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Svetllana [295]
2 years ago
12

A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha

t is the magnitude of its average acceleration from the third to the eighth second?
Physics
1 answer:
Gelneren [198K]2 years ago
8 0

Answer:

a_{acceleriation}=-4cm/s^{2}\\ or\\  a_{acceleriation}=-0.04m/s^{2}

Explanation:

Given data

velocity v₀=20 cm/s at time t=3s

velocity vf=0 at time t=8 s

To find

Average Acceleration at time=3s to 8s

Solution

As we know that acceleration is first derivative of velocity with respect to time

a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\  a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\  a_{acceleriation}=-4cm/s^{2}\\ or\\  a_{acceleriation}=-0.04m/s^{2}

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Vector A⃗ has magnitude 8.00 m and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axi
WINSTONCH [101]

Answer:

293.7 degrees

Explanation:

A = - 8 sin (37) i + 8 cos (37) j

A + B = -12 j

B = a i+ b j , where and a and b are constants to be found

A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

- 12 j = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

Comparing coefficients of i and j:

a = 8 sin (37) = 4.81452 m

b = -12 - 8cos(37) = -18.38908

Hence,

B = 4.81452 i  - 18.38908 j  ..... 4 th quadrant

Hence,

cos ( Q ) = 4.81452 / 12

Q = 66.346 degrees

360 - Q = 293.65 degrees from + x-axis in CCW direction

5 0
2 years ago
3) A defense football player on one team tackles the other team’s quarterback, who is running down the field. The quarterback is
alisha [4.7K]
Newton's first law says that an object at rest tends to stay at rest while an object in motion stays in motion at a constant velocity unless acted upon by an outside force so the amount of force behind the defensive football player (N) was greater than the quarterback's so he was able to over power him which is also called unbalanced forces
5 0
2 years ago
Combine Newton's 2nd law and Hooke's law for a spring to find the acceleration of the block a(t) as a function of time. Express
Inga [223]

Answer:

a=-\dfrac{k}{m}x(t)

Explanation:

From Newton's second law,

F = ma

where F is the force, m is the mass and a is the acceleration.

From Hooke's law,

F = -kx(t)

where k is the spring constant and x(t) is the displacement function measured from the origin. The negative sign indicates the force acts in opposite direction to the displacement. In fact, it is a restoring force; it acts to return the spring to its original undisturbed position.

Since both forces are the same,

F = ma= - kx(t)

a=-\dfrac{k}{m}x(t)

The implication of this is that the acceleration is proportional to the displacement but opposite to it. That last statement is the definition of a simple harmonic motion which this is.

The ratio \dfrac{k}{m} is a constant except in situations where the mass is varying (say, the mass on the spring is a decaying material).

4 0
2 years ago
If it takes an airplane 15 minutes to go from 30 mph to 330 mph, what is its acceleration?
zzz [600]
Using the a=vf-vi divided by tf-ti:
A is acceleration
Vf is final velocity- 330
Vi is intial velocity-30
Tf is final time-15
Ti is initial time-0
A = 330-30 divided by 15-0
A = 300 divided by 15
A= 20 m/s^2 
Hope this helps
3 0
2 years ago
A 13-cm-diameter cd has a mass of 25 g . part a what is the cd's moment of inertia for rotation about a perpendicular axis throu
faltersainse [42]
<span>Answer: For a disc, the moment of inertia about the perpendicular axis through the center is given by 0.5MR^2. where M is the mass of the disc and R is the radius of the disc. For the axis through the edge, use parallel axis theorem. I = I(axis through center of mass) + M(distance between the axes)^2 = 0.5MR^2 + MR^2 (since the axis through center of mass is the axis through the center) = 1.5 MR^2</span>
8 0
2 years ago
Read 2 more answers
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