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4 years ago

A 19.7 kg sled is pulled with a 42.0 N force at a

Physics

2 answers:

sertanlavr [38]4 years ago

5 0

The normal force on the sled is 190.9 N.

Explanation:

For this problem, we have to analyze the forces acting along the vertical direction on the sled.

We have:

- The weight of the sled, W = mg, acting downward, where

m = 19.7 kg is the mass of the sled

g = 9.8 m/s^2 is the acceleration of gravity

- The normal force on the sled, N, acting upward

- The vertical component of the pulling force, $F sin \theta$ , acting upward, where

F = 42.0 N is the magnitude of the force

$\theta=3.0^{\circ}$ is the angle

Since the sled is in equilibrium along the vertical direction, the equation of the force is

$F sin \theta + N - mg = 0$

And solving for N, we find

$N=mg -Fsin \theta = (19.7)(9.8) - (42.0)(sin 3)=190.9 N$

Learn more about forces and weight here:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

ZanzabumX [31]4 years ago

5 0

Answer:

164 N

Explanation:

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A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta

Vika [28.1K]

Answer:

$t_t=4.131\ s$

Explanation:

Given:

height above the horizontal form where the ball is hit, $y=1\ m$

angle of projectile above the horizontal, $\theta=30^{\circ}$

initial speed of the projectile, $u=40\ m.s^{-1}$

Firstly we find the vertical component of the initial velocity:

$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

$h=20.4082\ m$ (from the height where it is thrown)

Now we find the time taken to ascend to this height:

$v_y=u_y-g.t$

$0=20-9.8t$

$t=2.041\ s$

Time taken to descent the total height:

we've total height, $h'=h+y$ $=20.4082+1$

$h'=u_y'.t'+\frac{1}{2} g.t'^2$

during the course of descend its initial vertical velocity is zero because it is at the top height, so $u_y'=0\ m.s^{-1}$

$21.4082=0+4.9t'^2$

$t'=2.09\ s$

Now the total time taken by the ball to hit the ground:

$t_t=t'+t$

$t_t=2.09+2.041$

$t_t=4.131\ s$

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3 years ago

Which describes an image that can be produced by a concave lens?

mixer [17]

A concave is when the bump is pressed downwards; So if the image has a downward hole, then it's concave. If it's pressed upwards like a hill or mountain, then it's convex.

5 0

4 years ago

a tungsten and coil has a resistance of 12 ohm at 15 degree celsius is the temperature coefficient of resistance of tungsten is

Dennis_Churaev [7]

Answer:

The resistance of the tungsten coil at 80 degrees Celsius is 15.12 ohm

Explanation:

The given parameters are;

The resistance of the tungsten coil at 15 degrees Celsius = 12 ohm

The temperature coefficient of resistance of tungsten = 0.004/°C

The resistance of the tungsten coil at 80 degrees Celsius is found using the following relation;

R₂ = R₁·[1 + α·(t₂ - t₁)]

Where;

R₁ = The resistance at the initial temperature = 12 ohm

R₂ = The resistance of tungsten at the final temperature

t₁ = The initial temperature = 15 degrees Celsius

t₂ = The final temperature = 80 degrees Celsius

α = temperature coefficient of resistance of tungsten = 0.004/°C

Therefore, we have;

R₂ = 12×[1 + 0.004×(80 - 15)] = 15.12 ohm

The resistance of the tungsten coil at 80 degrees Celsius = 15.12 ohm.

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Answer:

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M g L / 2 = 1/3 M L^2 * ω^2 / 2

g = 3 L ω^2

ω = (g / (3 L))^1/2

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