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3 years ago

Calculate the net force in Newtons of the following situation. <—2N + →8 N =

Physics

1 answer:

vazorg [7]3 years ago

5 0

Answer:

F = + 6 [N] That is, this force points to the right

Explanation:

We have two forces, one of them pointing to the left with a value of 2 [N] while the other Force points to the right with a value of 8 [N].

In this way performing an arithmetic operation we can find the resulting force.

F = -2 + 8

F = + 6 [N]

That is, this force points to the right

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A machine part rotates at an angular speed of 0.07 rad/s; its speed is then increased to 3.4 rad/s at an angular acceleration of

34kurt

Answer:

In first case angular displacement will be 11.55 radian

In second case it will be 4 times the previous case

Explanation:

We have given that machine rotates at an angular velocity of 0.07 rad/sec

So initial angular velocity $\omega _i=0.07rad/sec$

And its speed is increases to 3.4 rad/sec

So final angular velocity is $\omega _f=3.4rad/sec$

Angular acceleration $\alpha =0.5rad/sec^2$

According to third equation of motion we know that

$\omega _f^2=\omega _i^2+2\alpha \Theta$

$3.4^2=0.07^2+2\times 0.5\times \Theta$

$\Theta =11.55radian$

Now initial angular velocity and final angular velocity both are doubled but angular acceleration is constant

So $\omega _i=2\times 0.07=0.14rad/sec$

And $\omega _f=2\times 3.4=6.8rad/sec$

So $6.8^2=0.14^2+2\times 0.5\times \Theta$

$\Theta =46.22radian$

So angular displacement will be 4 times the earlier displacement

4 0

3 years ago

The gravitional energy of an object is always measured realative to the ?

zaharov [31]

I would say that the answer would be MASS.

4 0

3 years ago

Read 2 more answers

Which year was pluto no longer considered a planet?.

katrin2010 [14]

2006, i hope this helps

6 0

2 years ago

A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.

enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, $KE_{0} = 0.5 mu^{2}$

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, $KE_{f} = 0.5mv^{2}$

a) Using the principle of energy conservation,

$KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s$

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0

4 years ago

Which of the following means that an<br> image is real?

Licemer1 [7]

Answer:_

Explanation:

7 0

3 years ago