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4 years ago

This is what scientists call Newton's cradle. It is a handy devise used to illustrate many concepts in physics.

2 answers:

7 0

According to Newton's third law of motion, "The single ball on the left will move as high as the ball on the right was dropped" is the expected result once the ball hits the ball closest to it.

Answer: Option A

Explanation:

Newton’s cradle is the best example for law of conservation of momentum. So that means that energy or momentum exhibited by a particle can be transferred to its nearer object with equal force. So in the given options the first option is correct. As when the right ball is dropped it will collide with its neighboring ball and transfer energy.

By this transferring of energy from one particle to another, it will reach the leftmost ball. But there will be no loss in the energy as the left ball will move as high position as the ball on the right was dropped from. So that means the energy is conserved in the cradle.

Dmitriy789 [7]4 years ago

7 0

The single ball on the left will move as high as the ball on the right was dropped. When you lift the ball on the right, it comes back down with a force and the energy is transmitted through the three balls in the center to the ball on the far left. The ball on the far left then moves with a motion that is equal and opposite to the motion of the ball that you lifted.

HOPED THIS HELPED!!!!!!

(T_T)

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A 1500 kg car traveling at 8 m/s slows to a stop when the driver hits the brakes. If the braking force between the tires and the

wariber [46]

Answer:

First you fond the total force the car initialy has which is F=ma so it is 1500 times 8 which leads you to get 12000N then you divide the force of the car by the breaks and the road (4200N) which gives you 2.85 seconds for the car to come to a stop.

5 0

4 years ago

A certain gas is compressed adiabatically. The amount of work done on the gas is 800 J. What is the change in the internal (ther

Black_prince [1.1K]

Answer:

800J

Explanation:

Using the formula for change in the internal energy of a system

∆U = Q - W

Q = heat added to the system

W =workdone by system.

We know the process is an adiabatic one then, there no addition/ removal of heat, then Q= 0

(∆U = -W )

Then substitute for W, we have

∆U = -[-800]

∆U= 800J

∆U = 800J

4 0

3 years ago

What is the primary mode of energy transfer from hot coffee inside a Thermos bottle to the environment? View Available Hint(s) P

Keith_Richards [23]

Answer Part A: radiation

Part B: conduction

Part C: conduction

Part D: conduction

Part E: conduction

Part F: radiation

Part G::radiation

Part H : convection

Explanation:conduction is heat transfer through direct contact

Convection is through molecules or particles of water or air current

Radiation is heat transfer through empty space .

8 0

4 years ago

A projectile is launched with a velocity of 25 m/s at an angle of 20 degrees with respect to the horizontal. What is its initial

irina [24]

Answer:

b. 23.49 m/s

d. 8.55 m/s

d. .87 s

a. 40.8 m

Explanation:

The initial along the x-axis is given by the following formula:

$V_{ox} = V_{o}Cos\theta\\\\V_{ox} = (25\ m/s)(Cos\ 20^{o})\\\\V_{ox} = 23.49\ m/s$

Hence, the correct option is:

b. 23.49 m/s

The initial along the x-axis is given by the following formula:

$V_{oy} = V_{o}Sin\theta\\V_{oy} = (25\ m/s)(Sin\ 20^{o})\\V_{oy} = 8.55\ m/s$

Hence, the correct option is:

d. 8.55 m/s

Now, for the time to reach maximum height:

$t = \frac{V_{o}Sin\theta}{g}\\\\t = \frac{(25\ m/s)Sin\ 20^{o}}{9.8\ m/s^{2}}\\\\t = 0.87\ s$

Hence, the correct option is:

d. .87 s

For the range of projectile:

$R = \frac{V_{o}^{2}\ Sin\ 2\theta}{g}\\\\R = \frac{(25\ m/s)^{2}(Sin\ 40^{o})}{9.8\ m/s^{2}}\\\\R = 40.9\ m$

Hence, the closest option is:

a. 40.8 m

5 0

3 years ago

ASAP

nikitadnepr [17]

Answer:

A. 59.4

Explanation:

The refractive index of the glass, n₁ = 1.50

The angle of incidence of the light, θ₁ = 35°

The refractive index of air, n₂ = 1.0

Snell's law states that n₁·sin(θ₁) = n₂·sin(θ₂)

Where;

θ₂ = The angle of refraction of the light, which is the angle the light will have when it passes from the glass into the air

Therefore;

θ₂ = arcsin(n₁·sin(θ₁)/n₂)

Plugging in the values of n₁, n₂ and θ₁ gives;

θ₂ = arcsin(1.50 × sin(35°)/1.0) ≈ 59.357551° ≈ 59.4°

The angle the light will have when it passes from the glass into the air, θ₂ ≈ 59.4°.

6 0

3 years ago