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stich3 [128]
2 years ago
6

Question 20 of 20

Physics
1 answer:
Sphinxa [80]2 years ago
6 0

Answer:

a

Explanation:

hI I

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O'Malley is riding on a bus which is moving at 10 m/s, and he throws a ball which he observes to be moving at 10 m/s relative to
Vikki [24]

Answer:

<em>20 m/s in the same direction of the bus.</em>

Explanation:

<u>Relative Motion </u>

Objects movement is always related to some reference. If you are moving at a constant speed, all the objects moving with you seem to be at rest from your reference, but they are moving at the same speed as you by an external observer.

If we are riding on a bus at 10 m/s and throw a ball which we see moving at 10 m/s in our same direction, then an external observer (called Ophelia) will see the ball moving at our speed plus the relative speed with respect to us, that is, at 20 m/s in the same direction of the bus.

3 0
3 years ago
(a) A light-rail commuter train accelerates at a rate of 1.35 m/s2 . How long does it take to reach its top speed of 80.0 km/h,
Mashcka [7]

Answer:

(a) Time t = 16.46 sec

(b) Time t =13.466 sec

(c) Deceleration = 2.677m/sec^2

Explanation:

(a) As the train starts from rest its initial velocity u = 0 m/sec

Acceleration a=1.35m/sec^2

Final speed v = 80 km/hr

80km/hr=\frac{80\times 1000}{3600sec}=22.22m/sec

From first equation of motion v =u+at

So t=\frac{v-u}{a}=\frac{22.22-0}{1.35}=16.46 sec

(b) Now initial speed u = 22.22 m/sec

As finally train comes to rest so final speed v=0 m/sec

Deceleration a=1.65m/sec^2

So t=\frac{v-u}{a}=\frac{0-22.22}{-1.65}=13.466 sec

(c) We have given that initial velocity = 80 km/hr = 22.22 m/sec

Final velocity v = 0 m/sec

Time t = 8.30 sec

So acceleration is given by

a=\frac{v-u}{t}=\frac{0-22.22}{8.3}=-2.6771m/sec^2

As acceleration is negative so it is a deceleration

7 0
4 years ago
a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
Are the two expressions equvalent 7(8x+5) and 48x + 35
Evgesh-ka [11]
Yes they are equivalent because 7x5=35 and 8x x 5=48x
4 0
3 years ago
What force, in newtons, must you exert on the balloon with your hands to create a gauge pressure of 62.5 cm H2O, if you squeeze
yulyashka [42]

Answer:54.70 N

Explanation:

Given

Gauge Pressure of 62.5 cm of H_2O

i.e. h=62.5 cm =0.625 m

Effective area A=51 cm^2

initial Pressure= 1 atm=101.325 kPa

Gauge Pressure P=\rho gh

\rho =density\ of\ water =1000 kg/m^3

P_{gauge}=1000\times 9.8\times 0.625=5.937 kPa

Force creates a pressure of P_1 which will be equal to Gauge Pressure

P_1=\frac{F}{A}

P_1=P_{gauge}

\frac{F}{A}=5.937 kPa

F=5.937\times 51\times 10^{-4}\times 10^3

F=30.27 N

6 0
4 years ago
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