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3 years ago

1. The Moon's mass is 7.34 x 1022 kg, and it is 3.8 x 105 km away from Earth. Earth's

Physics

1 answer:

sineoko [7]3 years ago

8 0

Answer:

2.03 x 10²⁴N

Explanation:

Given parameters:

Mass of moon = 7.34 x 10²²kg

Mass of the earth = 5.97 x 10²⁴kg

Distance = 3.8 x 10⁵km

Unknown:

Gravitational force of attraction = ?

Solution:

To find the gravitational force of attraction between the masses, we use the expression below;

F = $\frac{Gm_{1} m_{2} }{r^{2} }$

G is the universal gravitation constant

m is the mass

1 and 2 represents moon and earth

r is the distance

F = $\frac{6.67 x 10^{-11} x 7.34 x 10^{22} x 5.97 x 10^{24} }{(3.8 x 10^{5})^{2} }$

F = $\frac{2.92 x 10^{35} }{1.44 x 10^{11} }$ = 2.03 x 10²⁴N

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Based on the Law of Conservation of Energy, which of the below is true?(1 point)

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Answer:

Explanation:

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2 years ago

PERSONAL QUESTION <br> if a bird can fly why can't a fly just bird ?

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4 years ago

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For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substant

Scilla [17]

Consider the projectile launched at initial velocity V at angle θ relative to the horizontal.
Neglect wind or aerodynamic resistance.

The initial vertical velocity is Vsinθ.
When the projectile reaches its maximum height of h, its vertical velocity will be zero.
If the time taken to attain maximum height is t, then
0 = Vsinθ - gt
t = (Vsinθ)/g, where g = acceleration due to gravity.

The horizontal component of launch velocity is Vcosθ. This velocity remains constant because aerodynamic resistance is ignored.
The time to travel the horizontal distance D is twice the value of t.
Therefore
D = Vcosθ*[(2Vsinθ)/g]
= (2V²sinθ cosθ)/g
= (V²sin2θ)/g

In order for D (horizontal distance) to be maximum, $\frac{dD}{d \theta} =0$
That is,
$\frac{2V^{2}}{g} cos(2 \theta )=0$

Because $\frac{2V^{2}}{g} \neq 0$ , therefore cos(2θ) = 0.
This is true when 2θ = π/2 => θ = π/4.

It has been shown that the maximum horizontal traveled can be attained when the launch angle is π/4 radians, or 45°.

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4 years ago

Your best friend weighs 81.5 kg and is a rugby player. In one of his games, he slides to a stop in a phenomenal manner. The coef

Alex777 [14]

A. The acceleration during the slide is 6.86 m/s²

B. The time taken to slide until he stops is 1.2 s

<h3>How to determine the force of friction</h3>

Mass (m) = 81.5 Kg
Coefficient of friction (μ) = 0.7
Acceleration due to gravity (g) = 9.8 m/s²
Normal reaction (N) = mg = 81.5 × 9.8 = 798.7 N
Frictional force (F) =?

F = μN

F = 0.7 × 798.7

F = 559.09 N

<h3>A. How to determine the acceleration</h3>

Mass (m) = 81.5 Kg
Frictional force (F) = 559.09 N
Acceleration (a) =?

a = F / m

a = 559.09 / 81.5

a = 6.86 m/s²

<h3>B. How to determine the time </h3>

Initial velocity (u) = 8.23 m/s
Final velocity (v) = 0 m/s
Decceleration (a) = -6.86 m/s²
Time (t) =?

a = (v – u) / t

t = (v – u) / a

t = (0 – 8.23) / -6.86

t = 1.2 s

Learn more about acceleration:

brainly.com/question/491732

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2 years ago

The end diastolic volume of a heart is 140 mL Assume that it is a sphere. At end diastole, the intraventricular pressure is 7mmI

Vera_Pavlovna [14]

Answer:

Explanation:

We know that, V = 140 mL = 0.00014 m3

Assume that it is a sphere. so, we have

V = (4/3) \pir3

r3 = (0.00014 m3) (3) / (4) (3.14)

r = \sqrt[3]{}\sqrt[3]{}3\sqrt{}3.34 x 10-5 m3

r = 1.93 x 10-7 m

(a) The wall tension at end diastole will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end diastole = 7 mmHg = 933.2 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (933.2 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 8.18 x 10-3 N

(b) The wall tension at the end of isovolumetric contraction will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end of isovolumetric contraction = 80 mmHg = 10665.7 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (10665.7 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 9.35 x 10-2 N

(d) The wall stress from A and B which will be given as :

we know that, \sigma = T / w

For part A, we have

\sigmaA = (8.18 x 10-3 N) / (0.011 m)

\sigmaA = 0.743 N/m

For part B, we have

\sigmaB = (9.35 x 10-2 N) / (0.011 m)

\sigmaB = 8.5 N/m

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3 years ago