All categories

3 years ago

A 5000.0-kg car goes from 0 m/s to 70 m/s in 3600 seconds. What is the accelera

Physics

1 answer:

antiseptic1488 [7]3 years ago

4 0

given:

mass = 5000 kg

u (initial velocity) = 0 m/s

v (final velocity) = 70 m/s

time taken to change velocity = 3600 s

acceleration = v - u / t

a = 70 - 0 / 3600

a = 70 / 3600

a = 0.0194 m/s2 (approx)

given: mass = 5000 kg

acceleration = 0.0194 (found in 1st part)

force = mass * acceleration

f = ma

f = 5000*0.0194 = 97 N

therefore the acceleration will be 0.0194 m/s2

and the force involved in acceleration will be 97 N

You might be interested in

When placed in direct sunlight, which an object will absorb the most visible light energy

Vikentia [17]

I would guess anything that is black? black absorbs there most light energy. that's is why a black car in the sun will always be hotter inside than a white car.

3 0

3 years ago

A college student holds a pail full of water by the handle and whirls it around in a vertical circle at a constant speed. The ra

Pavlova-9 [17]

The minimum speed of the water must be 3.4 m/s

Explanation:

There are two forces acting on the water in the pail when it is at the top of its circular motion:

The force of gravity, mg, acting downward (where m is the mass of the water and g the acceleration of gravity)
The normal reaction, N also acting downward

Since the water is in circular motion, the net force must be equal to the centripetal force, so:

$N+mg=m\frac{v^2}{r}$

Where:

$g=9.8 m/s^2$

v is the speed of the pail

r = 1.2 m is the radius of the circle

The water starts to spill out when the normal reaction of the pail becomes zero:

N = 0

When this occurs, the equation becomes:

$mg=m\frac{v^2}{r}\\v=\sqrt{gr}$

And substitutin the values of g and r, we find the minimum speed that the water must have in order not to spill out:

$v=\sqrt{(9.8)(1.2)}=3.4 m/s$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

6 0

3 years ago

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

3 years ago

Which statement best describes frustration?

wlad13 [49]

C. when you can't achieve your goal due to events beyond your control

5 0

3 years ago

A basketball is tossed up into the air, falls freely, and bounces from the wooden floor. From the moment after the player releas

Tju [1.3M]

Answer:

Tha ball- earth/floor system.

Explanation:

The force acting on the ball is the force of gravity when ignoring air resistance. At the moment the player releases the ball, until it reaches the top of its bounce, the small system for which the momentum is conserved is the ball- floor system. The balls exerts and equal and opposite force on the floor. <u>Here the ball hits the floor, because in any collision the momentum is conserved. Moment of the ball -floor system is conserved</u>. Mutual gravitation bring the ball and floor together in one system. As the ball moves downwards, the earth moves upwards, although with an acceleration on the order of 1025 times smaller than that of the ball. The two objects meet, rebound and separate.

5 0

3 years ago