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3 years ago

Know that these units are for

1 answer:

4 0

Ohm for resistance
Volt for potential difference
ampere for current
watt for power
coulomb for charge
kilowatt hour is for energy !

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What is the escape velocity from a planet with a mass that is 10.8% Earths mass and a size that is 53% Earths radial size?

DENIUS [597]

The escape velocity from the planet is 5052 m/s

Explanation:

The formula to calculate the escape velocity from a planet is:

$v=\sqrt{\frac{2GM}{R}}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

M is the mass of the planet

R is the radius of the planet

For Earth,

$M_E=5.98\cdot 10^{24} kg$ is the mass

$R_E=6.37\cdot 10^6 m$ is the radius

Here we have a planet that:

- its mass is 10.8% of that of Earth, so

$M=0.108M_E$

- its radius is 53% of that of Earth, so

$R=0.53 R_E$

Substituting everything into the first equation, we find the escape velocity from this planet:

$v=\sqrt{\frac{2G(0.108M_E)}{0.53R_E}}=\sqrt{\frac{2(6.67\cdot 10^{-11})(0.108)(5.98\cdot 10^{24})}{(0.53)(6.37\cdot 10^6)}}=5052 m/s$

#LearnwithBrainly

4 0

3 years ago

When using the magnification equation, a value greater than 1 as the solution for M indicates that the image is_____________.

Alexxandr [17]

Larger than the object.

6 0

4 years ago

Read 2 more answers

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist

e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > $38 \frac{ft}{s^2} x \frac{1mi}{5280ft} x \frac{(3600s)^2}{(1h)^2} = 93272.27 \frac{mi}{h^2}$

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > $v_2^2=v_1^2 + 2as$

The initial velocity is 50mi/h $(v_1=50)$

When it stops the final velocity is $(v_2=0)$

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

$0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s = 2500/(185644.54)

s=0.0134
$

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > $0.0134 mi * \frac{5280ft}{1mi} = 70.8 ft$
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0

3 years ago

After t hours a freight train is s(t) = 18t2 − 2t3 miles due north of its starting point (for 0 ≤ t ≤ 9). (a) Find its velocity

BartSMP [9]

Answer:

Explanation:

Given the equation modelled by the height of the train given as:

s(t) = 18t²-2t³ for for 0 ≤ t ≤ 9

a) Velocity is the rate of change of displacement.

Velocity = dS(t)/dt

V = dS(t)/dt = 36t - 6t² miles

Velocity at t = 3hrs is determiner by substituting t = 3 into the velocity function.

V = 36(3) -6(3)²

V= 108 - 72

Velocity = 36mi/hr

b) for Velocity at time = 7hrs

V(7) = 36(7) - 6(7)²

V(7) = 252 - 294

V(7) = -42mi/hr

The velocity at t = 7hrs is -42mi/hr

c) Acceleration is the rate of change of velocity.

a(t) = dV(t)/dt

Given v(t) = 36t - 6t²

a(t) = 36 - 12t

Acceleration at t=1 is given as:

a(1) = 36 -12(1)

a(1) = 24mi/hr²

4 0

3 years ago

A truck accelerating at 0.0083 meters/second2 covers a distance of 5.8 × 104 meters. If the truck's mass is 7,000 kilograms, wha

anygoal [31]

<span>F = m*a = 7000kg * 9.8N/kg = 68,600 N.

68,600 N is your answer
</span>

3 0

3 years ago