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jarptica [38.1K]
3 years ago
6

What temperature will 1L of H20 at 200°F become when a piece of copper,0.25kg at 260.928K, comes into contact with water?

Physics
1 answer:
Lady_Fox [76]3 years ago
7 0

Answer:

Explanation:

mass of 1 L water = 1 kg .

200⁰F = (200 - 32) x 5 / 9 = 93.33⁰C .

260.928 K = 260.928 - 273 = - 12.072⁰C .

water is at higher temperature .

Let the equilibrium temperature be t .

Heat lost by water = mass x specific heat x  fall of temperature

= 1 x 4.2 x 10³ x ( 93.33 - t )

Heat gained by copper

= .25 x .385 x 10³ x ( t +  12.072 )

Heat lost = heat gained

1 x 4.2 x 10³ x ( 93.33 - t ) = .25 x .385 x 10³ x ( t +  12.072 )

93.33 - t = .0229 ( t + 12.072)

93.33 - t = .0229 t + .276

93.054 = 1.0229 t

t = 90.97⁰C .

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Explanation:

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2 years ago
A ball is moving with velocity 5 m/s in a direction which makes an angle of 30° with horizontal (i.e. with positive x-direction)
klasskru [66]

Answer:

Explanation:

(a)

From the given information:

The initial velocity v_1 = 5 m/s

The direction of the angle θ = 30°

Therefore, the component along the x-axis = v_1 \ cos \ \theta

v_{1 \ x } = 5 \ cos \ 30^0

v_{1x} = 4.33 \ m/s

The component along the y-axis = v_2 { \ sin \  \theta}

v_{1 \ y } = 5 \ sin \ 30^0

v_{1 \ y } = 2.5 \ m/s

To find the final velocity( reflected velocity)

using the same magnitude v_2 = 5 \ m/s

The angle from the x-axis can be \theta_r = 90^0+60^0

= 150°

Thus, the component along the x-axis = v_2  \ cos \theta _r

v_{2x} = - 0.433 \ m/s

The component along the y-axis = v_2 \ sin \theta_r

v_{2y} = 5 \ sin \ 150^0

v_{2y} = 2.5 \ m/s

(b)

The velocity v_1 can be written as in vector form.

v_1 ^{\to}  = v_1 x \hat {i} + v_1 y \hat {j}

v_1 ^{\to}  =4.33 \ \hat {i} +  2.5 \  \hat {j}  ----  (1)

The reflected velocity in vector form can be computed as:

v_2 ^{\to}  = v_2 x \hat {i} + v_2 y \hat {j}

v_2 ^{\to}  =-4.33 \ \hat {i} +  2.5 \  \hat {j}  --- (2)

The change in velocity = v_2 ^{\to} - v_1 ^{\to}

\Delta v ^{\to} = - 4.33 \hat i + 2.5 \hat j - 4.33 \hat i - 2.5 \hat j

\Delta v ^{\to} = - 8.66 \hat  { i }

(c)

The magnitude of change in velocity = | \Delta V |

| \Delta V | = 8.66 m/s

6 0
3 years ago
A car is traveling at 21.0 m/s. It slows to a stop at a constant rate over 5.00s. How far does the car travel during those 5.00
Doss [256]

Answer:

d = 105 m

Explanation:

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We need to find the distance traveled by the dar during those 5 s before it stops. Let the distance is d. It can be calculated as :

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5 0
3 years ago
Consider a rigid body that rotates but whose center of mass is at rest. a. Trueb. false: the rotational kinetic energy of the en
Fiesta28 [93]

Answer:

Explanation:

True

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The velocity of each particle is v = r ω

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IrinaVladis [17]

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4 0
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