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3 years ago

A 10 kg brick and a 1 kg book are dropped in a vacuum. The force of gravity on the 10 kg brick is what?

2 answers:

7 0

<span>10 times as much. Since F=m*a, and a is constant, the only thing that affects force is the mass.

In response to the below answer, the acceleration due to gravity does not change. The force due to gravity definitely DOES change depending on the mass of the object. Since the force is what the problem asks for, the answer is 10</span>

LuckyWell [14K]3 years ago

3 0

If this is happening on Earth, then the force of gravity between the Earth and the brick is 98 Newtons. It doesn't matter whether the brick is falling, rising, dropped in a vacuum, in water, or into chicken soup, or what else is dropped near it. None of that has any effect on the gravitational force between the Earth and the brick.

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Water flows through a first pipe of diameter 3 inches. If it is desired to use another pipe for the same flow rate such that the

Alborosie

Answer:

the diameter of the second pipe is 2.52 in

Explanation:

Given the data in the question;

We know that; the rate of flow is the same;

Av1 = Av2

v ∝ √h

$\frac{A1}{A2}$ = $\frac{V2}{V1}$

$\frac{A1}{A2}$ = √( $\frac{h2}{h1}$ )

( π/4.D1² / π/4.D2² ) = √( $\frac{h2}{h1}$ )

( D1² / D2² ) = √( $\frac{2h1}{h1}$ ) since second is double of first

( D1² / D2² ) = √( $\frac{2}{1}$ )

3² / D2² = √2

D2²√2 = 9

D2² = 9/√2

D2² = 9 / 1.4142

D2² = 6.364

D2 = √ 6.364

D2 = 2.52 in

Therefore, the diameter of the second pipe is 2.52 in

3 0

3 years ago

why does diving 30m below sea level affect our bodies more than being in a building 30m above sea level

Genrish500 [490]

Imagine you are in a swimming pool 30m deep. Assuming you know that water is denser than air, you would know that the 30m of water above you will carry more weight, and press down on your body. Say you were in a swimming pool 60m deep, you would be sandwiched between 30m of water pressing down on you, and the upthrust created by the 30m of water below you.

In a building 30m up, the pressure will be regulated, as you are in a building. The floor will be strong enough to support the weight of the body, and the body will not recoil into itself.

5 0

3 years ago

You wish to buy a motor that will be used to lift a 10-kg bundle of shingles from the ground to the roof of a house. The shingle

gogolik [260]

Answer:

$\tau=19.21\ N-m$

Explanation:

It is given that,

Mass of bundle of shingles, m = 10 kg

Upward acceleration of the shingles, $a=1.5\ m/s^2$

The radius of the motor of the pulley, r = 0.17 m

Let T is the tension acting on the shingles when it is lifted up. It can be calculated as :

$T-mg=ma$

$T=m(g+a)$

$T=10\times (9.8+1.5)$

T = 113 N

Let $\tau$ is the minimum torque that the motor must be able to provide. It is given by :

$\tau=r\times T$

$\tau=0.17\times 113$

$\tau=19.21\ N-m$

So, the minimum value of torque is 19.21 N-m. Hence, this is the required solution.

4 0

3 years ago

If the car passes point A with a speed of 20 m/s and begins to increase its speed at a constant rate of at = 0.5 m/s2 , determin

IceJOKER [234]

Answer:

$1.68 \frac{m}{s^2}$

Explanation:

Please find the image for the question as attached file.

Solution -

Given -

First of all we will calculate the velocity at point C,

As per newton's third law of motion-

$V_C^2 = V_A^2 + 2 a_t (S_C - S_A)\\$

Substituting the given values in above equation, we get -

$V_C^2 = 20^2 + 2*0.5*(100-0)\\V_C = 22.361 \frac{m}{s}$

Now we will determine the radius of curvature for the curve shown in the attached image

$Y = 16 - \frac{1}{625} X^2\\$

Differentiating on both the sides, we get -

$\frac{dy}{dx} = -3.2 (10^-3) X\\\frac{d^2y}{d^2x} = -3.2 (10^-3)\\Curve = \frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}} }{\frac{d^2y}{d^2x}} \\Curve = 312.5$ meter