Ask question

Ask question

All categories

3 years ago

9

If the gravitational force exists between all objects in the universe, why aren’t people affected by the gravitational force exe

rted by cars and buildings?

1 answer:

soldier1979 [14.2K]3 years ago

6 0

Because cars and buildings are small in mass and so exert very low gravitational fore which is dominated by the large gravitational force of earth

You might be interested in

6. (a) Suppose the earth is revolving round the sun in a circular orbit of radius one b astronomical unit (1.5% 10 km). Find the

kkurt [141]

Answer:

tough

ques

Explanation:

5 0

2 years ago

once you decide that it is safe to move out of traffic, what steps should you follow to ensure safe passage?

Solnce55 [7]

First put your turn signal on, next check for any ongoing traffic and wait until it is clear lastly start to drift into the lane you need to clear away from traffic

4 0

4 years ago

H H H O H O N C C N C C H O H H O H H C H H H ALANINE GLYCINE

Trava [24]

i’m not 100 percent sure but I need points to ask questions good luck th

6 0

2 years ago

A car with a mass of 1500 kg accelerates when the traffic light turns green. If the net force on the car is 3750 newtons, what i

timurjin [86]

Answer:

2.5 ms⁻²

Explanation:

By Newton's 2nd law,

The rate of change of momentum is directly proportional to the unbalance force applied on the object,

By that you can get the equation,

F = ma ⇒ a = F/m

where terms are in usual meaning

a = 3750/1500 = 2.5 ms⁻²

8 0

3 years ago

You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an

a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, $\mu=6\times 10^{-2}$

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

$KE=PE+W$

$\dfrac{1}{2}mv^2=mgh+W$

W is the work done by the friction.

$W=f\times d$

$f=\mu mg\ cos\theta$

$W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}$

$W=\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}$

$\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}$

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0

3 years ago