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3 years ago

I need to write an acrostic poem on Globalisation and i need help. Does anyone have any ideas?

Physics

1 answer:

Aleonysh [2.5K]3 years ago

5 0

Answer:

you could talk about the people and say something like change

Explanation:

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A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

Bess [88]

Answer:

Explanation:

given,

initial velocity of the ball = 20 m/s

angle of ramp = 22°

ball travel at a distance = 5 m

a) for friction less

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh$

$v^2 = u^2 - 2gh$

$v = \sqrt{u^2- 2 g h cos 22^0}$

$v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0 }$

v = 17.58 m/s

b) considering the friction

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh-\mu_kmgl$

$v^2 = u^2 - 2gh-2\mu_kmgl$

$v = \sqrt{u^2- 2 g h cos 22^0-2\mu_kgl}$

$v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0-2\times 0.15\times 9.8 \times 5 }$

v = 17.16 m/s

7 0

3 years ago

Where does the rest of the energy 90% go?

Aleonysh [2.5K]

Answer:the 90% is spent performing other functions

Explanation:

3 0

3 years ago

As shown above an adhesive has been applied to contacting faces of two blocks so that the blocks interact with an adhesive force

rusak2 [61]

The magnitude of the adhesive force allows the top block to remain

attached to the bottom when the blocks and the wire are balanced.

The option that must be true is; $\mathbf{W_{bottom} = F_{adhesion}}$

Reason:

The tension exerted by the wire attached to the top block = T

Magnitude of the adhesive = $F_{adhesion}$

Weight of the top block = $W_{top}$

Weight of the bottom block = $W_{bottom}$

Given that with the exertion of the tension, the two blocks remain at rest, we have;

T = $W_{top}$ + $W_{bottom}$

The adhesive causes the bottom block to remain attached to the top block, we have;

Therefore, the magnitude of the adhesive force adds the bottom weight, to the top, weight, which gives;

The magnitude of the adhesive force = The weight of the bottom block

Therefore;

$W_{bottom}$ = $F_{adhesion}$

Learn more here:

brainly.com/question/18907970

8 0

3 years ago

ball A of mass 5.0 kg moving at 20 meters per second collides with ball B of unknown mass moving at 10. meters per second in the

antiseptic1488 [7]

Answer:The mass of ball B is 10 kg.

Explanation;

Mass of ball A = $M_A=5 kg$

Velocity of the ball A before collision: $U_A=20 m/s$

Velocity of ball A after collision= $V_A=10 m/s$

Mass of ball B= $M_B$

Velocity of the ball B before collision: $U_B=10 m/s$

Velocity of ball B after collision= $V_B=15 m/s$

$M_AV_A+M_BV_B=M_AU_A+M_BU_B$

$5 kg\times 10 m/s+M_B\times 15=5 kg\times 20m/s+M_B\times 10m/s$

$M_B=10kg$

The mass of ball B is 10 kg.

8 0

3 years ago

How does radiation differ from conduction?

Natali5045456 [20]

Answer:

Explanation:

3 0

3 years ago