Answer:
5
Explanation:
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
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]
Ion how to do this at all
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.
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
1. Lifting an elevator 18m takes 100kJ. If doing so takes 20s, what is the average power of the elevator during the process?
2. How much work can a 0.4 hp electric mixer do in 15 s?
Answer:
1. Power = 5000 Watts
2. Workdone = 11185.5 Joules
Explanation:
Given the following data;
1. Distance = 18 m
Energy = 100 KJ = 100,000 Joules
Time = 20 seconds
To find the average power of the elevator;
Power = energy/time
Power = 100000/20
Power = 5000 Watts
2. Power = 0.4 HP
Time = 15 seconds
Conversion:
1 horsepower = 745.7 Watts
0.4 horsepower = 0.4 * 745.7 = 298.28 Watts
To find the amount of work done by the electric mixer;
Work done = power * time
Workdone = 745.7 * 15
Workdone = 11185.5 Joules
Give the number of protons and the number of neutrons in the nucleus of each of the following isotopes Aluminum 25 :13 protons and 12 neutrons
Answer:
No of proton is 13 and nucleus is 13
the speed of the bus is 40km/hr. what does it mean?
Answer:
The speed of the bus is 40 km/hr so this means the bus is travelling at a speed of 40 km per hour.
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.
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
During one trial, the acceleration is 2m/s^2 to the right. What calculation will give the tensions in actin filaments during this trial
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
Definition of Xenophobia
Answer:
dislike of or prejudice against people from other countries.
Explanation:
Answer: dislike of or prejudice against people from other countries.
Explanation:
suppose a 1 square meter panel of colar cells has an efficiency of 20% and recieves the equivlent of 6 hours of direct sunlight per day. What average power, in watts, does the panel produce
Answer:
The average power per day is 1008 kW.
Explanation:
Solar constant = 1.4 kW/m2
efficiency = 20 %
area, a = 1 square meter
time = 6 hours
Energy falling on the panel in 6 hours = 1.4 x 6 x 3600 kJ
The output is
= 20 % of 1.4 x 6 x 3600
= 0.2 x 1.4 x 6 x 3600
= 6048 kJ
Average power per day is
= 6048/6 = 1008 kW
when a temparature of a coin is 75°C, the coin's diameter increases. if the original diameter of a coin is 1.8*10^-2 m and its coefficient of linear expansion is 1.7*10^5/°C, what is the change in coins diameter?
Answer:
ΔD = 2.29 10⁻⁵ m
Explanation:
This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation
ΔA = 2α A ΔT
the area is
A = π r² = π D² / 4
we substitute
ΔA = 2α π D² ΔT/4
as they do not indicate the initial temperature, we assume that ΔT = 75ºC
α = 1.7 10⁻⁵ ºC⁻¹
we calculate
ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4
ΔA = 6.49 10⁻⁷ m²
by definition
ΔA = A_f- A₀
A_f = ΔA + A₀
A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4
A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴
A_f = 2,551 10⁻⁴ m²
the area is
A_f = π D_f² / 4
A_f = [tex]\sqrt{4 A_f /\pi }[/tex]
D_f = [tex]\sqrt{4 \ 2.551 10^{-4} /\pi }[/tex]
D_f = 1.80229 10⁻² m
the change in diameter is
ΔD = D_f - D₀
ΔD = (1.80229 - 1.8) 10⁻² m
ΔD = 0.00229 10⁻² m
ΔD = 2.29 10⁻⁵ m
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.
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
A two-slit interference experiment in which the slits are 0.200 mm apart and the screen is 1.00 m from the slits. The m = 1 bright fringe in the figure is 9.49 mm from the central bright fringe. Find the wavelength of the ligh
Answer:
λ = 1.90 10⁻⁶ m
Explanation:
The interference pattern for the two-slit case is
d sin θ = m λ
let's use trigonometry
tan θ = y / L
interference experiments angles are small
tan θ = sin θ /cos θ = sin θ
sin θ = y / L
we substitute
d y / L = m λ
λ = [tex]\frac{ d \ y}{m \ L}[/tex]
we calculate
λ = 0.2000 10⁻³ 9.49 10⁻³ / (1 1.00)
λ = 1.898 10⁻⁶ m
λ = 1.90 10⁻⁶ m
The wavelength of the light after calculation is find out to be λ = 1.90 *10⁻⁶ m
What is wavelength?
The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.The wavelength is also defined as the distance between two locations in a wave that have the same oscillation phase.
The interference pattern for the two-slit case is
d sin θ = m λ
let's use trigonometry
[tex]tan\theta=\dfrac{y}{L}[/tex]
interference experiments angles are small
[tex]sin\theta=\dfrac{y}{L}[/tex]
we substitute
[tex]\dfrac{dy}{L}=m\lambda[/tex]
[tex]\lambda=\dfrac{dy}{mL}[/tex]
we calculate
[tex]\lambda=\dfrac{0.2\times 10^{-3}\times 9.49\times 10^{-3}}{1\times 1}[/tex]
[tex]\lambda=1.90\times 10^{-6}\ m[/tex]
Hence the wavelength of the light after calculation is find out to be λ = 1.90 *10⁻⁶ m
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If you pitch a baseball with twice the kinetic energy you gave it in the
previous pitch, the magnitude of its momentum is
Answer:
the magnitude of momentum is √2≈ b
Explanation:
hope that helped
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
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
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?
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:
jonatha want to put ketchup on 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
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! :)
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
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
A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they stick together. What is the total momentum of the system after the collision? What is the total momentum of the system before the collision? What is the velocity of the cars after the collision?
Answer:
The total momentum of the cars before the collision is 61,000 kg.m/s
The total momentum of the cars after the collision is 61,000 kg.m/s
The velocity of the cars after the collision is 27.727 m/s
Explanation:
Given;
mass of the first car, m₁ = 1000 kg
initial velocity of the car, u₁ = 25 m/s
mass of the second car, m₂ = 1200 kg
initial velocity of the second car, u₂ = 30 m/s
The common velocity of the cars after collision = v
The total momentum of the cars before collision is calculated as;
P₁ = m₁u₁ + m₂u₂
P₁ = (1000 x 25) + (1200 x 30)
P₁ = 61,000 kg.m/s
The total momentum of the cars after collision is calculated as;
P₂ = m₁v + m₂v
where;
v is the common velocities of the cars after collision since they stick together.
P₂ = v(m₁ + m₂)
To determine "v" apply the principle of conservation of linear momentum for inelastic collision.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
(1000 x 25) + (1200 x 30) = v(1000 + 1200)
61,000 = 2,200v
v = 61,000/2,200
v = 27.727 m/s
The total momentum after collsion = v(m₁ + m₂)
= 27.727(1000 + 1200)
= 61,000 kg.m/s
Thus, momentum before and after collsion are equal.
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).
The question is incomplete. The complete question is :
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).
Mass of the ball : 16.3 g
Predicted range : 0.3503 m
Actual range : 1.09 m
Solution :
Given that :
The predicted range is 0.3503 m
Time of the fall is :
[tex]$t=\sqrt{\frac{2H}{g}}$[/tex]
[tex]v_1t= 0.35[/tex] ...........(i)
[tex]v_0t= 1.09[/tex] ...........(ii)
Dividing the equation (ii) by (i)
[tex]$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$[/tex]
∴ [tex]v_0=3.11 \ v_1[/tex]
Now loss of energy = change in the kinetic energy
[tex]$W=\frac{1}{2} m [v_0^2-v_1^2]$[/tex]
[tex]$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$[/tex]
[tex]$W=7.307\times 10^{-3} \ v_0^2$[/tex]
If f is average friction force, then
(f)(L) = W
(f) (1) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]
(f) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]
The Average force of friction is ( F ) = 7.307 * 10⁻³ v₀²
Given data:
Predicted range ( v₁t ) = 0.3503 m
Actual range ( v₀t ) = 1.09 m
mass = 16.3 g
First step : Determine the value of V₀
[tex]t = \sqrt{\frac{2H}{g} }[/tex] , v₁t = 0.3503 , ( v₀t ) = 1.09 m
To obtain the value of V₀
Divide ( v₀t ) by ( v₁t ) = 1.09 / 0.3503 = 3.11 v₁
∴ V₀ = 3.11 v₁
Next step : Determine the average force of friction ( f )
given that loss of energy results in a change in kinetic energy
W = [tex]\frac{1}{2} m ( vo^{2} - v1^{2} )[/tex]
= 1/2 * 16.3 * 10⁻³ * [ v₀² - [tex](\frac{v_{0} }{3.11} )^{2}[/tex] ]
∴ W = 7.307 * 10⁻³ v₀²
Average force of friction = W / Actual length
= 7.307 * 10⁻³ v₀² / 1
∴ Average force of friction ( F ) = 7.307 * 10⁻³ v₀²
Hence we can conclude that the average force of friction is 7.307 * 10⁻³ v₀²
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Your question has some missing data below are the missing data related to your question
Mass of the ball : 16.3 g
Predicted range : 0.3503 m
Actual range : 1.09 m
I’ve been stuck please help !!
Answer:
The slope of the position time graph gives the velocity.
Explanation:
The slope of the position time graph gives the value of velocity.
In first graph,
The slope is constant in both the parts but positive . So the velocity is also constant and positive for both the parts. and more than the second part, so the initial velocity is more than the final velocity.
In the second graph,
The slope is constant in both the parts but negative. So, the velocity is constant but negative for both the parts. Initial velocity is more negative than the final velocity.
How can i prove the conservation of mechanical energy?
Answer:
We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. W=ΔK. And ΔE=ΔK+ΔU. Hence the mechanical energy of the body is conserved
Explanation:
What is the SI unit for momentum?
0 kg.m
m
O kg
S
m
O kg
S
O kg •
m
At the start of a basketball game, a referee tosses a basketball straight into the air by giving it some initial speed. After being given that speed, the ball reaches a maximum height of 4.35 m above where it started. Using conservation of energy, find the height of the ball when it has a speed of 2.5 m/s.
Answer:
0.32 m.
Explanation:
To solve this problem, we must recognise that:
1. At the maximum height, the velocity of the ball is zero.
2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.
With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:
Mass (m) = constant
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) = 2.5 m/s
Height (h) =?
PE = KE
Recall:
PE = mgh
KE = ½mv²
Thus,
PE = KE
mgh = ½mv²
Cancel m from both side
gh = ½v²
9.8 × h = ½ × 2.5²
9.8 × h = ½ × 6.25
9.8 × h = 3.125
Divide both side by 9.8
h = 3.125 / 9.8
h = 0.32 m
Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.
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?
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.
A 4.76 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t)=(2.80m/s)t +(0.61 m/s3 )t3
What is the magnitude of the force F when 3.71 s ?
Answer:
The magnitude of the force is 64.634 newtons.
Explanation:
According to the statement, the crate is a constant mass system, whose upward force is described by the following expression:
[tex]F(t) = m\cdot \ddot{y} (t)[/tex] (1)
Where:
[tex]F(t)[/tex] - Force, in newtons.
[tex]m[/tex] - Mass, in kilograms.
[tex]\ddot {y}(t)[/tex] - Acceleration, in meters per square second.
The function acceleration is obtained by deriving the function position twice in time:
[tex]\dot y (t) = 2.80 + 1.83\cdot t^{2}[/tex] (2)
[tex]\ddot y(t) = 3.66\cdot t[/tex] (3)
And we expand (1) by applying (3):
[tex]F(t) = 3.66\cdot m \cdot t[/tex]
Where [tex]t[/tex] is the time, in seconds.
If we know that [tex]m = 4.76\,kg[/tex] and [tex]t = 3.71\,s[/tex], then the magnitude of the force is:
[tex]F = 3.66\cdot (4.76)\cdot (3.71)[/tex]
[tex]F = 64.634\,N[/tex]
The magnitude of the force is 64.634 newtons.
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?
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 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?
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
PLEASE HELP ME WITH THIS ONE QUESTION
A photon has 2.90 eV of energy. What is the photon’s wavelength? (h = 6.626 x 10^-19, 1 eV = 1.6 x 10^-19 J)
A) 677 nm
B) 218 nm
C) 345 nm
D) 428 nm
Answer:
The correct option is D.
Explanation:
The wavelength of the photon can be calculated with the following equation:
[tex] E = h\frac{c}{\lambda} [/tex]
Where:
E: is the energy of the photon = 2.90 eV
h: is the Planck's constant = 6.62x10⁻³⁴ J.s
c: is the speed of light = 3x10⁸ m/s
λ: is the wavelength
Hence, the photon's wavelength is:
[tex] \lambda = \frac{hc}{E} = \frac{6.62 \cdot 10^{-34} J*s*3.0 \cdot 10^{8} m/s}{2. 90 eV*\frac{1.6 \cdot 10^{-19} J}{1 eV}} = 428 nm [/tex]
Therefore, the correct option is D.
I hope it helps you!
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?
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