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3 years ago

Physics question about density

Physics

2 answers:

scZoUnD [109]3 years ago

7 0

So the right answer is 4kg/m^3.

Look at the attached picture

Hope it will help you

Good luck on your assignment

ohaa [14]3 years ago

6 0

Answer:

The answer is option 4.

Explanation:

Firstly, you have to convert the mass of tea bagto kilogram as the question wants the unit in kg/m³.

$1000g = 1kg$

$100g = 1kg \div 10$

$100g = 0.1kg$

So the mass of tea bag is 0.1 kg. Next you have to use density formula ρ = mass/volume. Then you have to substitute the values into the formula :

$ρ = \frac{mass}{volume}$

Let mass = 0.1kg,

Let volume = 0.025m³,

$ρ = \frac{0.1}{0.025}$

$ρ = 4kg \: per \: {m}^{3}$

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A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

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2 years ago

You have a spring-loaded air rifle. When it is loaded, the spring is compressed 0.3 m and has a spring constant of 150 N/m. In j

Feliz [49]

The potential energy of the spring is 6.75 J

The elastic potential energy stored in the spring is given by the equation:

$E= \frac{1}{2} kx^2$

where;

k is the spring constant

x is the compression/stretching of the string

In this problem, we have the spring as follows:

k = 150 N/m is the spring constant

x = 0.3 m is the compression

Substituting in the equation, we get

$E=\frac{1}{2} (150) (0.3)^2$

$E=6.75J$

Therefore. the elastic potential energy stored in the spring is 6.75J .

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3 years ago

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Jet001 [13]

Answer:

C) 3,000 kg m/s

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We can consider the horizontal velocity of the motorcycle to be zero, since it rolls off the edge of the cliff very slowly. So, we only need to find the vertical velocity at the time of the impact with the ground.

The vertical velocity of the motorcycle at time t is given by (free-fall motion):

$v(t)=v_0 -gt$

where

$v_0=0$ is the initial vertical velocity (zero, since the motorcycle is not moving)

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Since the motorcycle hits the ground after t = 3 seconds, we have

$v(3 s)=0-(9.8 m/s^2)(3 s)=-29.4 m/s$

And since we know its mass, m=100 kg, we can find its momentum:

$p=mv=(100 kg)(-29.4 m/s)=-2940 kg m/s \cdot -3000 kg m/s$

and the negative sign simply means downward direction.

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