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Scrat [10]
3 years ago
8

Which is an example of a non-contact force?

Physics
1 answer:
Arada [10]3 years ago
4 0
Examples of this force include: electricity, magnetism, radio waves, microwaves, infrared, visible light, X-rays and gamma rays.
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Are we the same age as the universe because matter cannot be created nor destroyed
wlad13 [49]
No. We aren't the same age as the universe.
4 0
3 years ago
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Evaluate (x +y)0 for x= -3 and y=5.<br> 0 1<br> 2<br> 01
frosja888 [35]

Answer:2.01201

Explanation:

8 0
3 years ago
A 580-turn solenoid is 18 cm long. The current in it is 36 A. A straight wire cuts through the center of the solenoid, along a 2
Karolina [17]

Answer:

F = 0.078N

Explanation:

In order to calculate the magnitude of the force on the wire you first calculate the magnitude of the magnetic field generated by the solenoid, by using the following formula:

B=\frac{\mu_oNi}{L}         (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

N: turns of the solenoid = 580

i: current in the solenoid = 36A

L: length of the solenoid = 18cm = 0.18m

You replace the values of all parameters in the equation (1):

B=\frac{(4\pi*10^{-7}T/A)(580)(36A)}{0.18m}=0.145T

Next, you calculate the force exerted on the wire, by using the following formula:

F=iLBsin\theta         (2)

i: current in the wire = 27A

L: length of the wire that perceives the magnetic field (the same as the radius of the solenoid) = 2.0 cm = 0.02m

θ: angle between wire and the direction of B

B: magneitc field in the solenoid = 0.145T

The direction of the wire are perpendicular to the direction of the magnetic field, hence, the angle is 90°.

You replace the values of the parameters in the equation (2):

F=(27A)(0.02m)(0.145T)sin90\°=0.078N

The magnitude of the force on the wire is 0.078N

8 0
3 years ago
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A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.
Anastasy [175]

Initially, the velocity vector is \langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by 4.9(0.2)^2, so the velocity is \langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle.

Converting back to direction and magnitude, we get \langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle

4 0
3 years ago
A spring stretches 1.68cm vertically when a 2.50kg object is suspended from it.Find the distance (in cm) the spring stretches if
Llana [10]

Answer:

2.96 cm

Explanation:

By Hook's law

Force(F) = Spring constant(k) × Extension(d)

 F = k × d

Force is the weight of the object, F = W = mg

So we get, mg = kd ⇒ m ∝ d

                                   2.5 ∝ 1.68  --------------(1)

                                   4.4 ∝ d'      --------------(2)

From (1) & (2),     4.4/2.5 = d'/1.68

                                   d' = 2.96 cm ⇒ the required extension.

7 0
3 years ago
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