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kap26 [50]
3 years ago
10

SOMEBODY PLEEASEEE HELP A STRUGGLING HIGHSCHOOLERRRR +(

Physics
1 answer:
CaHeK987 [17]3 years ago
5 0

Explanation:

What is the weight of a 2.00-kilogram object on the surface of Earth?

2.00 N

4.91 N

9.81 N

19.6 N

Given parameters:

Mass of the object = 2kg

Unknown:

Weight of the object  = ?

Solution:

The weight of an object is the force of gravity acting on the object;

 Weight  =  mass x acceleration due to gravity

Acceleration due to gravity  = 9.8m/s²

 Now insert the parameters and solve;

    Weight  = 2 x 9.8 = 19.6N

A person weighing 785 Newtons on the surface of the Earth would weigh 47 Newtons on the surface of Pluto. What is the magnitude of the gravitational acceleration on the surface of Pluto?

1.7 m/s²

0.59 m/s²

0.29 m/s²

9.8 m/s²

Given parameters:

Weight on Earth  = 785N

Weight on Pluto = 47N

Unknown:

Acceleration due to gravity on Pluto = ?

Solution

The mass of the body both on Earth and Pluto is the same.

  Weight = mass x acceleration due to gravity

Now find the mass on Earth;

  Acceleration due to gravity on Earth  = 9.8m/s²

    785   = mass x 9.8

         mass  = \frac{785}{9.8}   = 80.1kg

So;

  Acceleration due to gravity on Pluto = \frac{Weight on Pluto}{mass }  

  Acceleration due to gravity  = \frac{47}{80.1}   = 0.59m/s²

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Read 2 more answers
What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long water tank that can hold 50.0 kg of water by the weight
svetoff [14.1K]

Answer:

1088.9N/m2

Explanation:

Calculation for What pressure is exerted

First step is to find the area of bottom of the tank using formula

Area=Width*breadth

Let plug in the formula

Area=0.5*0.9

Area=0.45m2

Now let calculate what pressure is exerted using this formula

Pressure=Force/Area

Where,

Force=Mass *Gasoline

Area=Width of the tank* Length of the tank

Let plug in the formula

Pressure=50*9.8/0.5*0.9

Pressure=490/0.45

Pressure=1088.9N/m2

Therefore What pressure is exerted is 1088.9N/m2

3 0
3 years ago
Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi
trapecia [35]

Answer:

a)  F_p=882N

b)  P=4410W

c)  V_p'=24135 ,n=15.2\%

Explanation:

From the question we are told that:

Mass M=1500kg

Velocity v=4.9m/s

Coefficient of Rolling Friction \mu=0.06

a)

Generally the equation for The Propulsion Force is mathematically given by

 F_p=\mu*mg

 F_p=0.06*1500*9.81

 F_p=882N

b)

Therefore Power Required at

 V_p=5.0m/s

 P=F_p*V_p

 P=882*5

 P=4410W

c)

 V_p' =15mpg

 V_p'=15*\frac{1609}

 V_p'=24135

Generally the equation for Work-done is mathematically given by

 W=F_p*V_p'

 W=882*15*1609

 W=2.13*10^7

Therefore

Efficiency

 n=\frac{W}{E}*100\%

Since

Energy in one gallon of gas is

 E=1.4*10^8J

Therefore

 n=\frac{2.1*10^7}{1.4*10^8}*100\%

 n=15.2\%

7 0
3 years ago
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