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Setler [38]
3 years ago
8

A 2.44 kg block is pushed 1.55 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle o

f 69.6 ◦ with the horizontal. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of kinetic friction between the block and wall is 0.691, find the work done by F.
Physics
1 answer:
svlad2 [7]3 years ago
6 0

Answer:

The work done is L= 49.83 Joules

Explanation:

m= 2.44kg

d= 1.55m

α= 69.6°

g= 9.8 m/s²

μ= 0.691

F= ?

Fy= F*sin(α)

Fx= F*cos(α)

Fr= μ * Fx

Fr= μ * F*cos(α)

W= m*g

W= 23.91 N

Fy - W - Fr = 0

F*sin(α) - W -  μ * F*cos(α) = 0

F* ( sin(α) - μ *cos(α) ) = W

F= W /  ( sin(α) - μ *cos(α) )

F= 34.3 N

L= Fy * d

L= 49.83 J

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Answer:

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this average velocity is not necessarily the velocity of the extreme points,

Explanation:

To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.

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this average velocity is not necessarily the velocity of the extreme points, in the only case that it is so is when there is no acceleration.

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A sprinter set a high school record in track and field, running 200.0 m in 20.6 s . what is the average speed of the sprinter in
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Answer : The average speed of the sprinter is, 34.95 Km/hr

Solution :

Average velocity : It is defined as the distance traveled by the time taken.

Formula used for average velocity :

v_{av}=\frac{d}{t}

where,

v_{av} = average velocity

d = distance traveled = 200 m

t = time taken = 20.6 s

Now put all the given values in the above formula, we get the average velocity of the sprinter.

v_{av}=\frac{200m}{20.6s}\times \frac{3600}{1000}=34.95Km/hr

conversion :

(1 Km = 1000m)

(1 hr = 3600 s)

Therefore, the average speed of the sprinter is, 34.95 Km/hr

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What is an example of kinetic energy
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Momentum=mass X velocity
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Answer:

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A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes
Agata [3.3K]

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

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Substituting the values;

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Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

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