Answer:
If efficiency is .22 then W = .22 * Q where Q is the heat input
Heat Input Q = 2510 / .22 = 11,400 J
Heat rejected = 11.400 - 2510 = 8900 J of heat wasted
Also, 8900 J / (4.19 J / cal) = 2120 cal
An efficiency is the measure of productivity of an engine. The heat rejected by the engine is 8900 Joules.
What is efficiency?An efficiency of a heat engine is the ratio of the work done and heat supplied.
Given is the automobile engine has the efficiency 22% and Work done is 2510 Joules.
The efficiency is written as,
η= W / Qs.
The work done is W= Qs - Qr, where Qr is the rejected heat.
The heat rejected can be represented as
Qr = W ( 1/η -1)
Substituting the value into the equation, we get the rejected heat.
Qr = 2510 (1/0.22 -1)
Qr = 8900 Joules.
Thus, the heat rejected by the engine is 8900 Joules.
Learn more about efficiency.
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A uniformly charged thin rod of length L and positive charge Q lies along the x-axis with its left end at the origin as shown in Figure 1.
a. Set up a correct integral expression for the potential at point A,which lies a distance H above the right end of the rod. Point A has coordinates (L, H). You need to give appropriate limits of integration and expressions for r and dq
b. Set up a correct integral expression for the potential at point B on the x-axis, a distance D from the left end of the rod with the appropriate limits of integration.You need to give appropriate limits of integration and expressions for r and dq.
Answer:
b)
Explanation:
How much work is required to stretch an ideal spring of spring constant (force constant) 40 N/m from x
Answer:
The work done will be "0.45 J".
Explanation:
Given:
K = 40 N/m
x₁ = 0.20 m
x₂ = 0.25 m
Now,
The required work done will be:
= [tex]\frac{1}{2}k[x_2^2-x_1][/tex]
By putting the values, we get
= [tex]\frac{40}{2}[(0.25)^2-(0.20)^2][/tex]
= [tex]20\times 0.0225[/tex]
= [tex]0.45 \ J[/tex]
Riley, a student, notices that the protractor tool does not measure the angle just as the ball leaves the surface. She sees that the ball must travel some distance before it crosses the protractor, so the direction of travel may have changed as the ball moves upwards. She says that this is the cause of the discrepancy between her predicted angle and the measured angle. Does this reasoning explain the discrepancy between your predicted angle and your measured angle. Use evidence to support your claim.
Answer:
Riley's reasoning is correct
Explanation:
Her reasoning is correct because as the ball moves upwards, the acceleration due to gravity would be vertical and in downward position. Therefore at all points as the ball moves, the velocity of the ball is going to change in magnitude as well as in direction. given that the direction keeps changing at certain points, the angle made by the initial velocity just as the ball left the surface would also have to continuously change.
If Riley has to wait for this ball to move some inches before she uses the protractor to measure the angle, the angle of travel would have to change.
Therefore there is going to be discrepancies between the measured angle and the predicted angle. The predicted is the angle of velocity with the horizontal just as this ball moves from the surface.
1. An excited lithium atom emits a red light with wavelength a = 671nm. What is the corresponding photon energy? hc (6.63 x 10-34).S)(3.0 x 108m/s)
Answer:
E = 2,964 10⁻¹⁹ J
Explanation:
The energy of the photons is given by the Planck relation
E = h f
the speed of light is related to wavelength and frequency
c = λ f
we substitute
E = h c /λ
let's reduce the magnitude to the SI system
λ = 671 nm = 671 10⁻⁹ m
let's calculate
E = 6.63 10⁻³⁴ 3 10⁸ /671 10⁻⁹
E = 2,964 10⁻¹⁹ J
A horizontal force of P=100 N is just sufficient to hold the crate from sliding down the plane, and a horizontal force of P=350 N is required to just push the crate up the plane. Determine the coefficient of static friction between the plane and the crate, and find the mass of the crate.
"down/up the plane" suggests an inclined plane, but no angle is given so I'll call it θ for the time being.
The free body diagram for the crate in either scenario is the same, except for the direction in which static friction is exerted on the crate. With the P = 100 N force holding up the crate, static friction points up the incline and keeps the crate from sliding downward. When P = 350 N, the crate is pushed upward, so static friction points down. (see attached FBDs)
Using Newton's second law, we set up the following equations.
• p = 100 N
∑ F (parallel) = f + p cos(θ) - mg sin(θ) = 0
∑ F (perpendicular) = n - p sin(θ) - mg cos(θ) = 0
• P = 350 N
∑ F (parallel) = P cos(θ) - F - mg sin(θ) = 0
∑ F (perpendicular) = N - P sin(θ) - mg cos(θ) = 0
(where n and N are the magnitudes of the normal force in the respective scenarios; ditto for f and F which denote static friction, so that f = µn and F = µN, with µ = coefficient of static friction)
Solve for n and N :
n = p sin(θ) + mg cos(θ)
N = P sin(θ) - mg cos(θ)
Substitute these into the corresponding equations containing µ, and solve for µ :
µ = (mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ))
µ = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))
Next, you would set these equal and solve for m :
(mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ)) = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))
...
Once you find m, you back-substitute and solve for µ, but as you might expect the result will be pretty complicated. If you take a simple angle like θ = 30°, you would end up with
m ≈ 36.5 kg
µ ≈ 0.256
The coefficient of static friction between the plane and the crate is μ = 0.256 and the mass of the crate is m=36.4 kg.
From the given,
The force that opposes the crate by sliding is P = 100N
In X-axis, the sum of forces is zero.
ΣF = 0
Pcosθ - mgsinθ-Ff = 0
Ff = Pcosθ - mgsinθ
In Y-axis
Psinθ - mgcosθ - N = 0
N = Psinθ-mgcosθ
Frictional force, Ff = μN, μ is the coefficient of friction
Ff = μN
Pcos30- mgsin30 + μ( Psin30+mgcos30) = 0
μ = mgsin30-Pcos30/Psin30+mgcos30 ------1
The block is sliding with the horizontal force, F = 350N
X-axis
P₂cosθ - mgsinθ-Ff = 0
Y-axis
P₂sinθ - mgcosθ - N = 0
N = P₂sinθ-mgcosθ
μ = P₂cos30-mgsin30/P₂sin30-mgcos30 -----2
Equate equations 1 and 2
mgsin30-Pcos30/Psin30+mgcos30 =P₂cos30-mgsin30/P₂sin30-mgcos30
4.905m-86.6/50+8.49 = 303.1-4.905m/175+8.49
41.7m² + 123m - 1.516×10⁴ = 0
-41.7m² +2330m -1.516×10⁴(4.905-86.6)(175+8.49) =(303.1-4.905)(50+8.49)
83.4m² - 2207m -3.03×10⁴ = 0
m= 36.4 kg
Hence, the mass of the crate is 36.4 Kg.
Substitute the value of m in equation 1,
μ = 4.905(36.4) - 86.6 / 50 + 8.49
μ = 0.256
Thus, the coefficient of static friction is 0.256.
To learn more about friction and its types:
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Every object around you is attracted to you. In fact, every object in the galaxy is attracted to every other object in the galaxy.
a. True
b. False
Answer:
True
Explanation:
With the gravitational pull that our planets have, we are able to remain in orbit. This demonstrates how every object in the galaxy is attracted to every other object. Every object in the universe that has mass exerts a gravitational pull on every other mass. We as humans do it too, but since our force isn't strong, we don't have much of an effect. I hope this helped and please don't hesitate to reach out with more questions!
A body initially at rest travels a distance 100 m in 5 s with a constant acceleration. calculate
(i) Acceleration
(ii) Final velocity at the end of 5 s.
Answer:
(i)8m/s²(ii)40m/s
Explanation:
according to the formula
½at²=s.
then substituting the data
½a•5²=100
a=8m/s²
v=at=8•5=40m/s
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2} a {t}^{2} }} \\ 100 = (0 \times 5) + \frac{1}{2} \times a \times {5}^{2} \\ 200 = 25a \\ { \tt{acceleration = 8 \: m {s}^{ -2} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ v = 0 + (8 \times 5) \\ { \tt{final \: velocity = 40 \: m {s}^{ - 1} }}[/tex]
You are on an airplane that is landing. The plane in front of your plane blows a tire. The pilot of your plane is advised to abort the landing, so he pulls up, moving in a semicircular upward-bending path. The path has a radius of 450 m with a radial acceleration of 17 m/s^2.
Required:
What is the plane's speed?
Answer:
v = 87.46 m/s
Explanation:
The radial acceleration is the centripetal acceleration, whose formula is given as:
[tex]a_c = \frac{v^2}{r}[/tex]
where,
[tex]a_c[/tex] = centripetal acceleration = 17 m/s²
v = planes's speed = ?
r = radius of path = 450 m
Therefore,
[tex]17\ m/s^2 = \frac{v^2}{450\ m}\\\\v^2 = (17\ m/s^2)(450\ m)\\\\v = \sqrt{7650\ m^2/s^2}[/tex]
v = 87.46 m/s
A pilot drops a bomb from a plane flying horizontally. Where will the plane be located when the bomb hits the ground
Answer:
The plane will be located directly above the bomb because they both have the same horizontal speed.
What is the rate of the entropy change of the universe as heat leaks out a window, consisting of a single pane of glass that is 0.5 cm thick and 1.0 m2 in area, where the indoor temperature is 25°C and the outdoor temperature is -10°C?
Answer:
The change in entropy is 1.6 W/K.
Explanation:
Thickness, d = 0.5 cm
Area, A = 1 m^2
T = 25°C
T' = - 10°C
Coefficient of thermal conductivity of glass, K = 0.8 W/mK
The change in entropy is given by
S = Q/T
Here,
[tex]S =\frac{Q}{T}\\\\S = \frac{K A (T - T')}{d(T - T')}\\\\S = \frac{0.8\times 1}{0.5} = 1.6 W/K[/tex]
Two charged particles attract each other with a force of magnitude F acting on each. If the charge of one is doubled and the distance separating the particles is also doubled, the force acting on each of the two particles has magnitude
(a) F/2,
(b) F/4,
(c) F,
(d) 2F,
(e) 4F,
(f) None of the above.
Answer:
F/2
Explanation:
In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant
Hence;
F= KQ1Q2/r^2 ------(1)
Where the charge on Q1 is doubled and the distance separating the charges is also doubled;
F= K2Q1 Q2/(2r)^2
F2= 2KQ1Q2/4r^2 ----(2)
F2= F/2
Comparing (1) and (2)
The magnitude of force acting on each of the two particles is;
F= F/2
A 10 kg box is at static equilibrium and the downward pull of gravity acting on the box is 98 Newton’s what is the minimum force that would require to just pick up the box
Explanation:
static equilibrium means its on the floor or something
so slightly greater than 98 newtons in the upward direction
If 5kg Stone and 1kg stone throw the from the building which will land more fa ster and why?
Answer:
Both stones will land at the same time because both stones will fall with the same acceleration through the same height.
Explanation:
We are given that
Mass of stone ,m1=5 Kg
Mass of stone, m2=1 kg
We have to find which stone more faster will land and why.
[tex]h=u+\frac{1}{2}gt^2[/tex]
Initial velocity of both stones=0
[tex]h=\frac{1}{2}gt^2[/tex]
[tex]t^2=\frac{h}{g}[/tex]
[tex]t=\sqrt{\frac{h}{g}}[/tex]
[tex]t_1=t_2=\sqrt{\frac{h}{g}}[/tex]
Because both stones are thrown from the same height.
Both stones will land at the same time because both stones will fall with the same acceleration through the same height and the acceleration does not depend of its mass.
During a practice shot put throw, the 7.9-kg shot left world champion C. J. Hunter's hand at speed 16 m/s. While making the throw, his hand pushed the shot a distance of 1.4 m. Assume the acceleration was constant during the throw.
Required:
a. Determine the acceleration of the shot.
b. Determine the time it takes to accelerate the shot.
c, Determine the horizontal component of the force exerted on the shot by hand.
Answer:
a) a = 91.4 m / s², b) t = 0.175 s, c)
Explanation:
a) This is a kinematics exercise
v² = vox ² + 2a (x-xo)
a = v² - 0/2 (x-0)
let's calculate
a = 16² / 2 1.4
a = 91.4 m / s²
b) the shooting time
v = vox + a t
t = v-vox / a
t = 16 / 91.4
t = 0.175 s
c) let's use Newton's second law
F = ma
F = 7.9 91.4
F = 733 N
A black T-shirt is warmer in the summertime than a white T-shirt because the black T-shirt
A. Is reflecting all wavelengths of light.
B. Absorbs violet light, the highest energy wavelength.
C. Is absorbing all wavelengths of light. D. Doesn’t absorb red, the longest wavelength.
Answer:
c
Explanation:
darker colors absorb app light
Answer:
C. Is absorbing all wavelengths of light.
Explanation:
Black isn't a color, but rather the absence of color. We see a T-shirt as black because it isn't reflecting any light toward our eyes. A black T-shirt absorbs all of the wavelengths of light, causing it to absorb more energy and become warmer than white, which reflects light.
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by
The question is incomplete. The complete question is :
A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (4.7 cm)sin[(7.9 rad/s)πt].
Determine the following:
(a) frequency of the motion
(b) period of the motion
(c) amplitude of the motion
(d) first time after t = 0 that the object reaches the position x = 2.6 cm
Solution :
Given equation : x = (4.7 cm)sin[(7.9 rad/s)πt].
Comparing it with the general equation of simple harmonic motion,
x = A sin (ωt + Φ)
A = 4.7 cm
ω = 7.9 π
a). Therefore, frequency, [tex]$f=\frac{\omega}{2 \pi}$[/tex]
[tex]$=\frac{7.9 \pi}{2 \pi}$[/tex]
= 3.95 Hz
b). The period, [tex]$T=\frac{1}{f}$[/tex]
[tex]$T=\frac{1}{3.95}[/tex]
= 0.253 seconds
c). Amplitude is A = 4.7 cm
d). We have,
x = A sin (ωt + Φ)
[tex]$x_t=4.7 \sin (7.9 \pi t)$[/tex]
[tex]$2.6 = 4.7 \sin (7.9 \pi t)$[/tex]
[tex]$\sin (7.9 \pi t) = \frac{26}{47}$[/tex]
[tex]$7.9 \pi t = \sin^{-1}\left(\frac{26}{47}\right)$[/tex]
Hence, t = 0.0236 seconds.
Could you show detailed steps in how to solve this problem please
Answer: See attached pic. Hope this helps.
Explanation:
If the resistance in a circuit remains constant, what happens to the electric power when the current increases?
The power will increase.
B.
The power will decrease.
Ο Ο Ο Ο
There will be no power.
D
The current does not affect the power.
Answer:
Resistance is inversly proportional to the current.
V=I.R.
P=V.I
A gymnast falls from a height onto a trampoline. For a moment, both the gymnast’s kinetic energy and gravitational potential energy are zero. How is the gymnast’s mechanical energy stored for that moment? Question 12 options: rest energy chemical energy elastic energy thermal energy
Answer:
elastic energy
Explanation:
When a gymnast falls on a trampoline from a height, after coming in contact with the trampoline, both the gymnast and the trampoline start to move down due to the elastic property of the trampoline.
During this stretching of the trampoline there comes a maximum point up to which the trampoline is stretched. At this point, both the kinetic energy and the gravitational potential energy of the gymnast are zero due to zero speed and zero height, respectively.
The only energy stored in the gymnast's body at this point is the elastic potential energy due to stretching of the trampoline. Hence,the correct option is:
elastic energy
A baseball of mass 0.145 kg is thrown at a speed of 40.0 m/s. The batter strikes the ball with a force of 15,000 N; the bat and ball are in contact for 0.500 ms. The force is exactly opposite to the original direction of the ball. Determine the final speed of the ball.
The final speed of the ball is 91.72 m/s.
Given the following data:
Mass of baseball = 0.145 kgInitial speed = 40.0 m/sForce = 15,000 NewtonTime = 0.500 milliseconds (ms) to seconds = 0.0005 seconds.To find the final speed of the ball, we would use the following formula:
[tex]F = \frac{M(V - U)}{t}[/tex]
Where:
F is the force applied. u is the initial speed. v is the final speed. t is the time measured in seconds.Substituting the parameters into the formula, we have;
[tex]15000 = \frac{0.145(V \;- \;40)}{0.0005}\\\\15000(0.0005) = 0.145(V \;- \;40)\\\\7.5 = 0.145V - 5.8\\\\0.145V = 7.5 + 5.8\\\\0.145V = 13.3\\\\V = \frac{13.3}{0.145}[/tex]
Final speed, V = 91.72 m/s
Therefore, the final speed of the ball is 91.72 m/s.
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what are the dynamic properties of a nucleus
Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0 m and Ay 450.0 m. Suppose that the coordinate system is rotated counterclockwise by 35.0, but the magnitude (450.0 m) and direction of vector remain unchanged, as in drawing b. What are the scalar components, Ax and Ay, of the vector in the rotated x, y coordinate system
Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m
Based on the information in the table, what
is the acceleration of this object?
t(s) v(m/s)
0.0
9.0
1.0
4.0
2.0
-1.0
3.0
-6.0
A. -5.0 m/s2
B. -2.0 m/s2
C. 4.0 m/s2
D. 0.0 m/s2
Answer:
Option A. –5 m/s²
Explanation:
From the question given above, the following data were obtained:
Initial velocity (v₁) = 9 m/s
Initial time (t₁) = 0 s
Final velocity (v₂) = –6 m/s
Final time (t₂) = 3 s
Acceleration (a) =?
Next, we shall determine the change in the velocity and time. This can be obtained as follow:
For velocity:
Initial velocity (v₁) = 9 m/s
Final velocity (v₂) = –6 m/s
Change in velocity (Δv) =?
ΔV = v₂ – v₁
ΔV = –6 – 9
ΔV = –15 m/s
For time:
Initial time (t₁) = 0 s
Final time (t₂) = 3 s
Change in time (Δt) =?
Δt = t₂ – t₁
Δt = 3 – 0
Δt = 3 s
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Change in velocity (Δv) = –15 m/s
Change in time (Δt) = 3 s
Acceleration (a) =?
a = Δv / Δt
a = –15 / 3
a = –5 m/s²
Thus, the acceleration of the object is
–5 m/s².
1. A block of mass m = 10.0 kg is released with a speed v from a frictionless incline at height 7.00 m. The
block reaches the horizontal ground and then slides up another frictionless incline as shown in Fig. 1.1. If the
horizontal surface is also frictionless and the maximum height that the block can slide up to is 26.0 m, (a) what
is the speed v of the block equal to when it is released and (b) what is the speed of the block when it reaches
the horizontal ground? If a portion of length 1 2.00 m on the horizontal surface is frictional with coefficient
of kinetic friction uk = 0.500 (Fig. 1.2) and the block is released at the same height 7.00 m with the same
speed v determined in (a), (c) what is the maximum height that the block can reach, (d) what is the speed of the
block at half of the maximum height, and (e) how many times will the block cross the frictional region before
it stops completely?
1 = 2.00 m (frictional region)
Let A be the position of the block at the top of the first incline; B its position at the bottom of the first incline; C its position at the bottom of the second incline; and D its position at the top of the second incline. I'll denote the energy of the block at a given point by E (point).
At point A, the block has total energy
E (A) = (10.0 kg) (9.80 m/s²) (7.00 m) + 1/2 (10.0 kg) v₀²
E (A) = 686 J + 1/2 (10.0 kg) v₀²
At point B, the block's potential energy is converted into kinetic energy, so that its total energy is
E (B) = 1/2 (10.0 kg) v₁²
The block then slides over the horizontal surface with constant speed v₁ until it reaches point C and slides up a maximum height of 26.0 m to point D. Its total energy at D is purely potential energy,
E (D) = (10.0 kg) (9.80 m/s²) (26.0 m) = 2548 J
Throughout this whole process, energy is conserved, so
E (A) = E (B) = E (C) = E (D)
(a) Solve for v₀ :
686 J + 1/2 (10.0 kg) v₀² = 2548 J
==> v₀ ≈ 19.3 m/s
(b) Solve for v₁ :
1/2 (10.0 kg) v₁² = 2548 J
==> v₁ ≈ 22.6 m/s
Now if the horizontal surface is not frictionless, kinetic friction will contribute some negative work to slow down the block between points C and D. Check the net forces acting on the block over this region:
• net horizontal force:
∑ F = -f = ma
• net vertical force:
∑ F = n - mg = 0
where f is the magnitude of kinetic friction, a is the block's acceleration, n is the mag. of the normal force, and mg is the block's weight. Solve for a :
n = mg = (10.0 kg) (9.80 m/s²) = 98.0 N
f = µn = 0.500 (98.0 N) = 49.0 N
==> - (49.0 N) = (10.0 kg) a
==> a = - 4.90 m/s²
The block decelerates uniformly over a distance 2.00 m and slows down to a speed v₂ such that
v₂² - v₁² = 2 (-4.90 m/s²) (2.00 m)
==> v₂² = 490 m²/s²
and thus the block has total/kinetic energy
E (C) = 1/2 (10.0 kg) v₂² = 2450 J
(c) The block then slides a height h up the frictionless incline to D, where its kinetic energy is again converted to potential energy. With no friction, E (C) = E (D), so
2450 J = (10.0 kg) (9.80 m/s²) h
==> h = 25.0 m
(d) At half the maximum height, the block has speed v₃ such that
2450 J = (10.0 kg) (9.80 m/s²) (h/2) + 1/2 (10.0 kg) v₃²
==> v₃ ≈ 15.7 m/s
The block loses speed and thus energy as it moves between B and C, but its energy is conserved elsewhere. If we ignore the inclines and pretend that the block is sliding over a long horizontal surface, then its velocity v at time t is given by
v = v₁ + at = 22.6 m/s - (4.90 m/s²) t
The block comes to a rest when v = 0 :
0 = 22.6 m/s - (4.90 m/s²) t
==> t ≈ 4.61 s
It covers a distance x after time t of
x = v₁t + 1/2 at ²
so when it comes to a complete stop, it will have moved a distance of
x = (22.6 m/s) (4.61 s) + 1/2 (-4.90 m/s²) (4.61 s)² = 52.0 m
(e) The block crosses the rough region
(52.0 m) / (2.00 m) = 26 times
Two spheres are rolling without slipping on a horizontal floor. They are made of different materials, but each has mass 5.00 kg and radius 0.120 m. For each the translational speed of the center of mass is 4.00 m/s. Sphere A is a uniform solid sphere and sphere B is a thin-walled, hollow sphere. Part B How much work, in joules, must be done on the solid sphere to bring it to rest? Express your answer in joules. VO AE4D ? J WA Request Answer Submit Part C How much work, in joules, must be done on the hollow sphere to bring it to rest? Express your answer in joules. Wa Request
Answer:
Explanation:
Moment of inertia of solid sphere = 2/5 m R²
m is mass and R is radius of sphere.
Putting the values
Moment of inertia of solid sphere I₁
Moment of inertia of hollow sphere I₂
Kinetic energy of solid sphere ( both linear and rotational )
= 1/2 ( m v² + I₁ ω²) [ ω is angular velocity of rotation ]
= 1/2 ( m v² + 2/5 m R² ω²)
= 1/2 ( m v² + 2/5 m v²)
=1/2 x 7 / 5 m v²
= 0.7 x 5 x 4² = 56 J .
This will be equal to work to be done to stop it.
Kinetic energy of hollow sphere ( both linear and rotational )
= 1/2 ( m v² + I₂ ω²) [ ω is angular velocity of rotation ]
= 1/2 ( m v² + 2/3 m R² ω²)
= 1/2 ( m v² + 2/3 m v²)
=1/2 x 5 / 3 m v²
= 0.833 x 5 x 4² = 66.64 J .
This will be equal to work to be done to stop it.
1. A 20.0 N force directed 20.0° above the horizontal is applied to a 6.00 kg crate that is traveling on a horizontal
surface. What is the magnitude of the normal force exerted by the surface on the crate?
N = 52.0 N
Explanation:
Given: [tex]F_a= 20.0\:\text{N}=\:\text{applied\:force}[/tex]
[tex]m=6.00\:\text{kg}[/tex]
[tex]N = \text{normal force}[/tex]
The net force [tex]F_{net}[/tex] is given by
[tex]F_{net} = N + F_a\sin 20 - mg=0[/tex]
Solving for N, we get
[tex]N = mg - F_a\sin 20[/tex]
[tex]\:\:\:\:\:\:= (6.00\:\text{kg})(9.8\:\text{m/s}^2) - (20.0\:\text{N}\sin 20)[/tex]
[tex]\:\:\:\:\:\:= 52.0\:\text{N}[/tex]
Which of the following represents the velocity time relationship for a falling apple?
Answer "a" would be correct.
Answer:
d
Explanation:
There's an acceleration from gravity, thus the velocity is becoming faster and faster as it reaches the ground. Thus its D
Brainliest please~
A proton is held at rest in a uniform electric field. When it is released, the proton will gain:_________
a) electrical potential energy.
b) kinetic energy.
c) both kinetic energy and electric potential energy.
d) either kinetic energy or electric potential energy.
What's the speed of a sound wave through water at 25 Celsius?
A. 1,000 m/s
B. 1,500 m/s
C. 1,250 m/s
D. 750 m/s
Answer:
B) 1500m/s
Explanation:
Ans is 1500m/s
A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity increases to 57 rad/s. Assume that the angular acceleration was constant during this time interval. How many revolutions does the wheel turn through during this time interval
Answer:
The number of revolutions is 44.6.
Explanation:
We can find the revolutions of the wheel with the following equation:
[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]
Where:
[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s
t: is the time = 8 s
α: is the angular acceleration
We can find the angular acceleration with the initial and final angular velocities:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Where:
[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s
[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]
Hence, the number of revolutions is:
[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]
Therefore, the number of revolutions is 44.6.
I hope it helps you!
