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Semenov [28]
3 years ago
12

Which of the following objects would have the strongest gravitational pull on a third object positioned at an equal distance fro

m both boxes
Box A is 3 times bigger than Box B but Box A weighs 100g & Box B weighs 10,000 g
Physics
1 answer:
3241004551 [841]3 years ago
8 0
I believe Box B will have a greater gravitational pull because the gravitational pull of an object depends on its mass. The more mass an object has, the greater its gravitational pull will become.

For example, we can take planets. Naturally, they are round because once upon a time there was a larger piece of rock that attracted others. But the size of the rock won't matter, it's the weight that matters. If the rock weighed nothing, the other rocks would just rebound upon contact. But if the rock weighed a lot, then things wouldn't so easily rebound and might actually stick to it.
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For the different values given for the radius of curvature RRR and speed vvv, rank the magnitude of the force of the roller-coas
nirvana33 [79]

Explanation:

The force of the roller-coaster track on the cart at the bottom is given by :

F=\dfrac{mv^2}{R}, m is mass of roller coaster

Case 1.

R = 60 m v = 16 m/s

F=\dfrac{(16)^2m}{60}=4.26m\ N

Case 2.

R = 15 m v = 8 m/s

F=\dfrac{(8)^2m}{15}=4.26m\ N

Case 3.

R = 30 m v = 4 m/s

F=\dfrac{(4)^2m}{30}=0.54m\ N

Case 4.

R = 45 m v = 4 m/s

F=\dfrac{(4)^2m}{45}=0.36m\ N

Case 5.

R = 30 m v = 16 m/s

F=\dfrac{(16)^2m}{30}=8.54m\ N

Case 6.

R = 15 m v =12 m/s

F=\dfrac{(12)^2m}{15}=9.6m\ N

Ranking from largest to smallest is given by :

F>E>A=B>C>D

5 0
3 years ago
The primary job of a(n) ____ <br> is to increase the power of a modified radio wave.
amm1812

The primary job of a(n) ____

is to increase the power of a modified radio wave.

Answer:

<h2>amplifier</h2>

Explanation:

Hope it helps:)

#CarryOnLearning

4 0
3 years ago
Read 2 more answers
A 15.5 kg mass vibrates in simple harmonic motion with a frequency of 9.73 Hz. It has a maximum displacement from equilibrium of
algol [13]

Answer:

12.14 cm

Explanation:

mass, m = 15.5 kg

frequency, f = 9.73 Hz

maximum amplitude, A = 14.6 cm

t = 1.25 s

The equation of the simple harmonic motion

y = A Sin ωt

y =  A Sin (2 x π x f x t)

put, t = 1.25 s, A = 14.6 cm, f = 9.73 Hz

y = 14.6 Sin ( 2 x 3.14 x 9.73 x 1.25)

y = 14.6 Sin 76.38

y = 12.14 cm

Thus, the displacement of the particle from the equilibrium position is 12.14 cm.

6 0
4 years ago
A particle moves along the x axis according to the equation x = 6t², where x is in meters and t is in seconds. Therefore:
vladimir2022 [97]
Answer is E

time can be negative.
A is not true because <span>a=<span><span><span>d2</span>x</span><span>d<span>t2</span></span></span>=12 m/<span>s2</span></span>
C: question already said that particle move along x-axis, which is not parabola path.
D: velocity is <span><span><span>dx</span><span>dt</span></span>=12t</span>, therefore velocity changes by 12 m/s and not 9.8 m/s
So we are left with E. <span>
</span>
7 0
3 years ago
Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of th
sattari [20]

Answer:

48 m

Explanation:

Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of the impending Collision and hit their brakes. The eastbound train, initially moving at 97.0 km/h Slows down at 3.50ms^2. The westbound train, initially moving at 127 km/h slows down at 4.20 m/s^2.

The eastbound train

First convert km/h to m/s

(97 × 1000)/3600

97000/3600

26.944444 m/s

As the train is decelerating, final velocity V = 0 and acceleration a will be negative. Using third equation of motion

V^2 = U^2 - 2as

O = 26.944^2 - 2 × 3.5 S

726 = 7S

S = 726/7

S1 = 103.7 m

The westbound train

Convert km/h to m/s

(127×1000)/3600

127000/3600

35.2778 m/s

Using third equation of motion

V^2 = U^2 - 2as

0 = 35.2778^2 - 2 × 4.2 × S

1244.52 = 8.4S

S = 1244.52/8.4

S2 = 148.2 m

S1 + S2 = 103.7 + 148.2 = 251.86

The distance between them once they stop will be

300 - 251.86 = 48.14 m

Therefore, the distance between them once they stop is 48 metres approximately.

5 0
3 years ago
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