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faust18 [17]
2 years ago
15

You observe someone pulling a block of mass 41 kg across a low-friction surface. While they pull a distance of 5 m in the direct

ion of motion, the speed of the block changes from 5 m/s to 6 m/s. Calculate the magnitude of the force exerted by the person on the block.
Physics
2 answers:
arlik [135]2 years ago
8 0

Answer: 45.1N

Explanation:

Answer: 45.1N

Explanation:

The work done in the block is equal to change in energy is the system. We then go ahead to calculate W with the given values of initial and final speed.

W= Ef - Ei

W= 1/2MVf² - 1/2MVi²

W= 1/2M (Vf² - Vi²)

W= 1/2 * 41 (6²-5²)

W= 1/2 * 41 * 11

W= 225.5J

We note the definition of work and solve for F.

Taking into cognizance the displacement we have been given in the question. We then have

F = W/x

F = 225.5/5

F = 45.1N

Scilla [17]2 years ago
4 0

Answer:

45.1 N

Explanation:

Using Newton's Fundamental equation of Kinematics,

F = ma............. Equation 1

Where F = force exerted by the person on the block, m = mass of the block, a = acceleration of the block

We can look for the acceleration of the block by applying newton's equation of motion,

Using,

v² = u²+2as....................... Equation 2

Where v = final velocity, u = initial velocity, a = acceleration, s = distance.

make a the subject of the equation

a = (v²-u²)/2s................... Equation 3

Given: v = 6 m/s, u = 5 m/s, s = 5 m

Substitute into equation 3

a = (6²-5²)/(2×5)

a = 11/10

a = 1.1 m/s²

Also Given: m = 41 kg

Substitute into equation 1

F = 41(1.1)

F = 45.1 N

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Riders in a carnival ride stand with their backs against the wall of a circular room of diameter
Veseljchak [2.6K]

Answer:

option C

Explanation:

given,

diameter of circular room = 8 m

rotational velocity of the rider = 45 rev/min

                  = 45 \times \dfrac{2\pi}{60}

                  =4.712 rad/s

here in this case normal force is equal to centripetal force

N = m r ω²

N = m x 4 x 4.712²

N = 88.83m

frictional force = μ N

    = 88.83m x μ

now, for the body to not to slide

gravity force is equal to frictional force

m g = 88.83 m x μ

g = 88.83 x μ

9.8 = 88.83 x μ

 μ = 0.11

hence, the correct answer  is option C

6 0
3 years ago
A drowsy cat spots a flowerpot that sails first up and then down past an open window. the pot was in view for a total of 0.49 s,
Alika [10]

For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. <span>

An initial condition has been supplied by the problem: 

s=1.80m when t=0.245s 

<span>This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2) 

to solve for v, the initial velocity of the pot as it enters the cat's view through the window. Substituting and solving (note that gravitational acceleration is negative since this is opposite our coordinate orientation): 

(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2 

v=8.549m/s 

<span>Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

(v final)=(v initial)+(a)(t) 

to solve for the time it will take for the pot to reach the apex of its flight. Because (v final)=0, this equation will look like 

0=(v)+(a)(t) 

Substituting and solving for t: 

0=(8.549m/s)+(-9.81m/s^2)(t) 

t=0.8714s 

<span>Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

s=3.725m<span>

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window: 

3.725m – 1.8m=1.925m</span>

 

Answer:

<span>1.925m</span>

3 0
3 years ago
Which of the following air conditions would be least likely to have precipitation?
Lynna [10]
The best and most correct answer among the choices provided by the question is the first choice "warm, dry air"


In meteorology, precipitation<span> is any product of the condensation of atmospheric water vapor that falls under gravity. The main forms of </span>precipitation<span> include drizzle, rain, sleet, snow, graupel and hail.</span>

I hope my answer has come to your help. God bless and have a nice day ahead!
7 0
3 years ago
Read 2 more answers
You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner
omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

Q_1 = Q_2

V_1A_1=V_2A_2

Our values are given as,

A_1=\frac{1}{2}^2*\pi=0.785 in^2

A_2=\frac{5}{8}^2*\pi=1.227 in^2

Re-arrange the equation to find the first ratio of rates we have:

\frac{V_1}{V_2}=\frac{A_2}{A_1}

\frac{V_1}{V_2}=\frac{1.227}{0.785}

\frac{V_1}{V_2}=1.56

The second ratio of rates is

\frac{V2}{V1}=\frac{A_1}{A2}

\frac{V2}{V1}=\frac{0.785}{1.227}

\frac{V2}{V1}=0.640

3 0
3 years ago
an alligator crawls 25m to the left with an average velocity of -1.2m/s. how many seconds did the alligator crawl?
Crank
It will take 21s for yhe alligator to crawl that distance. Reported answer contains 2 Significant Figures.

5 0
3 years ago
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