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Gennadij [26K]
3 years ago
15

PLEASE HELP How much work is required to raise a 1200 kg rock 15 m?

Physics
1 answer:
Alex3 years ago
3 0

Answer:

18000

Explanation:

the is to raise how much work is required is 18000

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ExtremeBDS [4]
Ridges, mountains, and volcanoes!

3 0
3 years ago
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A bus moving along a level road increases its speed from 20 m/s to 35 m/s in 15.0s. What is the car's acceleration?​
Ksju [112]

Explanation:

initial velocity U = 20m/s

Final velocity V = 35m/s

time = 15.0 secs

change in velocity = 35 - 15

= 20m/s

acceleration a = change in velocity/time V/t

a = (35-20)/15

a= 15/15

Hence, your acceleration is 1m/s^2

5 0
2 years ago
What is the velocity of a snail that travels 0.20 meters in 1 minute
stepladder [879]

The speed is 0.2 meter per minute.

There is not enough information given in the question to determine the velocity.

5 0
3 years ago
How is Coulomb’s law similar to newton’s law of gravitational force? How is it different
natulia [17]

The similarities and the differences between gravitational and electric force are listed below

Explanation:

- The magnitude of the gravitational force between two objects is given by Newton's law of gravitation:

F=G\frac{m_1 m_2}{r^2}

where

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

m_1, m_2 are the masses of the two objects

r is the separation between them

- Coloumb's law gives instead the strength of the electrostatic force between two charged objects, which is

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

By comparing the two equations, we find the following similarities:

  • Both the forces are inversely proportional to the square of the distance between the two objects, F\propto \frac{1}{r^2}
  • Both the forces are proportional to the product between the "main quantity" of each force, which is the mass for the gravitational force (F\propto m_1 m_2) and the charge for the electric force (F\propto q_1 q_2

Instead, we have the following differences:

  • The gravitational force is always attractive, since the sign of m is always positive, while the electric force can be either attractive or repulsive, since the sign of q can be either positive or negative
  • The value of the gravitational costant G is much smaller than the value of the Coulomb's constant, so the gravitational force is much weaker than the electric force

Learn more about gravitational force and electric force:

brainly.com/question/1724648

brainly.com/question/12785992

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and
lawyer [7]

1 kilometre=1000 metre

      1 hour = 3600 second

       1\ km/hr=\frac{1000}{3600} m/s

       1\ km/hr=\frac{5}{18} m/s

The initial velocity of car A is 35.0 km/hr i.e

                                         35.0\ km/hr=35*\frac{5}{18} m/s

                                                                   = 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as  25\ km/hr^2

                                            =\ 25*\frac{1000}{3600*3600} m/s^2

                                            =0.00192901234 m/s^2

The time taken by car A = 15 min.

From equation of kinematics we know that-

                                 v= u+at      [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A,  v = 9.72 m/s +[0.00192901234×15×60]m/s

                                   =11.456111106 m/s

The acceleration of B is given as    15\ km/hr^2

                                    =0.00115740740740 m/s^2

The time taken by car B =20 min

The final velocity of B is -

                             v= u+at

                               = u-at    [Here a is negative due to deceleration]

                               =12.5 m/s +[0.0011574074074×20×60]

                               =13.8888888.....

                               =13.9

The acceleration of C is given as    40\ km/hr^2          

                                                            =\ 0.003086419753 m/s^2

The time taken by car C =30 min

The final velocity of C is-

                                v = u+at

                                   =8.89 m/s+[0.003086419753×30×60] m/s

                                   =14.4455555555..m/s

                                   =14.45 m/s

The car C is decelerating.The deceleration is given as-  60\ km/hr^2

                                                                      =0.0046296296296m/s^2

The time taken by car D= 45 min.

The final velocity of the car D is-

                     v =u+at

                        =30.56 -[0.00462962962962×45×60]m/s

                        =18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

                 


3 0
3 years ago
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