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3 years ago

The gravitational force between two objects that

Physics

1 answer:

leonid [27]3 years ago

5 0

Answer:

The answer to your question is m₂ = 38.5 kg

Explanation:

Data

distance = d = 2.1 x 10⁻¹ m

Force = 3.2 x 10⁻⁶ N

m₁ = 55 kg

m₂ = ?

G = 6.67 x 10 ⁻¹¹ Nm²/kg²

Process

1.- To solve this problem use Newton's law of Universal Gravitation.

F = G m₁m₂ / r²

-Solve for m₂

m₂ = Fr² / Gm₁

2.- Substitution

m₂ = (3.2 x 10⁻⁶)(2.1 x 10⁻¹)² / (6.67 x 10⁻¹¹)(55)

3.- Simplification

m₂ = 1.411 x 10⁻⁷ / 3.669 x 10⁻⁹

4.- Result

m₂ = 38.5 kg

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A 2000 g of C 14 is left to decay radioactively the half-life of Corbin 14 is approximately 5700 years what fraction of that sam

ahrayia [7]

Answer:

1/8

Explanation:

17,100 years is 3 times the half-life of 5,700 years. After each half-life, half remains, so the amount remaining after 3 half-lives is ...

(1/2)(1/2)(1/2) = 1/8

1/8 of the sample remains after 17,100 years.

8 0

3 years ago

Read 2 more answers

Which of the following is evidence that supports the idea of uniformitarianism?

Inga [223]

D. rates of soil erosion are much lower during droughts that last several years

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3 years ago

An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact wi

hoa [83]

Answer:

a) a = 3.72 m / s², b) a = -18.75 m / s²

Explanation:

a) Let's use kinematics to find the acceleration before the collision

v = v₀ + at

as part of rest the v₀ = 0

a = v / t

Let's reduce the magnitudes to the SI system

v = 115 km / h (1000 m / 1km) (1h / 3600s)

v = 31.94 m / s

v₂ = 60 km / h = 16.66 m / s

et's calculate

a = 31.94 / 8.58

a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

I = Δp

F Δt = m v_f - m v₀

F = $\frac{m ( v_f - v_o)}{t}$

F = m [16.66 - 31.94] / 0.815

F = m (-18.75)

Having the force let's use Newton's second law

F = m a

-18.75 m = m a

a = -18.75 m / s²

4 0

2 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

A 110-kg tackler moving at a speed of 3.5 m/s

IgorC [24]

What exactly do u want me to know step by step with me plz I am a slow person

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3 years ago