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maxonik [38]
4 years ago
10

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:
Kamila [148]4 years ago
3 0

The force of friction is 25 N

Explanation:

For this problem, we can apply Newton's second law of motion along the horizontal direction:

\sum F_x = ma_x

where

\sum F_x is the net force in the horizontal direction

m is the mass of the block

a_x is the horizontal acceleration

Here  the block is moving at constant speed, so its acceleration is zero, therefore:

a=0 \rightarrow \sum F_x = 0 (1)

The net force in the horizontal direction can be written as:

\sum F_x = Fcos \theta -F_f (2)

where

  • Fcos\theta is the horizontal component of the pulling force, with F = 50 N being the magnitude and \theta=60^{\circ} being the direction, acting forward
  • F_f is the force of friction, acting backward

Combining (1) and (2), we find the magnitude of the force of friction:

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

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

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A rod of length 35.50 cm has linear density (mass per length) given by λ = 50.0 + 23.0x where x is the distance from one end, an
hram777 [196]

Answer:

(a)20.65g

(b)0.19m

Explanation:

(a) The total mass would be it's mass per length multiplied by the total lenght

0.355(50 + 23*0.355) = 20.65 g

(b) The center of mass would be at point c where the mass on the left and on the right of c is the same

Hence the mass on the left side would be half of its total mass which is 20.65/2 = 10.32 g

c(50 + 23c) = 10.32

23c^2 + 50c - 10.32 = 0

c \approx 0.19m

7 0
4 years ago
A sled is pulled at a constant velocity across a horizontal snow surface. If a force of 100 N is being applied to the sled rope
tangare [24]

Answer:

60.18 N

Explanation:

Given that:

The force applied on the sled = 100 N

Suppose, the angle between the sled rope and the ground = 53°

The horizontal force which acts  in the horizontal direction can be expressed as:

F_x = F \ cos \theta

F_x = 100 \ cos (53)

F_x = 60.18 \ N

But if the angle between the sled rope is parallel to the ground. Then, we use an angle on a straight line which is = 180°

F_x = F \ cos \theta

F_x = 100 \ cos (180)

= 100 × -1

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3 0
3 years ago
HELP ME PLEASE!!!!!!!!!
Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

F = \frac{kq_1q_3}{d_{13}^2}

now from the above equation

F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}

F = 0.288 N

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

F_{net} = 2Fcos\theta

here we know that angle is

\theta = 37 degree

now we have

F_{net} = 2\times 0.288 cos37

F_{net} = 0.46 N

so net force on q3 is 0.46 N vertically upwards along +Y axis

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3 years ago
Can anyone help me with these questions? TIA!<br> (Don’t actually answer please! :) )
nataly862011 [7]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Question 7: </h2>

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<h2>_____________________________________ </h2><h2>Question 8: </h2>

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