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2 years ago

Need help guys please

Physics

1 answer:

goldenfox [79]2 years ago

5 0

Answer:

6.378 * 10^6 m

Explanation:

Radius = 6378 km

Changing in metres

6378 km = 6378000 m

Now in scientific notation

= 6.378 * 10 ^ 6 m

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The distance recorded for riding a motorcycle on its rear wheel without stopping is more than 320 km! Suppose the rider in this

antiseptic1488 [7]

Answer:

Explanation:

Given

Initial velocity of the car u = 8m/s

Distance travelled by the rider S = 40m

Acceleration a = 2m/s²

Required

rider's velocity after the acceleration v

Using the equation of motion

v² = u²+2as

v² = 8²+2(2)(40)

v² = 64+160

v² = 224

v = √224

v = 14.97m/s

Hence the rider's velocity after the acceleration is 14.97m/s

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2 years ago

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balu736 [363]

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3 years ago

Playing soccer on a beach will require more effort because the sand causes a great deal of _______ to the ball.

34kurt

Answer:

I believe the answer to be B

Explanation:

If you were playing on grass, the ball would be able to roll around much easier rather than it to be on sand. If it's wrong I am so sorry

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2 years ago

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A roller coaster track is 3000 meters long. It takes 100 seconds to travel once around the roller coaster. What is the average s

Svetllana [295]

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3 years ago

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A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and

garri49 [273]

Answer:

(a) t = 1.14 s

(b) h = 0.82 m

Explanation:

(b)

Considering the upward motion, we apply the third equation of motion:

$2gh = v_f^2 - v_i^2$

where,

g = - 9.8 m/s² (-ve sign for upward motion)

h = max height reached = ?

vf = final speed = 0 m/s

vi = initial speed = 4 m/s

Therefore,

$(2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\$

Now, for the time in air during upward motion we use first equation of motion:

$v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s$

(c)

Now we will consider the downward motion and use the third equation of motion:

$2gh = v_f^2-v_i^2$

where,

h = total height = 0.82 m + 1.8 m = 2.62 m

vi = initial speed = 0 m/s

g = 9.8 m/s²

vf = final speed = ?

Therefore,

$2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\$

Now, for the time in air during downward motion we use the first equation of motion:

$v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s$

(a)

Total Time of Flight = t = t₁ + t₂

t = 0.41 s + 0.73 s

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2 years ago