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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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Someone please help with this
SSSSS [86.1K]

Answer:

<em>The new force is 2/3 of the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:

\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}

Factoring out 2/3:

\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}

Substituting the original force:

F'=\frac{2}{3}F

The new force is 2/3 of the original force

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3 years ago
Which kind of force do you exert on an object when you pull it toward you?
Vilka [71]
It's gravity I believe? I'm not entirely sure.
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3 years ago
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Imagine that a hypothetical life form is discovered on our moon and transported to Earth. On a hot day, this life form begins to
almond37 [142]

Answer:

The heat of vaporization 580 cal/g times 602g = cal in human  and do the same for life form.

Explanation:

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A ship, carrying freshwater to a desert island in the caribbean, has a horizontal cross-sectional area of 2800 m2 at the waterli
Alex17521 [72]
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3 0
3 years ago
A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes
OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

y(t) = h +u_y t - \frac{1}{2}gt^2

where

h = 2.5 km = 2500 m is the initial height

u_y = 0 is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

y(t) = 0

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s

So the horizontal distance travelled is

d=v_x t

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

d=(333.3)(22.6)=7533 m = 7.5 km

7 0
3 years ago
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