All categories

4 years ago

How does inertia affect a person who is not wearing a seatbelt during a collision?

1 answer:

4 0

When someone fails to wear a seat belt the passenger becoming a projectile the force a person will be subjected to for a passenger weighting 100 pounds, and the car is traveling at 60 mph would be the same as 6000 pounds. thats like hitting a brick wall. so is other words put ur seat belt on.

hope this helped (:

You might be interested in

Using the rules for significant figures, what do you get when you multiply 67.6 by 1.2?

Anika [276]

Answer:

Explanation:

The problem here is to use the rules for significant figures in the multiplication below.

67.6 x 1.2:

This product will yield an answer that will produce the least number of significant figures.

67.6 has 3 significant figures; 6, 7 and 6

1.2 has 2 significant figures; 1 and 2

The product must give us an answer of two significant figures:

67.6 x 1.2 = 81.12 to 2significant figures gives 81

The solution to this problem is 81

5 0

3 years ago

What is the distance between a 900 kg compact car and a 1600 kg pickup truck if the gravitational force between them is about 0.

elena-14-01-66 [18.8K]

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

$r= \sqrt{\frac{Gm1m2}{F} }$

$r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m$

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required:

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

$r= \sqrt{\frac{Gm1m2}{F} }$

$r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m$

3 0

3 years ago

Read 2 more answers

A frictionless inclined plane is 6 m long and rests on a wall that is 2 m high.

tresset_1 [31]

Answer:

The force needed to push a block of ice is 100 N.

Explanation:

It is given that,

Length of the friction less plane, l = 6 m

Height of the wall, h = 2 m

Weight of the block of ice, W = 300 N

The length of the inclined plane and the height of the wall act as hypotenuse and the perpendicular of the right angled triangle. Using trigonometry to find the angle as :

$sin\theta=\dfrac{P}{H}$

$sin\theta=\dfrac{2}{6}$

$sin\theta=\dfrac{1}{3}$

Force needed to push a block of ice weighing 300 N up the plane is equal to the horizontal component of the force as :

$F_x=F\ sin\theta$

$F_x=300\times \dfrac{1}{3}$

$F_x=100\ N$

So, the force needed to push a block of ice is 100 N. Hence, this is the required solution.

4 0

4 years ago

A 0.0780 kg lemming runs off a

Sindrei [870]

Answer:

0.913

Explanation:

k.e=1/2mv square

k.e=1/2×0.078g×23.4256m/s square

k.e=0.913J

3 0

3 years ago

P1: Explain in detail - How can the motion of an object that is already moving change? (What are the different ways a moving obj

Tomtit [17]

Inertia: tendency of an object to resist changes in its velocity. An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force.

5 0

4 years ago