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vodka [1.7K]
3 years ago
10

Please help!!! Question is posted as a pic! Thank you

Physics
2 answers:
avanturin [10]3 years ago
6 0

The answer would be 30 J

stepan [7]3 years ago
4 0
The correct answer is 30 j





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Two samples of water are mixed together.
horrorfan [7]

Let the cold water go up x degrees.

Let the hot water go down 100 - x degrees.

The formula for heat exchange is m*c*delta t

Givens

Ice

deltat = x

m = 0.50 kg

c = 4.18

Hot water

deltat = 100 - x

m = 1.5 kg

c = 4.18

Formula

The heat up = heat down

0.50 * c * x = 1.5 * c * (100 - x)            Divide both sides by c

Solution

0.50 *x = 1.5*(100 - x)                          Remove the brackets.

0.5x = 150 - 1.5x                                  Add 1.5x to both sides.

0.5x + 1.5x = 150 - 1.5x + 1.5x             Combine like terms  

2x = 150                                               Divide by 2

x = 75

Answer

A

6 0
3 years ago
Find the potential energy of a 2 kg ball 15 m in the air.
mart [117]

Answer:

294.3 Joules

Explanation:

2kg*9.81m/s^2*Δ15=294.3J

8 0
3 years ago
1. A woman driving in a car at 21 m/s pushes down on the gas pedal. 6 seconds later, she is moving at 36 m/s. What was her avera
Furkat [3]

a=Vf-Vi/t=36-21/6=15/6=2.5m/s^2

5 0
3 years ago
PLEASE HELP!!!! The displacement vectors A and B, when added together, give the resultant vector R, so that R = A + B. Use the d
GuDViN [60]

Answer:

Ax = 0

Ay = 6 m

Bx = 8 cos phi = cos 34 = 6.63 m

By = 8 sin phi = 8 sin (-34) = -4.47 m

Rx = Ax + Bx = 0 + 6.63 = 6.63 m

Ry = Ay + By = 6 - 4.47 = 1.53 m

R = (6.63^2 + 1.53^2)^1/2 = 6.80 m

tan theta = Ry / Rx = 1.53 / 6.8 = ,225

theta = 12.7 deg

7 0
3 years ago
A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc
Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

s_m = s_c + 58

23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58

4 \Delta t^2 = 58

\Delta t^2 = 14.5

\Delta t = \sqrt{14.5} = 3.807s

(b)

s_m = 23 \Delta t + 4 \Delta t^2

s_m = 23*3.807 + 58 = 145.581 m

5 0
2 years ago
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