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3 years ago

Compared to its weight on Earth, a 10-kg object on the moon will weigh Question 9 options: less. the same amount. more.

Physics

2 answers:

Nostrana [21]3 years ago

3 0

it will weigh less on the moon.

iVinArrow [24]3 years ago

3 0

ANY object will weigh 83% <em> less</em> on the Moon than weighs on Earth.

Your 10kg rock weighs 98N (22.4 lbs) on Earth and 16N (3.6 lbs) on the Moon.

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Karo-lina-s [1.5K]

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2 years ago

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Can someone help me with these questions plz

Korolek [52]

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3 years ago

PLZ HELP

Nadya [2.5K]

1) The mass of the continent is $3.3\cdot 10^{21}kg$

2) The kinetic energy of the continent is 1683 J

3) The speed of the jogger must be 6.57 m/s

Explanation:

The continent can be represented as a slab of size

$d=5850 km = 5.85\cdot 10^6 m$

and depth

$t = 35 km = 3.5\cdot 10^4 m$

So its volume is

$V=d^2 t = (5.85\cdot 10^6)^2(3.5\cdot 10^4)=1.20\cdot 10^{18} m^3$

We also know that the density of the continent is

$\rho = 2750 kg/m^3$

Therefore, we can calculate its mass as:

$m=\rho V=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg$

The kinetic energy of the continent is given by

$K=\frac{1}{2}mv^2$

where

m is its mass

v is its speed

We have already calculate its mass, while the speed is

v = 3.2 cm/year

We have to convert into SI units first, as follows:

$v=3.2 \frac{cm}{year} \cdot \frac{1}{100 cm/m} \cdot \frac{1}{(365 d/y)(24h/d)(60min/h)(60 s/min)}=1.01\cdot 10^{-9} m/s$

The mass is

$m=3.3\cdot 10^{21} kg$

So, the kinetic energy of the continent is

$K=\frac{1}{2}(3.3\cdot 10^{21})(1.01\cdot 10^{-9})^2=1683 J$

Here we have a jogger having the same kinetic energy of the continent, so

$K=1683 J$

And the kinetic energy of the jogger can be expressed as

$K=\frac{1}{2}mv^2$

where

m = 78 kg is the mass of the jogger

v is his speed

We can therefore re-arrange the equation to find the speed of the man, and we get:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1683)}{78}}=6.57 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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2 years ago

Galileo attempted to measure the speed of light by measuring the time elapsed between his opening a lantern and his seeing the l

Dafna11 [192]

Answer:

The distance is $2.58\times10^{7}\ m$ .

Explanation:

Given that,

Time $\Delta t = 0.2\ s$

The velocity is no more than a 14 % error in the speed of light.

So,

Velocity $v= c\times 86\%$

We need to calculate the distance

Using formula of speed

$v = \dfrac{d}{t}$

$d = v\times t$

Where, v = speed

d = distance

t = time

Put the value into the formula

$d = 3\times10^{8}\times\dfrac{86}{100}\times0.2$

$d=516\times10^{5}\ m$

We know that,

The one side distance d' is

$d'=\dfrac{d}{2}$

$d'=\dfrac{516\times10^{5}}{2}$

$d'=2.58\times10^{7}\ m$

Hence, The distance is $2.58\times10^{7}\ m$ .

5 0

2 years ago

How long would it take a drag racer to increase her speed from 10m/s to 20 m/s if her car accelerates at a uniform rate of 15 m/

weeeeeb [17]

Answer:

t = 0.67 [s]

Explanation:

To solve this problem we must use the following kinematics equation.

$v_{f} =v_{i} +(a*t)\\where:$

Vf = final velocity = 20[m/s]

Vi = initial velocity = 10 [m/s]

a = aceleration = 15 [m/s^2]

Now replacing in the equation we have:

20 = 10 + (15*t)

t = (20-10)/15

t = 0.67 [s]

7 0

3 years ago