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Vitek1552 [10]
3 years ago
13

A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the hori

zontal. The magnitude of the friction force acting on the block is:
Physics
1 answer:
vladimir2022 [97]3 years ago
4 0

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

  • The horizontal component of the pulling force, F cos \theta, where F = 50 N is the magnitude and \theta=60^{\circ} is the angle between the direction of the force and the horizontal; this force acts in the  forward direction
  • The force of friction, F_f, acting in the backward direction

According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

\sum F_x = ma_x

where

m is the mass of the block

a_x is the horizontal acceleration

However, the block is moving at constant speed, so the acceleration is zero:

a_x = 0

So the equation becomes

\sum F_x = 0 (1)

The net force here is given by

\sum F_x = F cos \theta - F_f (2)

And so, by combining (1) and (2), we find the magnitude of the friction force:

F cos \theta - F_f = 0\\F_f = F cos \theta = (50)(cos 60^{\circ})=25 N

Learn more about  force of friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

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PIT_PIT [208]

Answer:

v' = 2.83 m/s

Explanation:

Velocity of wave in stretched string is given by the formula

v = \sqrt{\frac{T}{\mu}}

here we know that

T = 4 N

also we know that linear mass density is given as

\mu = 1 kg/m

so we have

v = \sqrt{\frac{4}{1}} = 2 m/s

now the tension in the string is double

so the velocity is given as

v' = \sqrt{\frac{8}{1}} = 2\sqrt2 m/s

v' = 2.83 m/s

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3 years ago
A greyhound's velocity changes from 10 meters per second to 15 meters per second in 0.5 second. What is the greyhound's average
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The answer would be 10ms 2

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3 years ago
Four identical particles of mass 0.980 kg each are placed at the vertices of a 4.14 m x 4.14 m square and held there by four mas
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Answer:

a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²

b) I = 50.39 kg m²

c) I = 16.8 kg m²

Explanation:

a) Given data:

m = 0.98 kg

a = 4.14 * 4.14

The moment of inertia is:

I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}

For 4 particles:

I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}

c)

I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}

8 0
3 years ago
In the first direct detection of gravitational waves by LIGO in 2015, the waves came from A. the collapse of a nearby star into
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In the first direct detection of gravitational waves by LIGO in 2015, the waves came from the merger of two black holes. Option B is correct. This is further explained below.

<h3>What are gravitational waves?</h3>

A gravitational wave is simply defined as a ripple in space that is unseen though extremely rapid. Gravitational waves move at light speed. As they pass past, these waves compress and stretch everything in their path.

In conclusion, the merger of two black holes is the first direct detection of gravitational waves.

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8 0
2 years ago
Vector A has y-component Ay= +15.0 m . A makes an angle of 32.0 counterclockwise from the +y-axis. What is the x component of A?
Gemiola [76]

Answer:

x-component=-9.3 m

Magnitude of A=17.7m

Explanation:

We are given that

A_y=+15 m

\theta=32^{\circ}

We have to find the x-component of A and magnitude of A.

According to question

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Substitute the values then we get

15=\mid A\mid cos32

\mid A\mid =\frac{15}{cos32}=\frac{15}{0.848}

\mid A\mid=17.7m

tan\theta=\frac{perpendicular\;side}{Base}

tan32=\frac{A_x}{A_y}=\frac{A_x}{15}

0.62\times 15=A_x

A_x=9.3

The value of x-component of A is negative because the vector A lie  in second quadrant.

Hence, the x- component of A=-9.3 m

6 0
3 years ago
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