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Stels [109]
2 years ago
9

Whitch two options are forms of kinetic energy?

Physics
1 answer:
valentina_108 [34]2 years ago
7 0

Answer:the witch has nothing to do with the problem

Explanation:

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The voltage across the input terminals of a transformer is 120 V. The primary has 25 loops and the secondary has 50 loops. The v
saw5 [17]

Answer:

240 V

Explanation:

Vp = 120 V

Np = 25

Ns = 50

Vs = ?

Vs / Vp = Ns / Np

Vs / 120 = 50 / 25

Vs / 120 = 2

Vs = 240 V

5 0
3 years ago
Adolf and Ed are wearing harnesses and are hanging at rest from the ceiling by means of ropes attached to them. Face to face, th
exis [7]

Answer:

0.78 m

Explanation:

By the conservation of energy, the energy that they gain from potential energy, must be equal to the kinetic energy. So, for Adolf:

Ep = Ek

ma*g*ha = ma*va²/2

Where ma is the mass of Adolf, g is the gravity acceleration (10 m/s²), ha is the height that he reached, and va is the velocity. So:

100*10*0.51 = 100*va²/2

50va² = 510

va² = 10.2

va = √10.2

va = 3.20 m/s

Before the push, both of them are in rest, so the momentum must be 0. The system is conservative, so the momentum after the push must be equal to the momentum before the push:

ma*va + me*ve = 0, where me and ve are the mass and velocity of Ed. So:

100*3.20 + 81ve = 0

81ve = 320

ve = 3.95 m/s

By the conservation of energy for Ed:

me*g*he = me*ve²/2

81*10*he = 81*(3.95)²/2

810he = 631.90

he = 0.78 m

5 0
3 years ago
Claim-Evidence-Reasoning assessment. Scenario - The Sun constantly undergoes explosions called solar eruptions. When a solar eru
SpyIntel [72]

Answer:

b

Explanation:

4 0
3 years ago
In the figure below, a block of 1.67 slides on a track with different levels, which has friction only at the highest point where
Natasha_Volkova [10]

Answer:

2.0 m

Explanation:

Energy is conserved.

Initial KE = Final PE + Work done by friction

½ mv² = mgh + Fd

½ mv² = mgh + mgμd

½ v² = gh + gμd

½ v² − gh = gμd

d = (½ v² − gh) / (gμ)

d = (½ (7.5 m/s)² − (10 m/s²) (2.1 m)) / ((10 m/s²) (0.35))

d = 2.0 m

7 0
3 years ago
A speed bike tops a hill at 3.50 m/s and accelerates steadily down the hill until reaching a speed of 11.4 m/s after 4.20 second
k0ka [10]

Answer:

The bike traveled 31.3 meters during this period

Explanation:

A speed bike tops a hill at 3.50 m/s and accelerates steadily down the

hill until reaching a speed of 11.4 m/s after 4.20 seconds

→ The initial speed u = 3.5 m/s

→ The final speed v = 11.4 m/s

→ The time of acceleration t = 4.2 seconds

→ a = \frac{v-u}{t}

Substitute the values above in the rule to find the acceleration a

→ a = \frac{11.4-3.5}{4.2} = 1.88 m/s²

We need to find the distance that the bike traveled during the

acceleration

→ v² = u² + 2 a s, where s is the distance

→ (11.4)² = (3.5)² + 2 (1.88)(s)

→ 129.96 = 12.25 + 3.76 s

Subtract 12.25 from both sides

→ 117.71 = 3.76 s

Divide both sides by 3.76

→ s = 31.3 meters

<em>The bike traveled 31.3 meters during this period</em>

8 0
3 years ago
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