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schepotkina [342]
3 years ago
11

please solve this for me ,A garden roller is pulled with a force of 200N acting at an angle of 50 degree with the ground level.f

ind the force pulling the roller along the ground...​
Physics
1 answer:
Bingel [31]3 years ago
8 0

Answer:

The force pulling the roller along the ground is 128.55 N

Explanation:

A force of 200 N acting at an angle of 50° with the ground level

This force is pulled a garden roller

We need to find the force pulling the roller along the ground

The force that pulling the roller along the ground is the horizontal

component of the force acting

→ The force acting is 200 N at direction 50° with ground (horizontal)

→ The horizontal component = F cosФ

→ F = 200 N , Ф = 50

→ The horizontal component = 200 cos(50) = 128.55 N

128.55 N is the horizontal component of the force that pulling the

roller along the ground

<em>The force pulling the roller along the ground is 128.55 N</em>

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Before the asteroid is threw, the total momentum is zero, since neither Superman nor the asteroid are moving.

Conservation of momentum commands the total momentum after the astronaut is threw must be zero too. This means that Superman's backward momentum afterward throwing the asteroid is equivalent to the asteroid forwards momentum, in size. 

Momentum is mass times velocity. We know the mass of the asteroid is 1000M and its velocity is 850 m/s, so its momentum is 
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A stationary charge is located between the poles of a horseshoe magnet. Is a magnetic force excerted of the charge? why?
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Thus, the magnetic force exerted by the charge is zero.

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7 0
2 years ago
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A 70kg car drives through a curve with radius 50m. What is the minimum angle the road should be tilted to keep the car on the ro
worty [1.4K]

when car moves along a circular path then centripetal force for taking turn will be provided by the component of normal force

Let the road is tilted by some angle

so force equations are given as

F_n cos\theta = mg

F_n sin\theta = \frac{mv^2}{R}

now divide two equations

tan\theta = \frac{v^2}{Rg}

now as it is given to us that

v = 80 km/h

v = 80* \frac{5}{18} = 22.2 m/s

R = 50 m

now we have

tan\theta = \frac{22.2^2}{50*9.8}

tan\theta = 1

\theta = tan^{-1}1

\theta = 45^0

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8 0
3 years ago
A wheel has a constant angular acceleration of 1.8 rad/s2. During a certain 4.0 s interval, it turns through an angle of 45 rad.
tankabanditka [31]

Answer:

4.25 s

Explanation:

Given:

angular acceleration'α'=  1.8 rad/s²

angle 'θ'= 45 rad

time 't'= 4s

initial angular velocity 'w_{o}'=0rad^{-1}

as we know that,

θ= w_{1}t + \frac{1}{2} \alpha t^{2}

45 = 4 w_{1} + (0.5 x 1.8 x 16)

45- 14.4 = 4 w_{1}

30.6 = 4 w_{1}

w_{1}= 7.65 rad^{-1}

Next is to find t by using the equation

w_{1} =  w_{o} + \alpha t_{1}

7.65= 0 + (1.8)t_{1}

t_{1}= 7.65/1.8

t_{1}= 4.25 s

Therefore, At the start of 4s interval the motion is at 4.25 second

7 0
4 years ago
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