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lianna [129]
1 year ago
8

The revolution of the earth around the sun demonstrate what motion?​

Physics
1 answer:
olga2289 [7]1 year ago
3 0

Answer:

Anticlockwise directions

Please mark me Brainliest to help me

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Please help on this one
mixas84 [53]
Ep=mgh
h= Ep/mg
h=57÷(3.3×9.8)
h= 57÷32.34
h= 1.8m
So; the answer is B. 1.8m
6 0
3 years ago
A small steel roulette ball rolls around the inside of a 30 cm diameter roulette wheel. It is spun at 150 rpm, but is slows to 6
liraira [26]

Solution :

Given

Diameter of the roulette ball = 30 cm

The speed ball spun at the beginning = 150 rpm

The speed of the ball during a period of 5 seconds = 60 rpm

Therefore, change of speed in 5 seconds = 150 - 60

                                                                      = 90 rpm

Therefore,

90 revolutions in 1 minute

or In 1 minute the ball revolves 90 times

i.e. 1 min = 90 rev

     60 sec = 90 rev

        1 sec = 90/ 60 rec

         5 sec = $\frac{90}{60}\times 5$

                   = 75 rev

Therefore, the ball made 75 revolutions during the 5 seconds.

7 0
3 years ago
A boy on a 1.9 kg skateboard initially at rest tosses a(n) 7.8 kg jug of water in the forward direction. if the jug has a speed
Tresset [83]
For this case we first think that the skateboard and the child are one body.
 We have then:
 1 = jug
 2 = skateboard + boy
 By conservation of the linear amount of movement:
 M1V1i + M2V2i = M1V1f + M2V2f
 Initial rest:
 v1i = v2i = 0
 0 = M1V1f + M2V2f
 Substituting values
 0 = (7.8) (3.2) + (M2) (- 0.65)
 0 = 24.96 + M2 (-0.65)
 -24.96 = (-0.65) M2
 M2 = (-24.96) / (- 0.65) = 38.4 kg
 Then, the child's mass is:
 M2 = Mskateboard + Mb
 Clearing:
 Mb = M2-Mskateboard
 Mb = 38.4 - 1.9
 Mb = 36.5 Kg
 answer:
 the boy's mass is 36.5 Kg
4 0
3 years ago
2. Tomas is hanging from a tree limb, that is inclined at a 65° angle. The force
LUCKY_DIMON [66]

Answer:

57 N

Explanation:

Were are told that the force

of gravity on Tomas is 57 N.

And it acts at an inclined angle of 65°

Thus;

The vertical component of the velocity is; F_y = 57 sin 65

While the horizontal component is;

F_x = 57 cos 65

Thus;

F_y = 51.66 N

F_x = 24.09 N

The net force will be;

F_net = √((F_y)² + (F_x)²)

F_net = √(51.66² + 24.09²)

F_net = √3249.0837

F_net = 57 N

4 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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