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Ghella [55]
2 years ago
15

If you stood on mars and lifted a 15kg pack, you would be pulling with a force greater than...?

Physics
1 answer:
DIA [1.3K]2 years ago
7 0

Answer:

See the answers below

Explanation:

In this problem, we must be clear about the concept of weight. Weight is defined as the product of mass by gravitational acceleration.

We must be clear that the mass is always preserved, that is, the mass of 15 [kg] will always be the same regardless of the planet where they are.

W=m*g

where:

W = weight [N] (units of Newtons)

m = mass = 15 [kg]

g = gravity acceleration [m/s²]

Since we have 9 places with different gravitational acceleration, then we calculate the weight in each of these nine places.

<u>Mercury</u>

<u />w_{mercury} = 15*3.8\\w_{mercury}= 57 [N]\\<u />

<u>Venus</u>

<u />w_{venus}=15*8.8\\w_{venus}= 132 [N]<u />

<u>Moon</u>

<u />w_{moon}=15*1.6\\w_{moon}=24[N]<u />

<u>Mars</u>

w_{mars}=15*3.7\\w_{mars}=55.5 [N]

<u>Jupiter</u>

<u />w_{jupiter}=15*23.1\\w_{jupiter}= 346.5[N]<u />

<u>Saturn</u>

<u />w_{saturn}=15*9\\w_{saturn}=135[N]<u />

<u>Uranus</u>

<u />w_{uranus}=15*8.7\\w_{uranus}=130.5[N]<u />

<u>Neptune</u>

<u />w_{neptune}=15*11\\w_{neptune}=165[N]<u />

<u>Pluto</u>

<u />w_{pluto}=15*0.6\\w_{pluto}=9[N]<u />

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A car traveling at 30 m/s drives off a cliff that is 50 meters high? How far away does it land?
Semenov [28]

Answer:

The maximum range R_{max}= 132. 72 m

Explanation:

Given,

The initial velocity of the car, u = 30 m/s

The height of the cliff, h = 50 m

Let the car drives off the cliff with a horizontal velocity of 30 m/s.

The formula for a projectile that is projected from a height h from the ground is given by the relation

                                R_{max}= \frac{u}{g}\sqrt{u^{2} + 2gh }  m

Where,

                          g - acceleration due to gravity

Substituting the values in the above equation

                   R_{max}= \frac{30}{9.8}\sqrt{30^{2} + 2X9.8X50 }  

                                          = 132.72  m

Hence, the car lands at a distance, R_{max}= 132. 72 m            

3 0
3 years ago
On a straight road (taken to be in the x direction) you drive for an hour at 60 km per hour, then quickly speed up to 120 km per
Luda [366]

Answer:

The average velocity is 180 km/hr

Explanation:

Given;

initial velocity, u = 60 km per hour

final velocity, v = 120 km per hour

initial time = 1 hour

final time = 2 hour

Initial position = 60 km/h x 1 hour = 60 km

final position = 120 km/h x 2 hour = 240 km

The average velocity is given by;

V_{avg} = \frac{Final \ position\  - \ Initial \ position}{final \ time\  - \ initial \ time}\\\\V_{avg} = \frac{240km \ - \ 60km}{2hr\  - \ 1hr} \\\\V_{avg} = \frac{180 \ km}{1hr} \\\\V_{avg}= 180 \ km/hr

Therefore, the average velocity is 180 km/hr

3 0
3 years ago
If the temperature of a volume of ideal gas increases for 100°C to 200°C, what happens to the average kinetic energy of the mole
mina [271]

Answer:\Delta E=2.0715\times 10^{-21} J

Explanation:

Given

Temperature of the gas is increased from 100 to 200

Also we know that average kinetic energy of the molecules is

E=\frac{3}{2}\cdot \frac{R}{N_A}T

Where

R=Gas constant

N_A=Avogadro's number

T=Temperature in kelvin

\frac{R}{N_A}=1.381\times 10^{-23}

So kinetic energy increases by

\Delta E=\frac{3}{2}\times 1.381\times 10^{-23}\left ( 200-100\right )

\Delta E=2.0715\times 10^{-21} J

8 0
3 years ago
When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the
KonstantinChe [14]

Answer:

The ratio is  KE : TM  =  0.75

Explanation:

from the question we are told that

  The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical  energy of the mass is mathematically represented as

        TM  =  \frac{1}{2}  *  k  *  A^2

Here  k is the spring constant  ,  A is the total displacement of the  the mass  from maximum  compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

        TM  =  KE  + PE

=>     KE = TM  - PE

Here the potential  energy of the mass is mathematically represented as

     PE   = \frac{1}{ 2}  *  k *  [ x ]^2

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is  

      x = \frac{A}{2}

So

     PE   = \frac{1}{ 2}  *  k *  [ \frac{A}{2}  ]^2

So

      KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{2}  *  k  *  [\frac{A}{2} ]^2

=>    KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{8}  *  k  *  A ^2

=>    KE =  0.375  *  k  *  A^2

So the ratio of  KE :  TM is  mathematically represented as

       \frac{KE}{TM} =  \frac{0.375  k A^2 }{0.5 k A^2}

=>    \frac{KE}{TM} = 0.75

3 0
2 years ago
A 10 kg ball is held above a building with a height of 30 m. What is the
Darina [25.2K]

Answer: 2940 J

Explanation: solution attached:

PE= mgh

Substitute the values:

PE= 10kg x 9.8 m/s² x 30 m

= 2940 J

6 0
3 years ago
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