Answer:
Explanation:
From the given information:
The torque produced due to the force can be expressed as:
[tex]\tau = F \times r[/tex]
where;
[tex]\tau[/tex] = torque
F = force exerted
r = lever's arm radius
[tex]\tau[/tex] = [tex]2.00 \times 10^3 \times 0.03 m[/tex]
[tex]\tau[/tex] = 60 N.m
However, equating the torque with the moment of inertia & angular acceleration, we use the equation:
[tex]\tau[/tex] = I∝
60 Nm = I × 120 rad/s²
I = 60 Nm/120 rad/s²
I = 0.5 kg.m²
the two factors that affect the amount of heat
Answer:
The two important factors that affect heat energy are specific heat and temperature. Specific heat is a heat-constant of a material per unit mass per degree of temperature change (in units of energy per mass and temperature), like Joules/Kg-°C .
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Have a good day.....
An astronaut throws a wrench in interstellar space. How much force is required to keep the wrench moving continuously with constant velocity?
A.
a force equal to its weight on Earth
B.
a force equal to zero
C.
a force equal to half of its weight on Earth
D.
a force equal to double its weight on Earth
Answer:
0 N
Explanation:
This is a trick question, the mass of the wrench would be 0 due to it being in space and has no gravitational pull to weight it down. And since acceleration is defined as the rate and change of velocity with no respect of time and the wrench is moving at a constant velocity, that means the velocity is 0. and since F = m*a it would be F = 0 * 0 = 0 N
Which phenomenon occurs when one wave is superimposed on another?
A. Interference
B. Refraction
C. Diffraction
D. Polarization
Answer:Alternativa A. Damos o nome de interferência a superposição de efeitos que ocorre ao ser produzido dois pulsos de onda, que serão propagados e acabarão inevitavelmente por se encontrar. No instante em que os pulsos se cruzarem, há então, uma superposição de efeitos individuais de cada um deles. Se durante o cruzamento, houver um reforço das ondas, estará ocorrendo a este fenômeno.
Use the simulation to compare the masses of the three colored and unlabeled weights of different sizes. To do so, set the spring constant of both springs to the same value. Hang known weights on the left spring and an unknown weight on the right spring, and compare the two. Use as many known weights as necessary to determine the unknown masses, and then place each into the appropriate mass bins in the ranking task below. A. M<50 g
B. M = 50 g
C. 50 g D. M = 100 g
E. 100 g F. M = 250 g
G. M> 250 g
1. Blue medium sized weight
2. Magenta small sized weight
3. Gold large sized weight
Answer:
The answer is given as follows,
Explanation:
Gold large-sized weight 100 g < M < 250g
50 g < Magenta small-sized weight < 100g
100g < Blue medium-sized weight < 250g
Hence,
100g < Blue medium-sized weight < 250g
50 g < Magenta small-sized weight < 100g
100 g < Gold large-sized weight < 250g.
A spring of spring constant k=8.25N/m is displaced from equilibrium by a distance of 0.150m. What is the stored energy in the form of spring potential energy? Show your work.
Answer:
0.0928J
Explanation:
the pulling force of spring F=-kx
where x is the displacement from equilibrium position.
energy stored:
[tex]\int\limits^x_0 {-F} \, dx \\=\int\limits^x_0 {kx} \, dx \\\\=\frac{kx^{2} }{2}[/tex]
*** Its fine if you know nothing about calculus. Just apply the equation
[tex]U=\frac{kx^{2} }{2}[/tex]
where U is the potential energy of the spring***
put x=0.150, [tex]U=\frac{8.25}{2}[/tex]×[tex]0.150^{2}[/tex] = 0.0928J (corr. to 3 sig. fig.)
The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at the moments of time: 0, T, 2T, 3T and 4T.
Explanation:
The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum
In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0342 s, during which time it experiences an acceleration of 186 m/s2. The ball is launched at an angle of 45.9 ° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.
Answer:
b) v_y = 4.57 m / s
a) vₓ = 4.43 m / s
Explanation:
This is an exercise in kinematics, where we assume that the acceleration is in the direction of the force and the initial body with zero velocity
v = v₀ + a t
v = 0 + a t
v = 186 0.0342
v = 6.36 m / s
let's use trigonometry to decompose this velocity
sin 45.9 = v_y / v
cos 45.9 = vₓ / v
v_y = v sin 45.9
vₓ = v cos 45.9
v_y = 6.36 sin 45.9
vₓ = 6.36 cos 45.9
v_y = 4.57 m / s
vₓ = 4.43 m / s
ₓ
semiconductor have negative temperature coefficient of resistance why
Answer:
As the number of free electrons increases, the resistance of this type of non-metallic material decreases with increasing temperature.
Explanation:
12. A car is travelling at 30 m/s when the driver sees a red light in the distance and immediately applies the brakes. The car comes to a stop 1.5 s later. How far did the car move from the time the driver applied the brake to when it came to a stop?
Answer:
22.5 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 30 m/s
Time (t) = 1.5 s
Final velocity (v) = 0 m/s
Distance (s) =?
The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:
s = (u + v)t/2
s = (30 + 0)1.5 / 2
s = (30 × 1.5) / 2
s = 45 / 2
s = 22.5 m
Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.
45. Pressure in air undergoes a decrease when the air
a) rises to higher altitudes.
b) accelerates to higher speed.
c) fills a greater space.
d) All of these.
If a sprinter ran a distance of 100 meters starting at his top speed of 11 m/s and running with constant spreed throughout. How long would it take him to cover the distance?
Answer:
9.09 s
Explanation:
If the sprinter ran the 100 meters at the constant speed of 11 m/s it would take him 9.09 s to cover the full distance.
We can find this number by dividing 100 meters (the distance covered) by 11 meters per second (the speed)
[tex]\frac{100}{11} =9.09[/tex]
A canoe has a velocity of 0.330 m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.540 m/s east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.
Answer:
The velocity of the canoe relative to the river is 0.385 m/s, S37.26⁰W
Explanation:
Given;
velocity of the canoe relative to the earth, [tex]V_{r/e} = 0.33 \ m/s[/tex]
velocity of the river relative to the earth, [tex]V_{r/e} = 0.54 \ m/s[/tex]
The velocity of the canoe relative to the river is calculated as;
[tex]V_{(c/r)x} = V_{(c/e)x}- V_{(r/e)x} \ \ ----(1)\\\\V_{(c/r)y} = V_{(c/e)y}- V_{(r/e)y} \ \ ----(2)[/tex]
The x - component of the velocity of the canoe relative to the earth;
[tex]V_{(c/e)x} = 0.33 \times cos \ 45^0\\\\V_{(c/e)x} = 0.2333 \ m/s[/tex]
The y-component of the velocity of the canoe relative to the earth;
[tex]V_{(c/e)y} = 0.33 \times sin \ 45^0\\\\V_{(c/e)y} = 0.2333 \ m/s[/tex]
Note: velocity of the river relative to the earth has only x-component = 0.54 m/s
Apply equation (1) and (2) to calculate the velocity of the canoe relative to the river;
[tex]V_{(c/r)}x = 0.2333 - 0.54 = -0.3067 \ m/s\\\\V_{(c/r)}y = 0.2333 - 0 = 0.2333 \ m/s\\\\The \ resultant \ velocity;\\\\V_{c/r} = \sqrt{(-0.3067)^2 + (0.2333)^2} \\\\V_{c/r} = 0.385 \ ms/\\\\The \ direction:\\\\\theta = tan^{-1} (\frac{0.2333}{0.3067} ) = 37.26^0 \ south \ west \ of \ the \ river[/tex]
What is the energy equivalent of an object with a mass of 1.05g?
Answer:
The equivalent energy of an object given its mass is calculated through the equation,
E = mc²
where c is the speed of light (3 x 10^8 m/s)
Substituting the known values,
E = (1.05 g/ 1000) (3 x 10^8 m/s)²
E = 9.45x10^13 J
Explanation:
URGENTT
Which statement best defines the term "superconductivity"?
Answer:
the ability of certain substances at very low temperatures to conduct electricity with no resistance
Suppose that 2 J of work is needed to stretch a spring from its natural length of 32 cm to a length of 46 cm. (a) How much work (in J) is needed to stretch the spring from 37 cm to 41 cm
Answer:
the work required is 0.163 J
Explanation:
Given;
Energy applied to the spring, E = 2 J
initial length of the spring, x₀ = 32 cm
final length of the spring, x₁ = 46 cm
Extension of the spring, Δx = x₁ - x₀ = 46 cm - 32 cm = 14 cm = 0.14 m
The spring constant is calculated as follows;
E = ¹/₂kΔx²
[tex]k = \frac{2E}{(\Delta x)^2} = \frac{2\times 2}{(0.14)^2} = 204.1 \ N/m^2[/tex]
The extension of the spring when it is stretched from 37 cm + 41 cm:
Δx = 41 cm - 37 cm = 4 cm = 0.04 m
The work required:
W = ¹/₂kΔx²
W = ¹/₂ x (204.1) x (0.04)²
W = 0.163 J
Therefore, the work required is 0.163 J
Which term describes friction that acts on a stationary object?
O A. Static friction
B. Sliding friction
C. Kinetic friction
D. Resistance friction
Answer:
Static friction
Explanation:
Answer:
Static friction
Explanation:
List the 5 theoretical perspectives that underlie much of the research on human development. Also, name an individual strongly associated with each perspective.
Answer:
25
Explanation:
5 theroical name indvivdual perspective asssssoitive each persp
A lumberjack is trying to drag a small tree that he cut down. If the static
coefficient of friction of the tree on the ground is 0.5 and the tree weighs 430
N, what is the minimum amount of horizontal force that he will need to apply
so that the tree will start moving?
A. 215 N
B. 430 N
C. 365 N
D. 500 N
Answer:
A
Explanation:
weight of the tree =normal force
Horizontal force =coefficient of friction x Fnormal
0.5×430=215
A laser emits a single 3.0-ms pulse of light that has a frequency of 2.83E11 Hz and a total power of 65000 W. How many photons are in the pulse? Please provide all equations and work.
6.0E23
1.0E24
2.4E25
3.6E25
4.8E26
Answer:
The number of photons in the pulse is 1.04 x 10²⁴
Explanation:
Given;
frequency of the emitted photons, f = 2.83 x 10¹¹ Hz
duration of the incident light, t = 3 ms = 3 x 10⁻³ s
power of the incident light, P = 65,000 W
The energy of each photon emitted is calculated as;
E = hf
where;
h is Planck's constant, = 6.626 x 10⁻³⁴ Js
E = 6.626 x 10⁻³⁴ x 2.83 x 10¹¹
E = 1.875 x 10⁻²² J
let the number of photons in the pulse = n
n(E)= Power x time
[tex]n = \frac{Pt}{E} \\\\n = \frac{65,000 \times 3\times 10^{-3}}{1.875 \times 10^{-22}} \\\\n = 1.04 \times 10^{24} \ photons[/tex]
Two identical conductors have charge -1.8 C and 5.5 C on them, respectively. They are connected by a conducting wire for a short period of time and then disconnected. What is the net charge on each of the conductors after the interaction? g
Answer: 1.85 C
Explanation:
Given
charges on the conductors are [tex]-1.8\ C[/tex] and [tex]5.5\ C[/tex]
They are connected by a conducting wire for a short period of time and then disconnected. During this time charge flow from the wire and net charge becomes [tex]5.5-1.8=3.7\ C[/tex]
This charge will be equally distribute among the two conductors i.e. 1.85 C on each conductor.
The energy wasted in using a machine is 600j. if the machine is 70% efficient. calculate the volume of water pumb by the machine through a height of 15m.
Answer:
yhgigy6ftu5cg l8vbbnnbbgtccccvhklhaywje nc 62bbnzmakbdbvfvdbf93bdldmffmfkqhdv
A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the bucket is gradually increased by the addition of sand. At some point, the bucket will accumulate enough sand to set the block in motion. The coefficients of static and kinetic friction are 0.60 and 0.50 respectively.
Required:
a. Determine the mass of sand in [kg], including the bucket, needed to start the block moving.
b. Find the blocks acceleration, in [m/s^2] up the plane?
Answer:
a). M = 20.392 kg
b). am = 0.56 [tex]m/s^2[/tex] (block), aM = 0.28 [tex]m/s^2[/tex] (bucket)
Explanation:
a). We got N = mg cos θ,
f = [tex]$\mu_s N$[/tex]
= [tex]$\mu_s mg \cos \theta$[/tex]
If the block is ready to slide,
T = mg sin θ + f
T = mg sin θ + [tex]$\mu_s mg \cos \theta$[/tex] .....(i)
2T = Mg ..........(ii)
Putting (ii) in (i), we get
[tex]$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$[/tex]
[tex]$M=2(m \sin \theta + \mu_s mg \cos \theta)$[/tex]
[tex]$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$[/tex]
M = 20.392 kg
b). [tex]$(h-x_m)+(h-x_M)+(h'+x_M)=l$[/tex] .............(iii)
Here, l = total string length
Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so
[tex]$-\ddot{x}-2\ddot x_M=0$[/tex]
[tex]$\ddot x_M=\frac{\ddot x_m}{2}$[/tex]
[tex]$a_M=\frac{a_m}{2}$[/tex] .....................(iv)
We got, N = mg cos θ
[tex]$f_K=\mu_K mg \cos \theta$[/tex]
∴ [tex]$T-(mg \sin \theta + f_K) = ma_m$[/tex]
[tex]$T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$[/tex] ................(v)
Mg - 2T = M[tex]a_M[/tex]
[tex]$Mg-Ma_M=2T$[/tex]
[tex]$Mg-\frac{Ma_M}{2} = 2T$[/tex] (from equation (iv))
[tex]$\frac{Mg}{2}-\frac{Ma_M}{4}=T$[/tex] .....................(vi)
Putting (vi) in equation (v),
[tex]$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$[/tex]
[tex]$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$[/tex]
[tex]$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$[/tex]
[tex]$a_m= 0.56 \ m/s^2$[/tex]
Using equation (iv), we get,
[tex]a_M= 0.28 \ m/s^2[/tex]
Your hand and wrist curl in toward the center of your body (chest and stomach) to prepare to throw the frisbee.
O True
O False
True
Hope this helps! :)
Answer:
true because when trow the frisbee gives u level
The correct formula for finding the relative velocity of an object is:
WILL MARK BRAINLIEST TO THE CORRECT ANSWER!!
Answer:
[tex]V_{a/c} = V_{a/b} + V_{b/c}[/tex]
Explanation:
The relative velocity of an object is the velocity of the object relative to the observer or frame of reference.
The velocity of particle "A" with respect to particle "B" is written as [tex]V_{A/B} = V_{A} - V_{B}[/tex]
From the given options, the second option is the correct answer.
[tex]V_{a/c} = V_{a/b} + V_{b/c}\\\\Re-arrange \ the \ above \ equation;\\\\V_{a/c} - V_{b/c}= V_{a/b}\\\\or\\\\V_{a/b}= V_{a/c} - V_{b/c}[/tex]
A positively charged plastic ruler is brought close to a piece paper resting on the desk. The piece of paper was initially neutral. When the ruler was brought closer, the paper is attracted to the ruler. The surface of the paper became charged through:_________
Answer: static electricity
Explanation:
When the plastic ruler is rubbed, friction opposes the motion and causes the transfer of electron from one surface to another such that plastic becomes negatively charged. When ruler is brought nearer to the paper, it induces the positive charge in the piece of paper.
define a system who's momentum is observed
product of force and perpendicular distance
If the distance between the center of two objects is quadrupled. The gravitational
force between the two objects will change by a factor of:
1) 16
2) 0.25
3) 4
4) 0.0625
Answer:
F' = F/16
Explanation:
The gravitational force between masses is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]
If the distance between the center of two objects is quadrupled, r' = 4r
New force will be :
[tex]F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(4r)^2}\\\\F'=\dfrac{Gm_1m_2}{16r^2}\\\\F'=\dfrac{1}{16}\times \dfrac{Gm_1m_2}{r^2}\\\\F'=\dfrac{F}{16}[/tex]
So, the new force will change by a factor of 16.
Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.2 s and was brought jarringly back to rest in only 1 s. Calculate his (a) magnitude of acceleration in his direction of motion and (b) magnitude of acceleration opposite to his direction of motion. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity. g g
Answer:
a = 5.53 g , a = -15g
Explanation:
This is an exercise in kinematics.
a) Let's look for the acceleration
as part of rest v₀ = 0
v = v₀ + a t
a = v / t
a = 282 / 5.2
a = 54.23 m / s²
in relation to the acceleration of gravity
a / g = 54.23 / 9.8
a = 5.53 g
b) let's look at the acceleration to stop
va = 0
0 = v₀ -2 a y
a = vi / y
a = 282/2 1
a = 141 m /s²
a / G = 141 / 9.8
a = -15g
A sack of groceries with a mas of 22 kg is lifted off the floor with a velocity of 6 m/s. What is the kinetic energy of the sack
of groceries?
the answer is 396 joules :D
What is the source of almost all energy on Earth?
Earth’s hot core
the Sun
stored carbon
moving water
Answer:
the sun
Explanation:
The sun is the source of almost all energy on Earth beacuse both plants and animal on Earth derive their energy from the sun.
Source of energy on EarthThe Earth is one of the planets that make up the solar system. The Sun is the center of the universe and the Earth revolves round the sun.
The source of almost all energy on Earth is from the sun. The energy from the sun is callled solar energy.
This energy from the sun can be used by the planet to manufatcure its own food. The plants are consumed by animals to provide energy their metabolic activities.
Thus, we can conclude that the sun is the source of almost all energy on Earth beacuse both plants and animal on Earth derive their energy from the sun.
Learn more about solar energy here: https://brainly.com/question/17711999