I need help on this question !

Calculate the speed of a body covering a distance of 320 km in 4h

Mice21 [21]

We have that the speed of a body covering a distance of 320 km in 4h is mathematically given as

V=22.22m/s is

<h3 /><h3>
Speed</h3>

From the question we are told

calculate the speed of a body covering a distance of 320 km in 4h

Generally the equation for the Speed is mathematically given as

$V=\frac{distance }{time}\\\\Therefore\\\\V=\frac{320*1000}{4*60*60}\\\\$

V=22.22m/s

Hence

The speed of a body covering a distance of 320 km in 4h is

V=22.22m/s

For more information on Speed visit

brainly.com/question/7359669

4 0

2 years ago

How do exons and introns differ?

Norma-Jean [14]

Exons And Introns differ because Exons code for protein and Introns do not.

so your answer would be: Exons code for Proteins

Are you in K12?

7 0

2 years ago

Can someone help me due at 1:00 PM plus 100 points <br><br>questions B,C,D

vagabundo [1.1K]

<h3>Hope it helps...</h3>

6 0

3 years ago

Read 2 more answers

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:

1)

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

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

2)

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

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

3)

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

imagine riding a single-speed bicycle. why do you have to push harders on the pedals to start the bicycle moving than to keep it

Flura [38]

A large force is required to accelerate the mass of the bicycle and rider. Once the desired constant velocity is reached, a much smaller force is sufficient to overcome the ever-present frictional forces.

3 0

2 years ago