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lyudmila [28]
3 years ago
7

A 52.0-kg sandbag falls off a rooftop that is 22.0 m above the ground. The collision between the sandbag and the ground lasts fo

r a total of 17.0 ms. What is the magnitude of the average force exerted on the sandbag by the ground during the collision?
Physics
1 answer:
sergiy2304 [10]3 years ago
4 0

Answer:

F = 7,916,955.0N

Explanation:

According to newtons second law

Force = mass * acceleration

Given

mass = 52.0kg

distance S = 22.0m

time t = 17.0 ms = 0.017s

We need to get the acceleration first using the formula;

S = ut+ 1/2at²

22 = 0 + 1/2 a(0.017²)

22 = 0.0001445a

a = 22/0.0001445

a = 152,249.13m/s²

The magnitude of the average force exerted will be;

F = ma

F = 52 * 152,249.13

F = 7,916,955.0N

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The explanation for the following answer is explained below.

Explanation:

The sun is at an average distance of about 93,000,000 miles(150 million kilometers) away from the earth.It is so far away that light from the Sun,travelling at a speed of 186,000 miles (300,000 kilometers) per second, takes about 8 minutes to reach the earth.Earth does not travel around the Sun in a perfect circle.Instead its orbit is elliptical,like a stretched circle,with the sun just off the center of the orbit. At its closest,the Sun is 91.4 million miles (147.1 million kilometers ) away us.At its farthest ,the Sun is 94.5 million miles (152.1 million km) away.The Earth is closest to the Sun during winter in the northern hemisphere

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3 years ago
Suppose a radar gun sends out radio waves with a frequency of 2,000,000.0 Hz. The waves bounce off a moving car and return with
Igoryamba

Answer:

300,000,030m/s

Explanation:

Using the relationship between frequency and speed of a wave;

F ∝ V

F = kV

k = F/V

F1/V1 = F2/V2 = k

Let F1 be the frequency of the radio wave = 2,000,000.0Hz

V1 be the speed of light = 300,000,000m/s

F2 = frequency produced by the car = 2,000,000.2Hz

V2 be the velocity of the moving car

Substituting this values in the equation above;

2,000,000/300,000,000 = 2,000,000.2/V2

Cross multiplying

2,000,000V2 = 300,000,000×2,000,000.2

2V2 = 300×2,000,000.2

V2 = 150×2,000,000.2

V2 = 300,000,030m/s

The velocity of the car is 300,000,030m/s

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A 100-turn, 3.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60%u2218 away from vertical increase
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Answer:

Induced emf in the coil, E = 0.157 volts

Explanation:

It is given that,

Number of turns, N = 100

Diameter of the coil, d = 3 cm = 0.03 m

Radius of the coil, r = 0.015 m

A uniform magnetic field increases from 0.5 T to 2.5 T in 0.9 s.

Due to this change in magnetic field, an emf is induced in the coil which is given by :

E=-NA\dfrac{\Delta B}{\Delta t}

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Minus sign shows the direction of induced emf in the coil. Hence, the induced emf in the coil is 0.157 volts.

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