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3 years ago

PLEASE HELP IT'S THE END OF THE QUARTER AND I NEED HELP ON THIS TEST TO HELP MY FINAL GRADE !!!

Physics

2 answers:

PolarNik [594]3 years ago

5 0

Answer:

Gamma rays and X–rays

Explanation:

The electromagnetic spectrum consists of all types of electromagnetic radiation. The EM spectrum is arranged from the longest wavelength with the lowest frequency and lowest energy which are the radio waves and microwaves to the shortest wavelength with the highest frequency and highest energy which are the x-rays and gamma rays. Visible light that can be seen by a humans is a small portion in between the ultraviolet and infrared rays.

Brainliest

harina [27]3 years ago

4 0

Answer:

radio waves

Explanation:

if u need further help try looking it up on Google to help u out!

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What is an insurance premium?*

aalyn [17]

Answer:

Explanation:

In a nutshell, an insurance premium is the payment or installment you agree to pay a company in order to have insurance. You enter into a contract with an insurance company that guarantees payment in case of damage or loss and, for this, you agree to pay them a certain, smaller amount of money

8 0

3 years ago

3 simple machines that make up a can opener

Yanka [14]

Hello There

Answer: The wheel and axle, the lever, and the wedge

Reason: The two long arms that clamp onto the can are levers. The handle which is used to rotate the can is the wheel and axle. Finally the blade is the wedge

I hope I helped
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6 0

3 years ago

A force of 30 N is applied tangentially to the rim of a solid disk of radius 0.14 m. The disk rotates about an axis through its

jenyasd209 [6]

Answer:

3.8 kg

Explanation:

The torque, τ, due to the force, F, is given by

$\tau = F\times r$

where r is the radius.

This torque is also given by

$\tau = I\times\alpha$

where I and α are respectively the moment of inertia and the angular acceleration.

For a solid disk, its moment of inertia for an axis though its centre and perpendicular to its face is given by

$I = \frac{1}{2}mr^2$

where m is its mass and r is its radius.

Hence, we have

$\tau = F\times r = I\times\alpha=\frac{1}{2}mr^2\times\alpha$

$m = \dfrac{2F}{r\alpha} = \dfrac{2(30\ \text{N})}{(0.14\ \text{m})(115\ \text{rad/s}^2)} = 3.8\ \text{kg}$

5 0

3 years ago

Read 2 more answers

While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at

Mashcka [7]

Answer:

The speed of the moving walkway is 1.50 m/s

Explanation:

The position of the child can be calculated using the following equation:

x = x0 + v · t

Where :

x = position of the child at time t.

v = velocity of the child.

t = time.

When the child runs in the same direction as the walkway, the velocity of the child will be its velocity relative to the walkway plus the velocity of the walkway. Then, if we place the origin of the frame of reference at the start of the walkway:

x = x0 + v · t

25 m = 0 m + (2.8 m/s + v) · t₁

Where v is the velocity of the walkway

On its way back, the velocity of the child relative to the walkway is in the opposite direction to the velocity of the walkway. Then:

x = x0 + v · t

0 m = 25 m + (-2.8 m/s + v) · t₂

We also know that t₁ + t₂ = 25 s

Then: t₁ = 25 - t₂

So, we can write the following system of equations:

25 m = (2.8 m/s + v) · (25 s - t₂)

-25 m = (-2.8 m/s + v) · t₂

Let´s take the second equation and solve it for t₂

-25 m / (-2.8 m/s + v) = t₂

Now, let´s replace t₂ in the first equation:

25 m = (2.8 m/s + v) · (25 s + 25 m / (-2.8 m/s + v))

Let´s sum the fraction: 25 s + 25 m / (-2.8 m/s + v)

25 m = (2.8 m/s + v) · (25 s ·(-2.8 m/s + v) + 25 m) / (-2.8 m/s + v)

multiply by (-2.8 m/s + v) both sides of the equation:

25 m(-2.8 m/s + v) = (2.8 m/s + v) · (-70 m + 25 s · v + 25 m)

Apply distributive property:

-70 m²/s +25 m·v = -196 m²/s +70 m·v +70 m²/s -70 m·v +25 s ·v² + 25 m v

56 m²/s = 25 s · v²

56 m²/s / 25 s = v²

v = 1.50 m/s

The speed of the moving walkway is 1.50 m/s

7 0

3 years ago

suppose that the man pictured on the front side is orbiting the earth (mass=5.98 x 10^24kg at a distance of 310 miles above the

wlad13 [49]

Answer:

a = 9.81[m/s^2]; v = 18683.5[m/s]

Explanation:

The stament of the problem is:

Suppose that the man pictured on the front side is orbiting the earth (mass = 5.98 x 1024kg) at a distance of 310 miles (1600 meters = 1 mile) above the surface of the earth (radius = 4000 miles).

a. What acceleration does he experience due to the earth's pull?

b. What tangential velocity must he possess in order that he orbit safely (in m/s)?

First we need to convert all the initial data to units of the SI

Rs = 310 [mil] = 498897 [m]

RT= 400 [mil] = 643738 [m]

r = Rs + RT = 1142635 [m] "distance from the center of the earth to the man"

$F=G*\frac{M*m}{r^{2} } \\where:\\$

G = universal gravitational constant

M = mass of the earth [kg]

m = mass of the man [kg]

r = distance from the center of the earth to the man [m]

The acceleration he is experimenting is the same acceleration given by the gravity, therefore:

a = g = 9.81[m/s^2]

To find the tangential velocity, we must determinate the force exerted by the earth.

Now we will find the force exerted by the gravity when the man is orbiting the earth at distance r.

$G = 6.67*10^{-11} [\frac{N*m^{2} }{kg^{2} } ]\\M=5.98*10^{24}[kg]\\ m=100 [kg]\\Replacing:\\F = G*\frac{M*m}{r^{2} }$

$F = 6.67*10^{-11}*\frac{5.98*10^{24}*100 }{1142635^{2} } \\ F= 30550 [N]$

And this force will be equal to the following expression:

$F = m*\frac{v^{2} }{r} \\where:\\v= tangential velocity [m/s]\\v=\sqrt{\frac{F*r}{m} } \\v=\sqrt{\frac{30550*1142635}{100} } \\v=18683.5[m/s]$

8 0

3 years ago