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3 years ago

100 Points !!!!!!!! I WILL GIVE BRAINLIEST

Physics

2 answers:

melisa1 [442]3 years ago

6 0

Answer:

Answer: 1. 0.25 m/s squared

2. A = 0.4 m/s^2

3.=20kg.

Explanation:

brainliest? ill answer more of your recent questions!!

kumpel [21]3 years ago

3 0

Answer: 1. 0.25 m/s squared

2. A = 0.4 m/s^2

3.=20kg. I did this before.

Explanation:

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If a net force acting on a moving object causes its speed to increase, what

Klio2033 [76]

Answer:

It increases.

Explanation:

If a force acts in the same direction as the object's motion, then the force speeds the object up. Either way, a force will change the velocity of an object. And if the velocity of the object is changed, then the momentum of the object is changed.

7 0

3 years ago

PLEASE ANSWER ASAP!!! ITS DUE IN 10 MINUTES!!!!!!

forsale [732]

Explanation:

2. Aerobic exercise. ...

Strength building. ...

Balance Training. ...

Endurance. ...

Flexibility. ...

Moderate intensity exercise. ...

Vigorous exercise.

3.Brisk walking.

Running.

Jogging

Bear crawls.

Swimming.

Water aerobics.

Cycling/bicycling.

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Hope it helps..

5 0

3 years ago

You are driving your 1700 kg car at 21 m/s down a hill with a 5.0∘ slope when a deer suddenly jumps out onto the roadway. You sl

podryga [215]

Answer:

d = 27.7 m

Explanation:

Here the car is driving on the inclined plane

So here we can say that work done by the gravity and work done by friction is equal to change in kinetic energy of the system

So here we can write it as

$mgsin\theta \times d - F_f \times d = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

now we have

m = 1700 kg

$v_f = 0$

$v_i = 21 m/s$

$F_f = 1.5 \times 10^4 N$

$(1700)(9.81)sin5 (d) - (1.5\times 10^4)d = 0 - \frac{1}{2}(1700)(21^2)$

$1453.5 d - (1.5\times 10^4)d = -374850$

$d = 27.7 m$

6 0

3 years ago

A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc

Harman [31]

Answer:

$E=88200\ J$

Explanation:

Given:

mass of car, $m=1500\ kg$

distance of skidding after the application of brakes, $d=15\ m$

coefficient of kinetic friction, $\mu_k=0.4$

So, the energy dissipated during the skidding of car:

Frictional force:

$f=\mu_k.N$

where N = normal reaction by ground on the car

$f=0.4\ties 1500\times 9.8$

$f=5880\ N$

Now from the work-energy equivalence:

$E=f.d$

$E=5880\times 15$

$E=88200\ J$ is the dissipated energy.

3 0

3 years ago

Think about a girl on a swing. When is the kinetic energy of the girl zero?

Eduardwww [97]

Kinetic energy is energy of motion. Pick choice-A, at the top of the swing, where she stops moving & then goes the other way.

7 0

3 years ago

Read 2 more answers