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garri49 [273]
3 years ago
7

Identical guns fire identical bullets horizontally at the same speed from the same height above level planes, one on the Earth a

nd one on the Moon. Which of the following three statements is/are true? I. The horizontal distance traveled by the bullet is greater for the Moon
Physics
1 answer:
olga nikolaevna [1]3 years ago
3 0

Answer:

horizontal distance on moon > horizontal distance on earth

Explanation:

Let the time of fall is t and the height from which they fall is h.

The time of fall is

t=\sqrt{\frac{2h}{g}}

Time of fall on earth

t=\sqrt{\frac{2h}{g_{e}}

Time of fall on moon

t'=\sqrt{\frac{2h}{g_{m}}

As the gravity on moon is less than the gravity on earth

So, t' > t

Horizontal distance = horizontal velocity x time

So, the horizontal distance on moon is more than the horizontal distance on earth.

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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;
Kinetic Energy= 1/2x0.01kgx(400)^2
K.E=800J

Input value of the ball:
Kinetic Energy=1/2x80kgx(6.5)^2
K.E=1690J

Which means that th Energy of the ball is more than the bullet.
7 0
3 years ago
A car travels 10.0 m/s. What is its velocity in km/h?
Vitek1552 [10]
Since 1m/s=3.6 km/h, we can conclude that 10.0m/s = 36 km/h
5 0
3 years ago
Geologists have evidence that the continents were once a single giant landmass that land mass eventually split apart and the ind
mars1129 [50]
The part that it played was in the changing of the environment/climate for the organisms that live on those continents. And the part it could've played was the way that the organisms had to adapt to that climate and it stayed that way over generations.

7 0
3 years ago
A 750 g air-track glider attached to a spring with spring constant 14.0 N/m is sitting at rest on a frictionless air track. A 20
alexandr402 [8]

Answer:

the amplitude of  the subsequent oscillations is 0.11  m

the period of the subsequent oscillations is 1.94 s

Explanation:

given Information:

the mass of air-track glider, m_{1} = 750 g = 0.75 kg

spring constant, k = 13.0 N/m

the mass of glider, m_{2} = 200 g = 0.2 kg

the speed of glider,  v_{2} = 170 cm/s = 1.7 m/s

the amplitude of  the subsequent oscillations is A = 0.11  m

according to mechanical enery equation, we have

A = \sqrt{\frac{m_{1} +m_{2} }{k} }v_{f}

where

A is the amplitude and  v_{f} is the final speed.

to find v_{f}, we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.

P_{f} = P_{i}

(m_{1} +m_{2} )v_{f} = m_{1} v_{1} +m_{2}v_{2}

v_{1} = 0, thus

(0.75+0.2)v_{f} = (0.75)(0)+(0.2)(1.7)

0.95 v_{f} = 0.34

v_{f} = 0.36 m/s

Now we can calculate the amplitude

A = \sqrt{\frac{0.75 +0.2 }{10} }0.36

A = 0.11  m

the period of the subsequent oscillations is T = 1.94 s

the equation for period is

T = 2π\sqrt{\frac{m_{1}+m_{2}  }{k} }

T = 2π\sqrt{\frac{0.75+0.2  }{10} }

T = 1.94 s

7 0
3 years ago
To practice problem-solving strategy 9.1 a strategy for conservation of momentum problems. an 80-kg quarterback jumps straight u
kherson [118]
Refer to the diagram shown below.

By definition momentum =  mass *  velocity.

Before throwing the ball:
The initial momentum is
P₁ = 0.

After throwing the ball:
Let u = the backward velocity of the quarterback.
The momentum is 
P₂ = (0.43 kg)*(15 m/s) + (80 kg)*(- u m/s)

Conservation of momentum requires that
P₂ = P₁
6.45 - 80u = 0
u = 6.45/80 = 0.0806 m/s

Answer: 0.08 m/s backward

6 0
3 years ago
Read 2 more answers
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