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11111nata11111 [884]
4 years ago
7

(NEED HELP BADLY)

Physics
1 answer:
spin [16.1K]4 years ago
8 0

Answer:

The answer to your question is below

Explanation:

Data 1

mass 1 = 250

mass 2 = 250 kg

gravity constant = 6.67 x 10⁻¹¹ Nm²/kg²

distance = 8 m

Formula

F = G\frac{m1m2}{r^{2} }

Substitution

F = 6.67 x 10^{-11} \frac{250 x 250}{8^{2} }

Result

F = 0.000000065 N

Data 2

mass 1 = 1000 kg

mass 2 = 1000 kg

distance = 5 m

Substitution

F = 6.67 x 10^{-11} \frac{1000 x 1000 }{5^{2} }

Result

F = 0.000002667 N      

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A 1000 kg box is being pushed with a force of 3500 N. What acceleration is the
WARRIOR [948]
Net Force = mass x acceleration
3500=1,000a
So a= 3500/1000
a=35/10
a=3.5 m/s^2
4 0
3 years ago
A beam of light, traveling in air, strikes a plate of transparent material at an angle of incidence of 56.0°. It is observed tha
tankabanditka [31]

Answer:

the index of refraction for the transparent material is 1.48

Explanation:

let n1 = 1.0 be the refractive index of air and n2 be the refractive index of the trasnparent material.

then:

Snell's Law state that:

n1×sin(∅1) = n2×sin(∅2)

where ∅1 = 56.0° is the angle of incidence and  ∅2 = 90 - 56.0 = 34° be the angle refraction.

n1×sin(∅1) = n2×sin(∅2)

          n2 = n1×sin(∅1)/sin(∅2)

               =  (1.0)×sin(56.0°)/sin(34.0°)

               =  1.48

Therefore, the index of refraction for the transparent material is 1.48

3 0
4 years ago
Why was the idea of plate tectonics difficult for many scientists to accept for many years after it was first introduced?
pentagon [3]
I think the correct answer from the choices listed above is the second option. The <span> idea of plate tectonics was difficult for many scientists to accept for many years after it was first introduced because there </span><span>was no explanation yet for how it was happening. It was only to the recent times that these were proven. </span>
3 0
4 years ago
Starting from rest, the distance a freely-falling object will fall in 0.50 second is?
FinnZ [79.3K]

Answer:

1.23 m

Explanation:

The vertical distance covered by a free-falling object starting from rest in a time t is

y=\frac{1}{2}gt^2

where

g = 9.8 m/s^2 is the acceleration due to gravity

In this problem, we have

t = 0.50 s

So the distance covered is

y=\frac{1}{2}(9.8 m/s^2)(0.50 s)^2=1.23 m

8 0
4 years ago
During a long jump, an Olympic champion's center of mass rose about 1.2 m from the launch point to the top of the arc. 1) What m
Alecsey [184]

Answer:

1) v_{A} \approx 8.272\,\frac{m}{s}

Explanation:

1) Let assume that the campion begins running at a height of zero. The movement of the Olympic champion is modelled after the Principle of Energy Conservation:

K_{A} = K_{B} + U_{g,B}

\frac{1}{2}\cdot m \cdot v_{A}^{2} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + m \cdot g \cdot h_{B}

\frac{1}{2} \cdot v_{A}^{2} = \frac{1}{2} \cdot v_{B}^{2} + g \cdot h_{B}

The minimum speed is obtained herein:

v_{A}=\sqrt{v_{B}^{2} + 2 \cdot g \cdot h}

v_{A} = \sqrt{(6.7\,\frac{m}{s} )^{2}+2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (1.2\,m)}

v_{A} \approx 8.272\,\frac{m}{s}

3 0
3 years ago
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