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3 years ago

32. Only a small percentage of the energy

Physics

1 answer:

nordsb [41]3 years ago

5 0

Only a small Percentage of the energy emitted by the sun strikes earth because, since the earth is going in circles round and round, the sun only hits part of the earth and not fully. So for example if the earth is going around and the sun hits Africa then in a couple minutes it will go to the next country and that country will have sunlight. Let me know if you need anything else or if this isn't partially correct. GLAD TO HELP! :)

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A satellite in outer space is moving at a constant velocity of 20.5 m/s in the +y direction when one of its on board thruster tu

Katyanochek1 [597]

Answer:

a) V_f = 25.514 m/s

b) Q =53.46 degrees CCW from + x-axis

Explanation:

Given:

- Initial speed V_i = 20.5 j m/s

- Acceleration a = 0.31 i m/s^2

- Time duration for acceleration t = 49.0 s

Find:

(a) What is the magnitude of the satellite's velocity when the thruster turns off?

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.

Solution:

- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:

V_f = V_i + a*t

V_f = 20.5 j + 0.31 i *49

V_f = 20.5 j + 15.19 i

- The magnitude of the velocity vector is given by:

V_f = sqrt ( 20.5^2 + 15.19^2)

V_f = sqrt(650.9861)

V_f = 25.514 m/s

- The direction of the velocity vector can be computed by using x and y components of velocity found above:

tan(Q) = (V_y / V_x)

Q = arctan (20.5 / 15.19)

Q =53.46 degrees

- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.

4 0

3 years ago

How much force is needed to accelerate a 15kg bowling ball at 2 m/s^2

olga_2 [115]

Golf no doing what I iuroeurir to do my homework and my homework

7 0

3 years ago

Read 2 more answers

If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?

SVETLANKA909090 [29]

Assume north and east as two sides of a right angled triangle. magnitude of the distance is then given by the length of the hypotenuse which is $\sqrt{a^2 + b^2}$

where a = 1.2 km north

and b = 1.6 km east

magnitude = 2 km

Direction is given by the angle between them, that is atan(a/b) = 36.86 deg north of east = 53.1 deg east of north.

8 0

3 years ago

A bullet of mass 11 g strikes a ballistic pendulum of mass 1.9 kg. The center of mass of the pendulum rises a vertical distance

Ymorist [56]

Answer:

The bullet's initial speed is 243.21 m/s.

Explanation:

Given that,

Mass of the bullet, $m_b=11\ g=0.011\ kg$

Mass of the pendulum, $m_p=19\ kg$

The center of mass of the pendulum rises a vertical distance of 10 cm.

We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,

$\dfrac{1}{2}(m_b+m_p)V^2 =(m_b+m_p)gh\\\\V=\sqrt{2gh} \ .................(1)$

V is the speed of the bullet and pendulum at the time of collision

Now using conservation of momentum as :

$m_bv=(m_p+m_b)V$

Put the value of V from equation (1) in above equation as :

$v=\dfrac{(m_p+m_b)}{m_b}\sqrt{2gh} \\\\v=\dfrac{(1.9+0.011)}{0.011}\sqrt{2\times 9.8\times 0.1}\\\\v=243.21\ m/s$

So, the bullet's initial speed is 243.21 m/s.

7 0

3 years ago

Please send me solution of the question pls

ELEN [110]

Answer:

20m

Explanation:

P.E=mgh

2000=10×10×h

2000=100h

Divide both side by 100

2000/100=20

4 0

3 years ago