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bija089 [108]
3 years ago
11

What is occurring in the umbra?

Physics
1 answer:
matrenka [14]3 years ago
3 0

Answer:

A. a total solar eclipse

Explanation:

During a solar eclipse,the moon comes between the sun and Earth, casting its shadow over the Earth

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PLEASEEE HELP ME WITH THIS ALSO. I DONT WANT TO FAIL. You push a merry-go-round on which Kim and Katie are riding. Kim weighs 45
Serjik [45]

Answer:

The body weight

Explanation:

5 0
3 years ago
A bike first accelerates from 0.0m/s to 4.5m/s in 4.5 s, the continues at this constant speed for another 6.0 s. What is the tot
Alex_Xolod [135]

Answer:

37.125 m

Explanation:

Using the equation of motion

s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration

<u>Distance during acceleration</u>

Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.

Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be

a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}

Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion

s=0*4.5+ 0.5*1*4.5^{2}=0+10.125 =10.125 m

<u>Distance at a constant speed</u>

At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time

Distance=4.5 m/s*6 s=27 m

<u>Total distance</u>

Total=27+10.125=37.125 m

3 0
3 years ago
An Airbus A350 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the
alexdok [17]

Answer:

t=67.7s

Explanation:

From this question we know that:

Vo = 6m/s

a = 1.8 m/s2

D = 1500m

And we also know that:

X=V_{o}*t + \frac{a*t^{2}}{2}   Replacing the known values:

1500=6t+0.9*t^{2}    Solving for t we get 2 possible answers:

t1 = -44.3s   and t2 = 67.7s    Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:

t = 67.7s

8 0
3 years ago
A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base
umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0
3 years ago
A satellite is in a circular orbit around an unknown planet. The satellite has a speed of 1.75 104 m/s, and the radius of the or
Arisa [49]

Answer:

v = 1.32 10² m

Explanation:

In this case we are going to use the universal gravitation equation and Newton's second law

    F = G m M / r²

    F = m a

In this case the acceleration is centripetal

    a = v² / r

The force is given by the gravitational force

    G m M / r² = m v² / r

    G  M/r =  v²

Let's calculate the mass of the planet

    M = v² r / G

    M = (1.75 10⁴)² 5.00 10⁶ / 6.67 10⁻¹¹

    M = 2.30 10²¹ kg

With this die we clear the equation to find the orbit of the second satellite

    v = √ G M / r

    v = √ (6.67 10⁻¹¹ 2.30 10²¹ / 8.75 10⁶)

    v = 1.32 10² m

8 0
3 years ago
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