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3 years ago

What upward gravitational force does a 5600kg elephant exert on the earth?

2 answers:

8 0

Force is determined by multiplying mass and gravity (F= mg). To determine the answer, the mass of the elephant (5600 kg) is multiplied with the gravity (9.8 m/s²). The answer is 5800 N. This is the upward gravitational force that the elephant exerts on the earth.

Inessa [10]3 years ago

6 0

<u>Answer</u>

54,880 N

<u>Explanation</u>

The Gravitational force is the force exited by an object towards the center of the earth. It is called the weight of an object. The Newton's 2nd law of motion says that for every action there is equal and opposite reaction. So, if the elephant exerts a force on the earth, the earth exerts an equal force to the elephant.

Weight = mass × gravity

= 5600 × 9.8

= 54,880 N

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if a ball with an original velocity of 0 is dropped from a tall structure and takes 7 Seconds to hit the ground what velocity do

krok68 [10]

$a_y=\dfrac{v_y-v_{0y}}t\implies-9.81\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{v_y-0}{7\,\mathrm s}\implies v_y=-68.7\,\dfrac{\mathrm m}{\mathrm s}$

5 0

3 years ago

What can you say about the magnitudes of the forces that the balloons exert on each other?

maxonik [38]

Answer:

$F_G=G. \frac{m_1.m_2}{R^2}$ gravitational force

$F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}$ electrostatic force

Explanation:

The forces that balloons may exert on each other can be gravitational pull due to the mass of the balloon membrane and the mass of the gas contained in each. This force is inversely proportional to the square of the radial distance between their center of masses.

The Mutual force of gravitational pull that they exert on each other can be given as:

$F_G=G. \frac{m_1.m_2}{R^2}$

where:

$G=$ gravitational constant $=6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}$

$m_1\ \&\ m_2$ are the masses of individual balloons

$R=$ the radial distance between the center of masses of the balloons.

But when there are charges on the balloons, the electrostatic force comes into act which is governed by Coulomb's law.

Given as:

$F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}$

where:

$\rm \epsilon_0= permittivity\ of\ free\ space$

$q_1\ \&\ q_2$ are the charges on the individual balloons

R = radial distance between the charges.

3 0

3 years ago

Two cylindrical containers are placed on a turntable. canister A is empty; canister B contains lead shots. each canister is the

devlian [24]

Answer:

c) both containers slide off the turntable at the same turntable speed.

Explanation:

The greater the mass of the cylinder, the greater the centrifugal force acting on it. That makes up for the greater force needed to move the heavier cylinder.

5 0

3 years ago

An object is propelled straight up from ground level with an initial velocity of 48 feet per second. Its height at time t is mod

Ket [755]

Answer:

Explanation:

For a. its max height and when it occurs. First the max height. That's a y-dimension thing, and in the y-dimension we have this info:

v₀ = 48 ft/s

a = -32 ft/s/s

v = 0 (the max height of an object occurs when the final velocity of the object is 0). Use the following equation for this part of the problem:

v² = v₀² + 2aΔx and filling in:

$0=48^2+2(-32)$ Δx and

0 = 2300 - 64Δx and

-2300 = -64Δs so

Δx = 36 feet.

Now for the time it takes to get to this max height. Final velocity is still 0 here, but the equation is a different one for this part of the problem. Use:

v = v₀ + at and filling in:

0 = 48 - 32t and

-48 = -32t so

t = 1.5 sec. That's part a. Onto part b:

The object hits the ground when its displacement, Δx, is 0. Use this equation for this problem:

Δx = v₀t + $\frac{1}{2}at^2$ and filling in:

$0=48t+\frac{1}{2}(-32)t^2$ and

$0=48t-16t^2$ and

0 = 16t(3 - t) so

t = 0 and t = 3. t = 0 is before the object is propelled, so it makes sense that at 0 seconds, the object was still on the ground, right? Then at 3 seconds, it's back on the ground. (Isn't math just perfectly, beautifully sensible!?) Now onto part c:

We are looking for the time interval when the object is >32 feet. So we use the same equation we just used, but with an inequality instead of an equals sign:

$48t+\frac{1}{2}(-32)t^2 >32$ and get everything on one side and factor it again:

$-16t^2+48t-32>0$ and we find that

1 < t < 2 so the time interval is between 1 and 2 seconds that the object is over 32 feet in the air.

8 0

3 years ago

According to Care-stream's operating manual, what is the acceptable range of exposure indicators?

Aleks [24]

Answer:

correct option is C. 1700 - 2300

Explanation:

solution

according to Care stream, exposure indicators provide the technical adequancy incident radiations on the x ray

and we muse care stream formula that is

Exposure Index EI = 1000 × $log_{10} (mR)$ + 2000

here 1 mR = 2000 EI

and 2 mR = 2300 EI

and 3 mR = 1700 EI

so here acceptable range of exposure indicators is 1700 - 2300

correct option is C. 1700 - 2300

5 0

3 years ago