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3 years ago

Which term describes the image of an object located between a concave mirror and its focal point?

Physics

1 answer:

labwork [276]3 years ago

6 0

Answer: I think it is b.

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How much force does it take to bring a 1,375 N car from rest to a velocity of 26 m/s in 6 seconds?

V125BC [204]

Explanation:

<u>Mass of car</u> = 137.5 kg

<u>Acceleration</u> = v - u / t = 26 - 0 / 6 = 4.33 m/sec^2

Force = m * a = 137.5 * 4.33 = 595.3 N

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3 years ago

A copper block rests 30.0 cm from the center of a steel turntable. The coefficient of static friction between the block and the

PIT_PIT [208]

Answer:

refer to the above attachment

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2 years ago

PLS HELPPP! Why are ions harmful and why are they not harmful?

Fittoniya [83]

Answer:

This is because we are surrounded by positive ions from electromagnetic fields generated by computers, cell phones, and other electronic devices which can impair brain function and suppress the immune system causing symptoms such as: anxiety, breathing difficulty, fatigue, headaches, irritability, lack of energy, poor concentration, nausea, and vertigo,

Explanation:

5 0

3 years ago

Read 2 more answers

The plane of a rectangular coil, 7.2 cm by 3.7 cm, is perpendicular to the direction of a uniform magnetic field B. If the coil

netineya [11]

Answer:

The rate of change of magnetic field is 2.23 T/s.

Explanation:

Given that,

Dimension of rectangular coil is 7.2 cm by 3.7 cm.

Number of turns in the coil, N = 104

Resistance of the coil, R = 12.4 ohms

Current, I = 0.05 A

We need to find the rate of change of magnetic field in the coil. The induced emf is given by the rate of change of magnetic flux. So,

$\epsilon=-\dfrac{d\phi}{dt}$

Ohm's law is :

$\epsilon=IR$

So,

$IR=-\dfrac{d\phi}{dt}\\\\IR=-\dfrac{d(NBA)}{dt}\\\\IR=-NA\dfrac{dB}{dt}\\\\\dfrac{dB}{dt}=\dfrac{IR}{NA}\\\\\dfrac{dB}{dt}=\dfrac{0.05\times 12.4}{104\times 7.2\times 10^{-2}\times 3.7\times 10^{-2}}\\\\\dfrac{dB}{dt}=2.23\ T/s$

So, the rate of change of magnetic field is 2.23 T/s.

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3 years ago

Evaluate the gravitational potential energy between two 5.00-kg spherical steel balls separated by a center-tocenter distance of

navik [9.2K]

Answer:

U = 8.30×10-⁹J

Explanation:

m1 = m2 = 5.00kg masses of the spheres

d = 15.0cm = 15×10-²m

r = 5.10cm = 5.10×10-²m

R = d + r = 15×10-² + 5.10×10-²

R = 20.10 ×10-²m = 0.201m

G = 6.67×10-¹¹Nm²/kg²

U = Gm1×m2/R = potential energybetween the spheres

U = 6.67×10-¹¹×5.00×5.00/0.201

U = 8.30×10-⁹J

7 0

3 years ago