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Anarel [89]
3 years ago
6

A rocket, blasting into space, accelerates at 2 m/s^2 for 20 seconds. What is its speed at the end of 20 seconds?​

Physics
1 answer:
OLEGan [10]3 years ago
4 0

Answer:

  Acceleration  =  (change in speed)  /  (time for the change)

Change in speed = (ending speed) - (beginning speed)

                           =     (3,600 mi/hr) - ( 0 )

                           =       3,600 mi/hr

                           =     same as  1 mile/second.

Acceleration  =  (1 mi/sec) / (10 sec)

                     =     0.1 mi/sec² .

But 1 mile = 5,280 ft,

so

                      0.1 mi/s²  =  528 ft/s²          

Explanation:

have a great day

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ruslelena [56]

Answer:

x_1= 0.0425m

Explanation:

Using the tension in the spring and the force of the tension can by describe by

T = kx

, T = mg

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m*g = k*x

With two springs, let, T1 be the tension in each spring,  x1 be the extension of each spring.  The spring constant of each spring is 2k so:

T_1 = 2k*x_1

2T_1 = m*g=4k x_1

Solve to x1

x_1=\frac{m*g}{4k}

x_1=\frac{k*x}{4*k}

x_1=\frac{x}{4}

x_1 = 0.170 / 4

x_1= 0.0425m

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Answer: 8.6

Explanation:

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A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is hold
lubasha [3.4K]

Answer:

to overcome the out of friction we must increase the angle of the plane

Explanation:

To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.

X axis

       fr - Wₓ = m    a                      (1)

Y axis  

       N- W_{y} = 0

       N = W_{y}

let's use trigonometry to find the components of the weight

        sin θ = Wₓ / W

        cos θ = W_{y} / W

        Wₓ = W sin θ

        W_{y} = W cos θ

the friction force has the formula

         fr = μ N

         fr = μ Wy

         fr = μ mg cos θ

from equation 1

at the point where the force equals the maximum friction force

in this case the block is still still so a = 0

           F = fr

           F = (μ  mg) cos θ

We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.

This is the force that balances the friction force, any force slightly greater than F initiates the movement.

Consequently, to overcome the out of friction we must increase the angle of the plane

the correct answer is to increase the angle of the plane

4 0
3 years ago
Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
3 years ago
Read 2 more answers
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