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Mumz [18]
2 years ago
12

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas

t in neutral so as to just reach the top of the 150 m high hill at B with vB = 0? Neglect friction.
Physics
1 answer:
SVEN [57.7K]2 years ago
8 0

Answer:

the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s

Explanation:

given information

car's mass, m = 1200 kg

h_{A} = 100 m

v_{A} = v_{A}

h_{B} = 150 m

v_{B} = 0

according to conservative energy

the distance from point A to B, h = 150 m - 100 m = 50 m

the initial speed v_{A}

final speed  v_{B} = 0

thus,

v_{B}² = v_{A}² - 2 g h

0 = v_{A}² - 2 g h

v_{A}² = 2 g h

v_{A} = √2 g h

    = √2 (9.8) (50)

    = 31.3 m/s

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Solution: The correct order is: C, A, B

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If the sun casts shadows at different angles at the same time of day in different places, we can determine how much Earth curves.

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2 years ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

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\Delta t_{o} = \Delta t

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2 years ago
Tiana jogs 1.5 km along a straight path and then turns and jogs 2.4 km in the opposite direction. She then turns back and jogs 0
vichka [17]

Answer:

Distance: 4.6km Displacement= -0.2km

Explanation:

Total distance: 1.5+2.4+0.7= 4.6 km

Displacement: 1.5-2.4+0.7= -0.2km

The displacement may also be 0.2km, it just depends on if it wants it negative or not.

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2 years ago
In what state must matter exist for fusion reactions to take place
steposvetlana [31]

Answer:

Plasma

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2 years ago
A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of a cliff 75.0 m high.How much later does it reach t
Katarina [22]

Answer:

5.72 seconds

848.27 m/s

97.94 m

Explanation:

t = Time taken

u = Initial velocity = 15 m/s

v = Final velocity

s = Displacement

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v=u+at\\\Rightarrow 0=15-9.81\times t\\\Rightarrow \frac{-15}{-9.81}=t\\\Rightarrow t=1.52 \s

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s=ut+\frac{1}{2}at^2\\\Rightarrow s=15\times 1.52+\frac{1}{2}\times -9.81\times 1.52^2\\\Rightarrow s=11.47\ m

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Total distance travelled by the stone would be 75+11.47+11.47 = 97.94 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 86.47=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{86.47\times 2}{9.81}}\\\Rightarrow t=4.2\ s

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Speed just before hitting the water is 848.27 m/s

3 0
2 years ago
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