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VashaNatasha [74]
2 years ago
9

A bird is flying due east. Its distance from a tall building is given by x(t) = 28.0 m + (12.4 m/s)t – (0.0450 m/s3)t3. What is

the instantaneous velocity of the bird when t = 8.00s?
Physics
1 answer:
Kruka [31]2 years ago
8 0

Answer:

3.76 m/s

Explanation:

Instantaneous velocity: This can be defined as the velocity of an object in a non uniform motion. The S.I unit is m/s.

v' = dx(t)/dt..................... Equation 1

Where v' = instantaneous velocity, x = distance, t = time.

Given the expression,

x(t) = 28.0 m + (12.4 m/s)t - (0.0450 m/s³)t³

x(t) = 28 + 12.4t - 0.0450t³

Differentiating x(t) with respect to t.

dx(t)/dt = 12.4 - 0.135t²

dx(t)/dt = 12.4 - 0.135t²

When t = 8.00 s.

dx(t)/dt = 12.4 - 0.135(8)²

dx(t)/dt = 12.4 - 8.64

dx(t)/dt = 3.76 m/s.

Therefore,

v' = 3.76 m/s.

Hence, the instantaneous velocity = 3.76 m/s

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A locomotive approaches its next stop and accelerates at -0.12 m/s^2, coming to a complete stop in 30 seconds. This motion could
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Answer:

<em>Answer: positive velocity & negative acceleration</em>

Explanation:

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4 0
2 years ago
A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:
madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take g = 10\; \rm m \cdot s^{-2}.

a. The momentum of the ball would be approximately 60\;\rm kg \cdot m \cdot s^{-1} two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately \rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right) three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum p of an object is equal its mass m times its velocity v. That is: \vec{p} = m \cdot \vec{v}.

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. In other words, its velocity would become approximately 10\; \rm m \cdot s^{-1} more negative every second.

The initial velocity of the ball is 60\; \rm m \cdot s^{-1}. After two seconds, its velocity would have become 60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}. The momentum of the ball at that time would be around p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}.

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be 0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}. Accordingly, the ball's momentum at that moment would be p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right).

3 0
3 years ago
Which has the larger kinetic energy, a 10 g bullet fired at 400 m/s or a 80 kg bowling ball rolled at 6.5 m/s ?
mixer [17]
Formulae for Kinetic energy is:
Kinetic Energy= 1/2xmassx(velocity)^2

For comparison we need to have same units,thus we convert 10g into Kg.
10g/1000=0.01Kg

Input the value of bullet in the formulae;
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Input value of the ball:
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K.E=1690J

Which means that th Energy of the ball is more than the bullet.
7 0
3 years ago
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