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3 years ago

What type of electromagnetic radiation was used to make this picture?

Physics

1 answer:

goldenfox [79]3 years ago

5 0

Answer:

Hey there

as you can see in the image, you only see the bones of the hand. not the skin or muscles. That is because only bones absorb X-rays while skin and other tissue let them pass-through for the most part. So your answer is Xrays.

Can i please have brainly

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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

4 years ago

Read 2 more answers

Calculate the kinetic energy of a 5.2 kg object moving at 2.4 m/s.

Lilit [14]

Answer:

Kinetic energy, K = 14.97 J

Explanation:

We have,

Mass of the object, m = 5.2 kg

Velocity of the object, v = 2.4 m/s

It is required to find the kinetic energy of the object. As it is moving with some velocity, it will have kinetic energy. It is given by :

$K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 5.2\times (2.4)^2\\\\K=14.97\ J$

So, the kinetic energy of the object is 14.97 J.

7 0

3 years ago

How to solve <br>9.8 = (x-0) =?<br> 4.3

Nadya [2.5K]

You want to know how to solve it?

7 0

3 years ago

A cyclist traveling at 30.0m/s along a straight road comes uniformly to stop in 5.00s. Determine the stopping acceleration, the

Amiraneli [1.4K]

Answer:

I. Stopping acceleration = 6 m/s²

II. Stopping distance, S = 75 meters

Explanation:

Given the following data;

Final velocity = 30 m/s

Time = 5 seconds

To find the stopping acceleration;

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Substituting into the equation;

$a = \frac{30 - 0}{5}$

$a = \frac{30}{5}$

Acceleration = 6 m/s²

II. To find the stopping distance, we would use the third equation of motion;

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

30² = 0² + 2*6*S

900 = 12S

S = 900/12

S = 75 meters

7 0

3 years ago

A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.4 m/

Bad White [126]

Answer:

1.06 secs

Explanation:

Initial speed of sled, u = 8.4 m/s

Final speed of sled, v = 5.8 m/s

Coefficient of kinetic friction, μ = 0.25

Using the impulse momentum theory, we know that the impulse applied to the sled is equal to change in momentum of the sled:

FΔt = mv - mu

where m = mass of the object

Δt = time interval

F = force applied

The force applied on the sled is the frictional force, which is given as:

F = -μmg

where g = acceleration due to gravity

Therefore:

-μmgΔt = mv - mu

-μmgΔt = m(v - u)

-μgΔt = v - u

Making Δt subject of formula:

Δt = (v - u) / -μg

Δt = (5.8 - 8.4) / (-0.25 * 9.8)

Δt = -2.6/ -2.45

Δt = 1.06 secs

It took the sled 1.06 secs to travel from A to B.

7 0

3 years ago