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BARSIC [14]
3 years ago
7

A ball is thrown forward at 5 m/s how would the path of the ball differ on earth than on the moon

Physics
2 answers:
Levart [38]3 years ago
6 0
In this problem, we are given that a ball is thrown upward and that it has an initial velocity of 5 m/s. Its path both on Earth and on the moon will be trajectory in nature. This is because while its horizontal velocity is maintained, its vertical velocity increases. 
Degger [83]3 years ago
6 0

Answer:

The ball on the moon would have a more straight path than on the Earth, and it would have a longer range

Explanation:

In both cases, the motion of the ball is the motion of a projectile, which has a parabolic trajectory. In fact, the path is the resultant of two indipendent motions:

- A horizontal uniform motion, with constant horizontal velocity 5 m/s

- An accelerated vertical motion, with constant acceleration downward equal to the gravitational acceleration of the planet

On the Moon, the value of the gravitational acceleration is about 1/6 that of the Earth. This means that the downward acceleration of the ball on the Moon is smaller than on the Earth: so, the ball will take more time to reach the ground on the Moon, and so it will have a longer range, and a more straight trajectory.

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In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron
-BARSIC- [3]

Answer:

a) 11 m/s

b) 0.0564 s

Explanation:

Given:

m = 2100 kg

vi = 22 ..... m/s before collision

vf = 0 ......after collision to stop

Δs = 0.62   distance traveled after collision .. crumpling of truck

Part a

V_{avg} = \frac{vi- vf}{2}\\\\V_{avg} = \frac{22- 0}{2}\\\\ V_{avg} = 11 m/s

Part b

vf = vi + a*t\\vf^2 = vi^2 + 2*a*s\\\\0 = 22 + a*t\\t = -22 /a\\\\0 = 484 +2*a*(0.62)\\a = - 390.3232 m/s^2\\\\t = -22/(-390.3232)\\\\t = 0.0564 s

7 0
3 years ago
Read 2 more answers
List 10 uses of metals​
algol [13]

Answer:

Shiny metals such as copper, silver, and gold are often used for decorative arts, jewelry, and coins.

Strong metals such as iron and metal alloys such as stainless steel are used to build structures, ships, and vehicles including cars, trains, and trucks.

Some metals have specific qualities that dictate their use. For example, copper is a good choice for wiring because it is particularly good at conducting electricity. Tungsten is used for the filaments of light bulbs because it glows white-hot without melting.

Nonmetals are plentiful and useful. These are among the most commonly used:

Oxygen, a gas, is absolutely essential to human life. Not only do we breathe it and use it for medical purposes, but we also use it as an important element in combustion.

Sulfur is valued for its medical properties and as an important ingredient in many chemical solutions. Sulfuric acid is an important tool for industry, used in batteries and manufacturing.

Chlorine is a powerful disinfectant. It is used to purify water for drinking and fill swimming pools.

Explanation:

8 0
2 years ago
When the spring, with the attached 275.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the
topjm [15]

Answer:

The period of oscillation is 1.33 sec.

Explanation:

Given that,

Mass = 275.0 g

Suppose value of spring constant is 6.2 N/m.

We need to calculate the angular frequency

Using formula of angular frequency

\omega=\sqrt{\dfrac{k}{m}}

Where, m = mass

k = spring constant

Put the value into the formula

\omega=\sqrt{\dfrac{6.2}{275.0\times10^{-3}}}

\omega=4.74\ rad/s

We need to calculate the period of oscillation,

Using formula of time period

T=\dfrac{2\pi}{\omega}

Put the value into the formula

T=\dfrac{2\pi}{4.74}

T=1.33\ sec

Hence, The period of oscillation is 1.33 sec.

4 0
3 years ago
A glider with mass m = 0.230 kg sits on a frictionless horizontal air track, connected to a spring with force constant k = 4.50
loris [4]

Answer

given,

mass of glider = 0.23 Kg

spring constant = k = 4.50 N/m

spring stretched to 0.130 m

The springs potential energy =

 U = \dfrac{1}{2}kx^2

 U = \dfrac{1}{2}\times 4.5 \times 0.13^2

        U = 0.038 J

at x = 0,the only energy will be kinetic .

 \dfrac{1}{2}mv^2=0.038

 \dfrac{1}{2}\times 0.23 \times v^2=0.038

         v² = 0.3304

         v = 0.575 m/s

displacement of the glider

      using conservation of energy

 \dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2

 x =v\sqrt{\dfrac{m}{k}}

 x =3\times \sqrt{\dfrac{0.23}{4.5}}

        x = 0.678 m

8 0
3 years ago
On a calm day with no wind, you can run a 1500-m race at a velocity of 4.0 m/s. If you run the same race on a day when you have
lesantik [10]

Answer:

The time taken to finish the race is 750 s.

Explanation:

The velocity of the person on the day of wind is slowed down by 2.0 m/s. So the person's velocity on the day of wind is 4-2=2 m/s.

The relation between time, speed and distance is t=v/d

Given d=1500 m and calculated v= 2 m/s.

t=1500/2

t=750 s.

Learn more about distance formula.

brainly.com/question/11954435

#SPJ10

3 0
1 year ago
Read 2 more answers
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