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2 years ago

8

A 50.0 kg object is moving at 18.2 m/s when a 200 N force is applied opposite the direction of the objects motion, causing it to

slow down to 12.6 m/s. How long was this force applies?

1 answer:

Gekata [30.6K]2 years ago

3 0

Answer:

t = 1.4[s]

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

$P=m*v\\or\\P=F*t$

where:

P = impulse or lineal momentum [kg*m/s]

m = mass = 50 [kg]

v = velocity [m/s]

F = force = 200[N]

t = time = [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

$(m_{1}*v_{1})-F*t=(m_{1}*v_{2})$

where:

m₁ = mass of the object = 50 [kg]

v₁ = velocity of the object before the impulse = 18.2 [m/s]

v₂ = velocity of the object after the impulse = 12.6 [m/s]

$(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]$

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What is the acceleration of a 150 kg object pushed with a force of 1000 N?

kirill115 [55]

Answer:

<h2>6.67 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{1000}{150} \\ = 6.66666..$

We have the final answer as

<h3>6.67 m/s²</h3>

Hope this helps you

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2 years ago

The basal metabolic rate is the rate at which energy is produced in the body when a person is at rest. A 75-kg (165-lb) person o

CaHeK987 [17]

Answer:

(a) Q = 142.67 W

(b) Basal Metabolic Rate = 178.33 W

Explanation:

(a)

We can find the heat radiated by the person by using Stefan-Boltzman's law:

$Q = \sigma A (T^4 - T_{s}^4)\\$

where,

Q = heat radiated per second = ?

σ = Stefan-Boltzman Constant = 5.6703 x 10⁻⁸ W/m².k⁴

A = Surface Area = 2 m²

T = Temperature of Skin = 30° C + 273 = 303 k

Ts = Temperature of room = 18° C +273 = 291 k

Therefore,

$Q = (5.6703\ x\ 10^{-8}\ W/m^2.k^4)(2\ m^2)[(303\ k)^4-(291\ k)^4]$ <u></u>

<u>Q = 142.67 W</u>

<u></u>

(b)

Since the heat calculated in part (a) is 80 percent of basal metabolic rate. Therefore,

$Q = (0.8)(Basal\ Metabolic\ Rate)\\Basal\ Metabolic\ Rate = \frac{Q}{0.8}\\\\Basal\ Metabolic\ Rate = \frac{142.67\ W}{0.8}$

<u>Basal Metabolic Rate = 178.33 W</u>

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2 years ago

A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of

andre [41]

Answer:

The angle of incidence is greater than the angle of refraction

Explanation:

Refraction occurs when a light wave passes through the boundary between two mediums.

When a ray of light is refracted, it changes speed and direction, according to Snell's Law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where :

$n_1$ is the index of refraction of the 1st medium

$n_2$ is the index of refraction of the 2nd medium

$\theta_1$ is the angle of incidence (the angle between the incident ray and the normal to the boundary)

$\theta_2$ is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

In this problem, we have a ray of light passing from air into clear plastic. We have:

$n_1=1.00$ (index of refraction of air)

$n_2=1.50$ approx. (index of refraction in clear plastic)

Snell's Law can be rewritten as

$sin \theta_2 =\frac{n_1}{n_2}sin \theta_1$

And since $n_2>n_1$ , we have

$\frac{n_1}{n_2}$

And so

$\theta_2$

Which means that

The angle of incidence is greater than the angle of refraction

6 0

2 years ago

What provides the vertical force to balance the force of gravity on the pendulum bob?

Zepler [3.9K]

Answer:

the string and metre rule

Explanation:

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1 year ago

If the lens diameter is doubled, what happens to the resolvable angle of images produced at fixed wavelength?a. The angle decrea

Sergeu [11.5K]

Answer:

Option A

Explanation:

Angular resolution for any optical equipment can be defined as the ability of that tool to differentiate the smallest details of the image formed.

The angular resolution is given by:

$\theta_{R} = \frac{1.22\lambda}{d}$ (1)

where

$\theta_{R}$ = Angular Resolution

$\lambda$ = wavelength

d = diameter of the lens

Now,

As per the question:

If the diameter of the lens is doubled, i.e., d' = 2d

Then

From eqn (1):

$\theta'_{R} = \frac{1.22\lambda}{d'} = \frac{1.22\lambda}{2d}$

$\theta'_{R} = \frac{1}{2}\frac{1.22\lambda}{d} = \frac{1}{2}\theta_{R}$

Thus when the diameter is doubled the angular resolution becomes half of its original value.

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2 years ago