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BlackZzzverrR [31]
2 years ago
11

A varying force is given by F=Ae ^-kx, where x is the position;A and I are constants that have units of N and m^-1 , respectivel

y. What is the work done when x goes from 0.10 m to infinity?
Physics
1 answer:
Burka [1]2 years ago
4 0
W = ∫ (x from 0.1 to +oo) F dx

= ∫ (x from 0.1 to +oo) A e^(-kx) dx

= A/k x [ - e^(-kx) ](between 0.1 and +oo)

= A/k x [ 0 + e^(-k * 0.1) ]

<span> = A/k x e^(-k/10) </span>
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A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi
OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

8 0
2 years ago
Wile E. Coyote wants to launch Roadrunner into the air using a long lever asshown below. The lever starts at rest before the Coy
algol13

Answer:

Explanation:

Find attached the solution

4 0
2 years ago
You are wallpapering two walls of a room. One wall measures 15 ft by 12 ft and the other measures 9 ft by 12 ft. The wall paper
enyata [817]

Answer:

5.76 round off to 6

Explanation:

wall 1 = 15 × 12 = 180

wall 2 = 9 × 12 = 108

now 1 roll covers 50 square feet

formula = wall 1 + wall 2 / 50

= 180 + 108 / 50

= 288÷ 50

= 5.76

4 0
2 years ago
Scotesia swims from the north end to the south end of a 50.0 m pool in 20.0 s. As she begins to make the return trip , Sean, who
slega [8]

Answer:

a) 2.5m/s

b) 0.91m/s

c) 0m/s

Explanation:

Average velocity can be said to be the ratio of the displacement with respect to time.

Average speed on the other hand is the ratio of distance in relation to time

Thus, to get the average velocity for the first half of the swim

V(average) = displacement of first trip/time taken on the trip

V(average) = 50/20

V(average) = 2.5m/s

Average velocity for the second half of the swim will be calculated in like manner, thus,

V(average) = 50/55

V(average) = 0.91m/s

Average velocity for the round trip will then be

V(average) = 0/75, [50+25]

V(average) = 0m/s

3 0
2 years ago
An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.
Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

\sin\theta=\dfrac{v'}{v}

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

\sin\theta=\dfrac{75}{310}

\theta=\sin^{-1}(\dfrac{75}{310})

\theta=14.0^{\circ}

We need to calculate the resultant speed

Using formula of resultant speed

\cos\theta=\dfrac{v''}{v}

Put the value into the formula

\cos14=\dfrac{v''}{310}

v''=\cos14\times310

v''=300.79\ km/h

Hence, The speed of the plane relative to the ground is 300.79 km/h.

6 0
2 years ago
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