All categories

3 years ago

The seatbelt across your chest should have about ______ fist width of slack

1 answer:

7 0

The correct answer is option (c) i.e. one

The seat belt across your chest should have about one fist width of the slack. Because the seat belt across your chest is required to have minimum fist width of the slack to hold the body in place and to avoid slamming of the body against the seat belt in case of a collision at high speeds. To have one fist width of the slack, makes it a safest to wear it in case of any collision.

You might be interested in

GIVING BRAINLIEST PLEASE HELP!!

erik [133]

Explanation:

the other 40% is used to power the 60% making it only capable of 60% efficiency

4 0

2 years ago

HELP I WILL GIVE BRAINLIEST!

Ann [662]

Answer:

Answer:B

Explanation:

Because it all stayed consistant

4 0

2 years ago

Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax

kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its x-component is

$F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its x-component is

$F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N$

So, the x-component of the resultant force is

$F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N$

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its y-component is

$F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its y-component is

$F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N$

So, the y-component of the resultant force is

$F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N$

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

$F=\sqrt{F_x^2+F_y^2}$

Where in this problem, we have:

$F_x=-7.14 N$ is the x-component

$F_y=2.70 N$ is the y-component

And substituting, we find:

$F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N$

6 0

3 years ago

Read 2 more answers

What is the net force on a bathroom scale when a 110 pound person stands on it? How do you know?

USPshnik [31]

The scale is a rest scale reads the support for is not enough net force.

8 0

2 years ago

An elevator motor provides 45.0 kW of power while lifting an elevator 35.0 m. If the elevator contains seven passengers each wit

mylen [45]

Find how much work ∆W is done by the motor in lifting the elevator:

P = ∆W / ∆t

where

• P = 45.0 kW = power provided by the motor

• ∆W = work done

• ∆t = 20.0 s = duration of time

Solve for ∆W :

∆W = P ∆t = (45.0 kW) (20.0 s) = 900 kJ

In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is M = (m + 490.0) kg, where m is the mass of the elevator alone. Then

∆W = M g h

where

• g = 9.80 m/s² = acceleration due to gravity

• h = 35.0 m = distance covered by the elevator

Solve for M, then for m :

M = ∆W / (g h) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg

m = M - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg

4 0

3 years ago