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castortr0y [4]
3 years ago
8

1. A 1.5 kg ball moves in a circle that is 0.5 m in radius at a speed of 5.1 m/s.

Physics
1 answer:
kolezko [41]3 years ago
3 0

Answer: a= 52.02 m/s²

Fc= 78.03 N

Explanation: Solution attached:

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E20E20: The word “virtual” refers to something that exists in effect but not in actual fact. How does this definition relate to
Lubov Fominskaja [6]

The correct answer (sample response) is:

The image seems to be behind the mirror, but nothing is really there.

Include the following in your response:

The image appears to be behind the mirror.

If someone looks behind the mirror, there is no image there.

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5 0
3 years ago
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A bar magnet is dropped toward a conducting ring lying on the floor. As the magnet falls toward the ring, does it move as a free
REY [17]

Explanation:

According to the Faraday-Lenz law, a conductive ring generates an induced current due to the change in the magnetic flux caused by the motion of the bar magnet. This induced current generates a magnetic field opposite to the magnetic field of the bar, generating an upward force that opposes the weight of the bar magnet, Therefore, it does not move as a freely  falling object.

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3 years ago
A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k = 160 W/m-K). The fin diameter is D = 4 mm, and
frozen [14]

Answer:

Given that

D= 4 mm

K = 160 W/m-K

h=h = 220 W/m²-K

ηf = 0.65

We know that

m=\sqrt{\dfrac{hP}{KA}}

For circular fin

m=\sqrt{\dfrac{4h}{KD}}

m=\sqrt{\dfrac{4\times 220}{160\times 0.004}}

m = 37.08

\eta_f=\dfrac{tanhmL}{mL}

0.65=\dfrac{tanh37.08L}{37.08L}

By solving above equation we get

L= 36.18 mm

The effectiveness for circular fin given as

\varepsilon =\dfrac{2\ tanhmL}{\sqrt{\dfrac{hD}{K}}}

\varepsilon =\dfrac{2\ tanh(37.08\times 0.03618)}{\sqrt{\dfrac{220\times 0.004}{160}}}

ε = 23.52

5 0
3 years ago
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If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
zvonat [6]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

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H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

3 0
2 years ago
Give an equation for kinetic energy<br>(supplement)​
Leya [2.2K]

Answer:

The equatiom of kinetic energy is 1/2×m×v² where m represents mass and v is velocity :

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The unit for kinetic energy is measured in Joules (J).

6 0
3 years ago
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