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shtirl [24]
4 years ago
7

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

Physics
2 answers:
Vikentia [17]4 years ago
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

<u>Explanation:</u>

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

A)

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 car runs for about 10 years during its average useful life. If the odometer reads 120,000 miles at the end, what was the avera
Eva8 [605]

Answer:

12000 mph

Explanation:

Given that,

Distance read by the odometer = 120,000 miles

Duration of car, t = 10 years

We need to find the average speed of the car. Speed of an object is equal to the total distance covered divided by total time taken. So,

v=\dfrac{120000}{10}\\\\=12000\ mph

So, the average speed of the car is 12000 mph.

7 0
3 years ago
An aircraft flies at sea level at a speed of 220 m/s. What is the highest pressure that can be acting on the surface of the airc
goldenfox [79]

Answer:

An aircraft flying at sea level with a speed of 220 m/s, has a highest pressure of 29136.8 N/m²

Explanation:

Applying Bernoulli's equation, we determine the highest pressure on the aircraft.

P = \frac{1}{2} \rho V^2

where;

P is the highest pressure on the aircraft

\rho is the density of air = 1.204 kg/m³ at sea level temperature.

V is the velocity of the aircraft = 220 m/s

P = 0.5*1.204*(220)² = 29136.8 N/m²

Therefore, an aircraft flying at sea level with a speed of 220 m/s, has a highest pressure of 29136.8 N/m²

6 0
3 years ago
Read 2 more answers
An object is located 50.0 cm from a concave mirror. The magnitude of the mirror focal length is 25.0 cm. What is the image dista
ioda

Answer:

Correct answer: C. 50 cm

Explanation:

Given data:

The distance of the object from the top of the concave mirror o = 50.0 cm

The magnitude of the concave mirror focal length 25.0 cm.

Required : Image distance d = ?

If we know the focal length we can calculate the center of the curve of the mirror

r = 2 · f = 2 · 25 = 50 cm

If we know the theory of spherical mirrors and the construction of figures then we know that when an object is placed in the center of the curve, there is also a image in the center of the curve that is inverted, real and the same size as the object.

We conclude that the image distance is 50 cm.

We will now prove this using the formula:

1/f = 1/o + 1/d => 1/d = 1/f - 1/o = 1/25 - 1/50 = 2/50 - 1/50 = 1/50

1/d = 1/50 => d = 50 cm

God is with you!!!

6 0
3 years ago
Read 2 more answers
A square-based shipping crate is being designed that must contain a volume of 16 ft3 . The material that is used for the base an
Vlada [557]

Answer:

Explanation:

Given

volume V=16 ft^3

Suppose base is square with side L

height of crate is h

Volume V=L^2\times h

16=L^2\times h

Cost of top and bottom area c_1=3L^2

Cost of Side area c_2=4Lh\times 2=8Lh=8L\times \frac{16}{L^2}=\frac{128}{L}

Total Cost C=c_1+c_2

Total Cost C=3L^2+\frac{128}{L}

Differentiate C w.r.t Length

\frac{dC}{dL}=6L-\frac{128}{L^2}

L^3=\frac{128}{6}

L=2.75 ft

h=\frac{16}{2.75^2}=11.46 ft

Dimensions are L\times L\times h=2.75\times 2.75\times 11.46    

6 0
3 years ago
A student placed a ladder up against a wall as shown below. The normal force applied by the wall in the ladder will be directed:
solmaris [256]
The normal force is always perpendicular to the surface. So it would be straight to the left of the wall
6 0
3 years ago
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