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3 years ago

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Robin is standing terrified at the end of a diving board, which is high above the water. If Robin has a mass of 76 kg and is sta

nding 1.6 m from the board's pivot point, what torque is Robin exerting on the board

1 answer:

masya89 [10]3 years ago

5 0

Answer:

Torque = 1191.68 N-m

Explanation:

Given data

mass m = 76 kg

standingdistance r = 1.6 m

Solution

we get here torque that si express as

torque = force × distance ................1

torque = r × F sin(theta)

and we know that

F = mg .........2

and g = 9.8 m/s²

put here value in equation 1 we get

Torque = 76 × 1.6 × 9.8 × sin(90)

Torque = 1191.68 N-m

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The average distance from Earth to the Moon is 384,000 km. In the late 1960s, astronauts reached the Moon in about 3 days. How f

Anon25 [30]

Answer:

They must have been traveling at 5333.33 km/h to cover that distance in 3 days.

That speed are 6,66 times higher than the speed of an aircraft jet.

Explanation:

d= 384000 km

t= 3 days = 3*24hr = 72hr

V= 384000km/72hr

V= 5333.33 km/h

comparison:

V1/V2= 5333.33/800

V1/V2= 6.66

3 0

3 years ago

A car moves at speed v across a bridge made in the shape of a circular arc of radius r. (a) Find an expression for the normal fo

inna [77]

Answer:

(a) FN = m (g - $\frac{v^{2} }{r}$ )

(b) vmin = 17.146 m/s

Explanation:

The radius of the arc is

r = 30m

The normal force acting on the car form the highest point is

FN = m (g - $\frac{v^{2} }{r}$ )

If the normal force become 0 we have

m (g - $\frac{v^{2} }{r}$ ) = 0

or

g - $\frac{v^{2} }{r}$ = 0

This way, when FN = 0, then v = vmin, so

g - $\frac{vmin^{2} }{r}$ = 0

vmin = $\sqrt[.]{g*r}$ = $\sqrt[.]{9.8 m/s^{2} * 30m } = 17.146 m/s$

4 0

3 years ago

Calculate the force of gravity between planet X and planet y if both planets are 3.75 X 10^11 m apart, planet X has a mass of 1.

GenaCL600 [577]

So, the force of gravity that the asteroid and the planet have on each other approximately $\boxed{\sf{2.9 \times 10^{17} \: N}}$

<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :

$\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}$

With the following condition :

F = gravitational force (N)
G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_1}$ = mass of the first object (kg)
$\sf{m_2}$ = mass of the second object (kg)
r = distance between two objects (m)

<h3>Problem Solving</h3>

We know that :

G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_X}$ = mass of the planet X = $\sf{1.55 \times 10^{22}}$ kg.
$\sf{m_Y}$ = mass of the planet Y = $\sf{3.95 \times 10^{28}}$ kg.
r = distance between two objects = $\sf{3.75 \times 10^{11}}$ m.

What was asked :

F = gravitational force = ... N

Step by step :

$\sf{F = G \times \frac{m_X \times m_Y}{r^2}}$

$\sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}}$

$\sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}}$

$\sf{F \approx 2.9 \times 10^{39 - 22}}$

$\sf{F \approx 2.9 \times 10^{17} \: N}$

<h3>Conclusion</h3>

So, the force of gravity that the asteroid and the planet have on each other approximately

$\boxed{\sf{2.9 \times 10^{17} \: N}}$

<h3>See More</h3>

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2 years ago

Which best describes a chain reaction associated with a nuclear reaction?

mojhsa [17]

I think fission chain reaction is the correct answer.

8 0

3 years ago

What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?

Stolb23 [73]

Answer:

$1.1\cdot 10^{-10}C$

Explanation:

The electric field produced by a single point charge is given by:

$E=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

$q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C$

8 0

3 years ago