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3 years ago

1. A group of students were trying to find the greatest

Physics

1 answer:

deff fn [24]3 years ago

8 0

Answer:

D th

Explanation:

D B. The force applied to the ball is a balanced force.

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The volume of the moon is about 33% of the volume of the earth.true or false

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The moons volume is that of 2 percent of the earth.

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3 years ago

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Consider what will happen when a bar magnet is pushed toward the coil. when the coil "feels" the changing magnetic field caused

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The question just basically explained what happens

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4 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

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3 years ago

What causes charges to move in a circuit?

ipn [44]

A )voltage
(i.e differance in electricity field)

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3 years ago

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6. A .25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target. a. What is the final

Alenkasestr [34]

Answer:

(a) the final velocity of the combined mass is 9.43 m/s

(b) the decrease in kinetic energy during the collision is 386.1 J

Explanation:

Given;

mass of arrow, m₁ = 25 kg

initial velocity of arrow, u₁ = 12 m/s

mass of target, m₂ = 6.8 kg

initial velocity of the target, u₂ = 0

Part (a)

From the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final velocity of the combined mass

25 x 12 + 0 = v(25 + 6.8)

300 = v(31.8)

v = 300/31.8

v = 9.43 m/s

Part(b)

Kinetic Energy, K.E = ¹/₂mv²

Initial kinetic energy = ¹/₂m₁u₁² + ¹/₂m₂u₂² = ¹/₂ x 25 x (12)² + 0 = 1800 J

Final kinetic energy = ¹/₂m₁v² + ¹/₂m₂v² = ¹/₂v²(m₁ + m₂)

= ¹/₂ x (9.43)²(25+6.8)

= 1413.91 J

Decrease in kinetic energy = Initial K.E - Final K.E

Decrease in kinetic energy = 1800J - 1413.91 J = 386.1 J

4 0

3 years ago