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zhannawk [14.2K]
4 years ago
8

What happens when an object is dropped?

Physics
1 answer:
katrin2010 [14]4 years ago
3 0
When an object is dropped, tossed, or kicked, as long as it is not laying on the ground, it accelerates downward, because of the force of gravity acting on it.
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Visible light waves do not diffract as well as radio waves because?​
Ivenika [448]
They have more energy than radio waves.
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3 years ago
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QUESTION 1 If you do 1,000 J of work on a system and remove 500 J of heat from it, what is the change in its internal energy?
Mazyrski [523]
1. U = Q + W
    U = -500 + 1000
    U = 500 J

2. The first law of thermodynamic is about the law of conservation of energy where energy in should be equal to energy out.

3. It is the windmill that does not transform energy from heat to mechanical instead it is the transforms the opposite.

4. In a heat engine, work is used to transfer thermal energy from a hot reservoir to a cold one.

5.  5.00 × 10^4 J - 2.00 × 10^4 J = 3.00 × 10^4 J 

6. To increase the work done, we raise the temperature of the cold reservoir. 
7 0
3 years ago
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A bowling ball of mass 4 kg moves in a straight line at 3.93 m/s. How fast must a Ping-Pong ball of mass 2.293 g move in a strai
Nonamiya [84]

M = mass of the bowling ball = 4 kg

V = speed of bowling ball = 3.93 m/s

P = magnitude of momentum of bowling ball = ?

magnitude of momentum of bowling ball is given as

P = MV

inserting the values

P = 4 x 3.93

P = 15.72 kgm/s

m = mass of ping-pong ball = 2.293 g = 2.293 x 10⁻³ kg

v = speed of the ping-pong ball = ?

p = magnitude of momentum of ping-pong ball

Given that :

magnitude of momentum of ping-pong ball  =  magnitude of momentum of bowling ball

p = P

m v = 15.72

(2.293 x 10⁻³) v = 15.72

v = 6.86 x 10³ m/s


4 0
3 years ago
A car is traveling at 24.0 m/s when the driver suddenly applies the brakes, causing the car to slow down with constant accelerat
Elan Coil [88]

Answer:

The answer to your question is : vf = 15.18 m/s

Explanation:

Data

vo = 24 m/s

d = 120 m

vf = ? when d = 60.0 m

Formula

              vf² = vo²  + 2ad

For d =100m

            a = (vf² - vo²) / 2d

            a = (0 -24²) / 2(100)

            a = -576/200

             a = 2.88 m/s²

Now, when d = 60

        vf² = (24)² - 2(2.88)(60)

        vf² = 576 - 345.6

        vf² = 230.4

        vf = 15.18 m/s

4 0
3 years ago
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How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius
maria [59]

Answer:

897

Explanation:

Speed of the car, v = 126 km/h, converting to m/s, we have v = 35 m/s and

Radius of the curve, R = 150 mm = 0.15 m

The centripetal acceleration a(c) is given by the formula = v² / R so that

a(c) = 35² / 0.15

a(c) = 1225 / 0.15

a(c) = 8167 m/s²

The force that causes the acceleration is frictional force = µ m g, where

µ = coefficient of friction

m = the mass of the car and

g = acceleration due to gravity, 9.81

From Newton's law:

µ m g = m a(c) , so that

µ = a(c) / g

µ = 8167 / 9.81

µ = 897

Therefore, the coefficient of static friction must be as big as 897

5 0
3 years ago
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