The Jacksons are driving to the lake when a car in front of theirs slams on its brakes.

Your welcome! :)

If a magnifying glass has a power of 10.0 D, what is the magnification it produces when held 6.55 cm from an object?

Answers

Answer: The magnification of the magnifying glass is -2.9

Explanation:

The equation for power is given as:

[tex]P=\frac{1}{f}[/tex]

where,

P = Power = 10 D

f = focal length

Putting values in above equation, we get:

[tex]f=\frac{1}{10}=0.1m=10 cm[/tex]            (Conversion factor: 1 m = 100 cm)

The equation for lens formula follows:

[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}[/tex]

where,

v = image distance

u = Object distance = 6.55 cm

Putting values in above equation, we get:

[tex]\frac{1}{v}=\frac{1}{10}-\frac{1}{(6.55)}\\\\\frac{1}{v}=\frac{6.55-10}{10\times 6.55}\\\\v=\frac{65.5}{-3.45}=-18.98cm[/tex]

Magnification (m) can be written as:

[tex]m=\frac{-v}{u}[/tex]

Putting values in above equation, we get:

[tex]m=\frac{-18.98}{6.55}\\\\m=-2.9[/tex]

Hence, the magnification of the magnifying glass is -2.9

PLS HELP ME 100 POINTS PLS I NEED HELP QUICK PLS


For this project, you are expected to submit the following:
1. Your Student Guide with completed Student Worksheet
2. Your scale model of the solar system
Step 1: Prepare for the project.
a) Read through the guide before you begin so you know the expectations for this project.
b) If anything is not clear to you, be sure to ask your teacher.
Step 2: Conduct research on the actual sizes of the planets.
a) Do research to find the actual sizes of the Sun and the planets. This information is typically represented as diameter in kilometers (km). Recall that diameter is the length of the imaginary straight line from one side of a figure, such as a sphere, to the opposite side of the figure. This line passes through the center of the figure.
b) Record the actual diameters of the Sun and the planets in the first column of the table in the Student Worksheet.
c) Copy the link of the website you used into the space provided in the Student Worksheet.
Step 3: Determine the scaled sizes of the planets.
a) Go to a reliable website to find a solar system model calculator.
b) Decide how big you want the Sun in your model to be. For example, you could assign your Sun to be 300 mm. Input this figure in the calculator, and the calculator will determine the diameters of the eight planets for you. You want to make sure that the Sun is big enough so that the smallest planet will still be big enough to draw.
c) Record information from the calculator in the second column of the table in the Student Worksheet.
d) Copy the link of the website you used into the space provided in the Student Worksheet.
Step 4: Create a scale model of the solar system.
a) Draw and cut construction paper models of the Sun and the planets using the scaled measurements from the table.
b) Glue the models on the poster board. You can glue or tape poster boards together if necessary. Be sure to put the Sun in the center and to put the planets and a drawing of their orbits in order from nearest to farthest from the Sun.
Note: Remember that in this model, the diameter of the planets is scaled but the distance of the planets from the Sun is not. That means your model does not accurately represent the distances of the planets from the Sun so you need not worry about these measurements.
c) Label the Sun and the planets.
d) Put an attention-catching title above or below your model.
e) Write your name on the back of your poster board.
Step 5: Complete the Student Worksheet.
a) Make sure the table in the Student Worksheet is complete.
b) Answer the questions in the Student Worksheet.
c) Check to make sure you added the sources you used for this project in the Student Worksheet.
Step 6: Evaluate your project using this checklist.
If you can check each of the following boxes, you are ready to submit your project.
 Did you conduct research to find the actual size of the Sun and the planets? Did you record this information in the table in the Student Worksheet?
 Did you use a solar system model calculator to determine the scaled size of the Sun and planets? Did you record this information in the Student Worksheet?
 Did you add the links of the websites you used for this project to the Student Worksheet?
 Did you use the scaled sizes to create models of the Sun and the planets?
 Did you put your model together in a way that represents the solar system (Sun in the center and planets in order from nearest to farthest from the Sun)?
 Did you label each component of your model?
 Did you add an attention-catching title above or below your model?
 Did you write your name on the back of your poster board?
 Did you complete the Student Worksheet at the end of this guide?
Step 7: Revise and submit your project.
a) If you were unable to check off all the requirements on the checklist, go back and make sure that your project is complete. Save your project before submitting it.
b) Turn in your scale model of the solar system to your teacher. Be sure that your name is on it.
c) Submit your Student Guide through the virtual classroom.
d) Congratulations! You have completed your project.

Answers

Answer

I hope this help....

Explanation:

Answer:

Hope this helps

Explanation:

A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equilibrium point and released. Determine
a) the spring constant
b) the maximum velocity of the mass
c) the maximum acceleration of the mass
d) the total mechanical energy of the mass
e) the period and frequency of the mass and spring and
f) the equation of time-dependent vertical position of the mass

Answers

Answer:

a)  [tex]k=19.6N/m[/tex]

b)  [tex]V_m=0.81m/s[/tex]

c)  [tex]a_m=6.561m/s^2[/tex]

d)  [tex]K.E=0.096J[/tex]

e)  [tex]T=0.78sec[/tex] &[tex]F=1.29sec[/tex]

f)   [tex]mx'' + kx' =0[/tex]

Explanation:

From the question we are told that:

Stretch Length [tex]L=0.150m[/tex]

Mass [tex]m=0.30kg[/tex]

Total stretch length[tex]L_t=0.150+0.100=>0.25[/tex]

a)

Generally the equation for Force F on the spring is mathematically given by

[tex]F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}[/tex]

[tex]k=19.6N/m[/tex]

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

[tex]V_m=A\omega[/tex]

Where

A=Amplitude

[tex]A=0.100m[/tex]

And

[tex]\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s[/tex]

Therefore

[tex]V_m=A\omega\\\\V_m=8.1*0.1[/tex]

[tex]V_m=0.81m/s[/tex]

c)

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

[tex]a_m=\omega^2A[/tex]

[tex]a_m=8.1^2*0.1[/tex]

[tex]a_m=6.561m/s^2[/tex]

d)

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

[tex]K.E=\frac{1}{2}mv^2[/tex]

[tex]K.E=\frac{1}{2}*0.3*0.8^2[/tex]

[tex]K.E=0.096J[/tex]

e)

Generally the equation for  the period T is mathematically given by

[tex]\omega=\frac{2\pi}{T}[/tex]

[tex]T=\frac{2*3.142}{8.1}[/tex]

[tex]T=0.78sec[/tex]

Generally the equation for  the Frequency is mathematically given by

[tex]F=\frac{1}{T}[/tex]

[tex]F=1.29sec[/tex]

f)

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

[tex]mx'' + kx' =0[/tex]

Where

'= signify order of differentiation

The velocity of an object traveling in a circle is quadrupled and its radius is tripled The acceleration of this object will change by factor of?

Answers

Answer:

The process of solving a circular motion problem is much like any other problem in physics class. The process involves a careful reading of the problem, the identification of the known and required information in variable form, the selection of the relevant equation(s), substitution of known values into the equation, and finally algebraic manipulation of the equation to determine the answer. Consider the application of this process to the following two circular motion problems.

 

Sample Problem #1

A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.

The solution of this problem begins with the identification of the known and requested information.

Known Information:

m = 900 kg

v = 10.0 m/s

R = 25.0 m

Requested Information:

a = ????

Fnet = ????

To determine the acceleration of the car, use the equation a = v2 / R. The solution is as follows:

a = v2 / R

a = (10.0 m/s)2 / (25.0 m)

a = (100 m2/s2) / (25.0 m)

a = 4 m/s2

To determine the net force acting upon the car, use the equation Fnet = m•a. The solution is as follows.

Fnet = m • a

Fnet = (900 kg) • (4 m/s2)

Fnet = 3600 N

 

 

Sample Problem #2

A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback.

The solution of this problem begins with the identification of the known and requested information.

Known Information:

m = 95.0 kg

R = 12.0 m

Traveled 1/4-th of the circumference in 2.1 s

Requested Information:

v = ????

a = ????

Fnet = ????

To determine the speed of the halfback, use the equation v = d / t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows:

v = d / t

v = (0.25 • 2 • pi • R) / t

v = (0.25 • 2 • 3.14 • 12.0 m) / (2.1 s)

v = 8.97 m/s

To determine the acceleration of the halfback, use the equation a = v2 / R. The solution is as follows:

a = v2 / R

a = (8.97 m/s)2 / (12.0 m)

a = (80.5 m2/s2) / (12.0 m)

a = 6.71 m/s2

To determine the net force acting upon the halfback, use the equation Fnet = m•a. The solution is as follows.

Fnet = m*a

Fnet = (95.0 kg)*(6.71 m/s2)

Fnet = 637 N

 

In Lesson 2 of this unit, circular motion principles and the above mathematical equations will be combined to explain and analyze a variety of real-world motion scenarios including amusement park rides and circular-type motions in athletics.

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a 230 kg roller coaster reaches the top of the steepest hill with a speed of 6.2 km/h. It then descends the hill, which is at an angle of 45 and is 51.0 m long/ What will its kinetic energy be wehn it reaches the bottom

Answers

Answer: 81.619 kJ

Explanation:

Given

Mass of roller coaster is [tex]m=230\ kg[/tex]

It reaches the steepest hill with speed of [tex]u=6.2\ km/h\ or \ 1.72\ m/s[/tex]

Hill to bottom is 51 m long with inclination of [tex]45^{\circ}[/tex]

Height of the hill is [tex]h=51\sin 45^{\circ}=36.06\ m[/tex]

Conserving energy to get kinetic energy at bottom

Energy at top=Energy at bottom

[tex]\Rightarrow K_t+U_t=K_b+U_b\\\Rightarrow \dfrac{1}{2}mu^2+mgh=K_b+0\\\\\Rightarrow K_b=0.5\times 230\times 1.72^2+230\times 9.8\times 36.06\\\Rightarrow K_b=340.216+81,279.24\\\Rightarrow K_b=81,619.456\ J\\\Rightarrow K_b=81.619\ kJ[/tex]

What recommendations would you give to the global government to help Decrease the global effects of human impact on the environment mystery recommendations and how they will positively impact our planet

Answers

Answer:

We can help to keep it magnificent for ourselves, our children and grandchildren, and other living things besides us.

Explanation:

5 ways our governments can confront climate change

PROTECT AND RESTORE KEY ECOSYSTEMS

SUPPORT SMALL AGRICULTURAL PRODUCERS

PROMOTE GREEN ENERGY

COMBAT SHORT-LIVED CLIMATE POLLUTANTS

BET ON ADAPTATION, NOT JUST MITIGATION

Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up for 391 m with an acceleration of +18.9 m/s2. A parachute then opens, slowing the car down with an acceleration of -9.92 m/s2. How fast is the racer moving 332 m after the parachute opens?​

Answers

Answer:

V = 90.51 m/s

Explanation:

From the given information:

Initial speed (u) = 0

Distance (S) = 391 m

Acceleration (a) = 18.9 m/s²

Using the relation for the equation of motion:

v² - u² = 2as

v² - 0² = 2as

v² = 2as

[tex]v = \sqrt{2as}[/tex]

[tex]v = \sqrt{2*18.9*391}[/tex]

v = 121.57 m/s

After the parachute opens:

The initial velocity = 121.57 m/ss

Distance S' = 332 m

Acceleration = -9.92 m/s²

How fast is the racer can be determined by using the relation:

[tex]V= \sqrt{v^2 + 2aS'}[/tex]

[tex]V = \sqrt{121.57^2+ 2 (-9.92)(332)}[/tex]

V = 90.51 m/s

A rotating fan completes 1150 revolutions every minute. Consider the tip of a blade, at a radius of 20.0 cm. (a) Through what distance does the tip move in one revolution

Answers

Answer:

125.6 cm

Explanation:

Applying,

S = 2πr................... Equation 1

Where S = distance moved by the tip in one revolution, r = radius of the rotating fan, π = pie

From the question,

Given: r = 20 cm,

Constant: π = 3.14

Substitute these values into equation 1

S = 2(3.14)(20)

S = 125.6 cm

Hence the distance moved by the tip in one revolution is 125.6 cm

two particles woth each charge magnitude 2.0×10^-7 c but opposite signs are held 15cm apart.what are the magnitude and direction of the electric field E at tge point midway between charges​

Answers

Answer:

The magnitude of the electric field strength is 6.4 x 10⁵ N/C, directed from positive particle to negative particle.

Explanation:

Given;

charge of each particle, Q = 2 x 10⁻⁷ C

distance between the two charges, r = 15 cm = 0.15 m

distance midway between the charges = 0.075 m

The magnitude of the electric field is calculated as;

[tex]E_{net} = E_{+q} + E_{-q}\\\\E_{net} = \frac{kQ}{r_{1/2}^2} + \frac{kQ}{r_{1/2}^2}\\\\E_{net} = 2(\frac{kQ}{r_{1/2}^2})\\\\E_{net} = 2 (\frac{9\times 10^9 \ \times 2\times 10^{-7}}{0.075^2} )\\\\E_{net} = 6.4\times 10^5 \ N/C[/tex]

The direction of the electric field is from positive particle to negative particle.

The diagram shows the molecular structure of ethane. What is the chemical
formula for ethane?
Ethane
H H
H-C-C-H
| |
H H

Answers

D
The C comes first and as there r 2 it would like like C2.
Then count how many h’s there r=6
So the overall formula should be C2H6

List and briefly explain the incidents leading to the occurrence of any five nuclear accidents that have taken place in different parts of the world.​

Answers

Answer:

Chernobyl Nuclear Disaster  Nuclear Disaster. Japan 2011  Kyshtym Nuclear Disaster. Russia 1957  Windscale Fire Nuclear Disaster. Sellafield, UK 1957 Three Mile Island Nuclear Accident. Pennsylvania, USA 1979

Explanation:

Hope this helps... pls vote as brainliest

Which statement accurately represents the arrangement of electrons in Bohr’s atomic model?


Electrons move randomly in the relatively large space surrounding the nucleus.

Electrons move around the nucleus in fixed orbits of equal levels of energy.

Electrons move around the nucleus in fixed orbits of increasing levels of energy.

Electrons vibrate in fixed locations around the nucleus.

Answers

Answer: Electrons move around the nucleus in fixed orbits of equal levels of energy

Explanation:

The statement that accurately represents the arrangement of electrons in Bohr’s atomic model is that the electrons move around the nucleus in fixed orbits of equal levels of energy.

It should be noted that the electrons have a fixed energy level when they travel around the nucleus in with energies which varies for different levels.

Higher energy levels are depicted by the orbits that are far from the nucleus. There's emission of light when the electrons then return back to a lower energy level.

What is the SI unit for momentum?
0 kg.m
m
O kg
S
m
O kg
S
O kg •
m

Answers

Mark brainliest please

Answer is : SI unit is kilogram meter per second (kg⋅m/s)

Momentum is a measurement of mass in motion: how much mass is in how much motion. It is usually given the symbol p. Where m is the mass and v is the velocity. The standard units for momentum are k g ⋅ m / s \mathrm{kg \cdot m/s} kg⋅m/sk, g, dot, m, slash, s, and momentum is always a vector quantity.

In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 70 kg, are pulled at a constant speed by a tow rope that maintains a constant tension of 350 N.

Required:
How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?

Answers

Answer:

21420 J

Explanation:

Given that:

mass of the rider = 70 kg

the tension of the rope = 350 N

Using the concept of conservation of energy;

Work done = change in the Total energy ( ΔT.E)

where;

Work done (W) = ΔK.E + ΔP.E + ΔThermal Energy

Recall that the man proceeds with a constant speed, thus the change in K.E energy will be zero.

As such:

W  = ΔP.E + ΔThermal Energy

We can now say that:

The thermal energy = W - ΔP.E

here;

W = force × displacement

The thermal energy = (350 × 120) - (70 × 9.8 × 30)

= 42000 - 20580

= 21420 J

Definition of Xenophobia​

Answers

Answer:

dislike of or prejudice against people from other countries.

Explanation:

Answer: dislike of or prejudice against people from other countries.

Explanation:

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