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masya89 [10]
3 years ago
9

Calculate the acceleration of a car in m/s that can go from rest to 28 m/s in 10 seconds.

Physics
1 answer:
mestny [16]3 years ago
7 0

Answer:

a = 2.8 m/s2

Explanation:

a = Δv ÷ Δt

a = 28 ÷ 10

a = 2.8 m/s2

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To drive a car at a constant velocity, you
kipiarov [429]

Answer:

the answer is C

Explanation:

The car, first is at rest and if you don't accelerate it won't move. When to hit the gas it will accelerate from rest

8 0
2 years ago
A child pushes a 75 N toy car across the floor. What is the mass of the car?
GaryK [48]

Answer:

7.6 kg

Explanation:

w=75N

w=mg

m=w÷g

m=75÷9.8

m=7.6kg

8 0
3 years ago
Two balls are dropped from rest and allowed to fall. If one ball is allowed to fall for 1 s and the other for 3 s compare the di
kirill115 [55]

The second ball traveled a greater distance when compared to the first ball because the second ball spent more time in motion.

The given parameters;

  • time of fall of the first ball, t = 1 s
  • time of fall of the second ball, t = 3 s

The distance traveled by each ball is calculated using the second equation of motion as shown below.

The distance traveled by the first ball is calculated as follows;

h = u_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 1^2)\\\\h = 4.9 \ m

The distance traveled by the second ball is calculated as follows;

h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 3^2)\\\\h = 44.1\ m

Thus, the second ball traveled a greater distance because it spent more time in motion.

Learn more here:brainly.com/question/5868480

3 0
1 year ago
A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul
Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2

also we know that

I = \frac{2}{5}mR^2

w= \frac{v}{R}

Now kinetic energy is given by

KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2

KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2

KE = \frac{7}{10}mR^2w^2

KE = \frac{7}{10}*150*(0.200)^2(50)^2

KE = 10500 J

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0
3 years ago
Why might the term red hot be misleading? (relating to stars)
Allushta [10]

red hot is hot, but other colours are even hotter. stars may be hotter than red hot.

there is also something calle the red shift.

3 0
2 years ago
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