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3 years ago

Emilio pushes a 100 kg freshman with 200 N of force. How much is the freshman accelerated?

Physics

1 answer:

ladessa [460]3 years ago

8 0

Explanation:

F = MA

200 = 100 * A

A = 200/100

A = 2m/sec^2

<h3><em>hope </em><em>it </em><em>helps </em><em>you </em></h3>

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Determine a formula for the acceleration of the system in terms of mA, mB, θ, and g. Ignore the mass of the cord and pulley. Exp

jekas [21]

Answer:

$a=\frac{mBg-mAgSin\theta}{mA+mB}$

Explanation:

Given two mass on an incline code $mA$ and $mB$ and an angle of inclination $\theta$ . $g$ . Assume that $mA$ is the weight being pulled up and $mB$ the hanging weight.

-The equations of motion from Newton's Second Law are:

$mBg-T=mBa$ where a is the acceleration.

#Substituting for $T$ (tension) gives:

$mBg-mAsin\theta-mAa=mBa$

#and solving for $a:$

$a=\frac{mBg-mAgSin\theta}{mA+mB}$ which is the system's acceleration.

8 0

3 years ago

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

7 0

2 years ago

If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front o

tangare [24]

The correct answer is letter A. 6 millimeters. <span>If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front of the lens, the height of the image is 6 millimeters.
</span>
Solution:
18 / x = 12 / 4
12x = 72
x = 6mm

7 0

3 years ago

Fraternity hazing is acceptable because it is an initational rite to the brotherhood

Fantom [35]

Answer:

that is the right answer

Explanation:

8 0

3 years ago

If both of these balls are traveling down a bowling lane at the SAME speed, which one would would have more kinetic energy?

Mariana [72]

<h2>Answer:</h2>

<u>Ball A has more kinetic energy</u>

<h2>Explanation:</h2>

As we know that Kinetic energy is given by

K. E = 1/2 mv²

Since K E is dependent upon both mass and velocity so increasing mass will produce more kinetic energy if the speed remains constant

As the mass of ball A is greater than ball B so we can say that the kinetic energy of ball is more than ball B

3 0

3 years ago