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3 years ago

6. An unknown material has a normal melting/freezing point of 225.0 °C, and the liquid

Physics

1 answer:

kkurt [141]3 years ago

7 0

Answer:

19.4 kJ/kg

Explanation:

I'm assuming you meant that the melting/freezing point is -25.0 °C.

Heat gained by the unknown material = heat lost by the oil and aluminum

mL + mCΔT = -(mCΔT + mCΔT)

First, let's find the amount of heat gained by the unknown material:

(0.100 kg) L + (0.100 kg) (160 J/kg/°C) (20.0°C − (-25.0°C))

(0.100 kg) L + 720 J

The heat lost by the oil:

(0.100 kg) (2430 J/kg/°C) (20.0°C − 27.0°C)

-1701 J

The heat lost by the aluminum:

(0.150 kg) (910 J/kg/°C) (20.0°C − 27.0°C)

-955.5 J

Therefore:

(0.100 kg) L + 720 J = -(-1701 J + (-955.5 J))

(0.100 kg) L + 720 J = 2656.5 J

(0.100 kg) L = 1936.5 J

L = 19365 J/kg

L = 19.4 kJ/kg

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a ball of mass 0.80 kg moving at a speed of 2.5 m/s along a straight line collided with a mass 2.5 kg which was initially statio

Likurg_2 [28]

The speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

The given parameters;

mass of the ball, m₁ = 0.8 kg
speed of the ball, u₁ = 2.5 m/s
mass of the object at rest, m₂ = 2.5 kg
final velocity of the object at rest, v₂ = 1 m/s

Let the final velocity of the 0.8 kg ball immediately after collision = v₁

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.8 x 2.5) + (2.5 x 0) = (0.8)v₁ + 2.5(1)

2 = 2.5 + (0.8)v₁

-0.5 = (0.8)v₁

$v_1 = \frac{-0.5}{0.8} \\\\v_1 = -0.625 \ m/s$

Thus, the speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

Learn more here: brainly.com/question/7694106

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2 years ago

A quarterback back pedals 3.3 meters southward and then runs 5.7 meters northward. For this motion, what is the distance moved?

mixas84 [53]

Answer:

The distance moved is 9 meters

The magnitude of the displacement is 2.4 meters

The direction of the displacement is northward

Explanation:

- Distance is the length of the actual path between the initial and the

final position. Distance is a scalar quantity

- Displacement is the change in position, measuring from its starting

position to the final position. Displacement is a vector quantity

The quarterback pedals 3.3 meters southward

That means it moves down 3.3 meters

Then runs 5.7 meters northward

That means it runs up 5.7 meters

The distance = 3.3 + 5.7 = 9 meters

The distance moved is 9 meters

It moves southward (down) for 3.3 meters and then moves northward

(up) for 5.7

It moves from zero to 3.3 down and then moves up to 5.7

The displacement = 5.7 - 3.3 = 2.4 meters northward

The magnitude of the displacement is 2.4 meters

The direction of the displacement is northward

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3 years ago

Answer the following question according to formal grammatical rules.

VARVARA [1.3K]

Answer:

Explanation:

just trying to follow basic grammar.

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3 years ago

Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

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$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

now, Force does the ring push on her at the top

$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

$N = 58\times (\dfrac{11.69^2}{8}- 9.8)$

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The force exerted by the ring to push her is equal to 422.36 N.

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3 years ago

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Luden [163]

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3 years ago