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Gwar [14]
3 years ago
13

15. A body moving with a velocity of 20 m/s begins to accelerate at 3 m/s2. How far does the body move in 5 seconds? A. 137.5 m

B. 196 m C. 260 m D. 162.5 m
Physics
2 answers:
Rudik [331]3 years ago
7 0

Answer is B. According to the equation of motion s = vt + 1/2 at2 Where s is distance covered, v is velocity, a is acceleration and t is time taken. So, by putting all the values, we get s = (20)(5) + 1/2 (3)(5)2 s = 100 + 1/2 (3)(25) s = 100 + 1/2 75 s = 100 + 37.5 s = 137.5 meters



adelina 88 [10]3 years ago
5 0

The Correct answer to this question for Penn Foster Students is: 137.5 m

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Answer: Because of the fine bore of the tube.

Explanation:

Temperature is the degree of hotness and coldness. And thermometer is the instrument use to measure temperature.

The two most common types of themometric fluids for thermometer are alcohol and mercury.

What makes a clinical thermometer suitable for measuring small changes in body temperature is because of the fine bore of the tube which makes it possible for small temperature changes to cause large changes in the length of mercury columns, making the thermometer very sensitive to temperature changes.

The most prominent feature of the thermometer is the kink or constriction of bore near the bulb.

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A research submarine has a 30-cm-diameter window that is 8.1 cm thick. The manufacturer says the window can withstand forces up
malfutka [58]

The pressure at a certain depth underwater is:

P = ρgh

P = pressure, ρ = sea water density, g = gravitational acceleration near Earth, h = depth

The pressure exerted on the submarine window is:

P = F/A

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A = π(d/2)²

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h = F/(ρgπ(d/2)²)

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5 0
3 years ago
A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.
SVETLANKA909090 [29]

A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

W=mg: weight of the ladder, with m = 20 kg (mass) and g=9.8 m/s^2 (acceleration of gravity)

W_M=Mg: weight of the person, with M = 54 kg (mass)

N_1: normal reaction exerted by the wall on the ladder

N_2: normal reaction exerted by the floor on the ladder

F_f = \mu N_2: force of friction between the floor and the ladder, with \mu (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}

where

Wsin 21^{\circ} is the component of the weight of the ladder perpendicular to the ladder

W_M sin 21^{\circ} is the component of the weight of the man perpendicular to the ladder

N_1 sin 69^{\circ} is the component of the normal  force perpendicular to the ladder

And solving for N_1, we find the force exerted by the wall on the ladder:

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of N_2.

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

\sum F_y = 0\\N_2 - W - W_M =0

And substituting and solving for N2, we find:

N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

\sum F_x = 0\\F_f - N_1 = 0

And re-writing the equation,

\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

\mu = \frac{N_1}{N_2}

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)

And substituting, we get

N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332

Learn more about torques and equilibrium:

brainly.com/question/5352966

#LearnwithBrainly

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Answer:

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