All categories

3 years ago

A 55.0 kg runner who weighs 539.0 N is accelerating at 3.2 m/s2. After 2

Physics

2 answers:

larisa86 [58]3 years ago

4 0

Answer:

The answer is C. 352.0 kg-m/s

omeli [17]3 years ago

3 0

Answer:

C. 352.0 kg-m/s

Explanation: Took the test on APEX

You might be interested in

The force of attraction between 2 objects is called gravity. Why do you think that an object, like your notebook, pulled towards

Gekata [30.6K]

This question is asking: Why does gravity occur? why will your notebook get pulled towards you?

6 0

3 years ago

Read 2 more answers

What kind of waves are responsible for all the damage an earthquake causes?

Liula [17]

Seismic waves are waves generated by Earthquakes.

6 0

3 years ago

two children having a disagreement pull on a 50N sled in opposite directions. one pulls with a force of 300N east, the other wit

prohojiy [21]

Since forces are vector quantities, we must indicate direction using positive and negative values. East will be assigned positive and west will be negative. Friction will act as a negative force since it impedes action. To calculate the net force we sum the vector quantities, as follows. Net force equals 50n which is derived by the following calculation: 300n-220n-30n.

7 0

3 years ago

Read 2 more answers

According to the _______ the amount of energy in the universe doesn't change.

balandron [24]

The answer is B, Law of Kinetic Energy

6 0

3 years ago

Read 2 more answers

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.