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andrew11 [14]
3 years ago
13

If it takes 32,000 joules of heat to warm 750 g of water, what was the temperature change?(the specific heat of water is 4.18 J/

goC)
Thank you!
Physics
1 answer:
Firdavs [7]3 years ago
7 0

Answer:

10.21°C

Explanation:

From the question,we are given;

  • Quantity of heat = 32,000 Joules
  • Mass of water = 750 g
  • Specific heat capacity of water = 4.18 J/g°C

We are required to calculate the change in temperature;

  • We need to know that quantity of heat is calculated by multiplying mass by specific heat then by change in temperature.
  • That is;

Q = m × c × ΔT

Rearranging the formula;

ΔT = (Q ÷ (m × c))

     = 32,000 J ÷ (750× 4.18 J/g°C)

     = 10.21°C

Therefore, the change in temperature  is 10.21°C

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A crane does 300 Joules of work in 6 seconds. How much power was used?
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Answer: P= mad/t or P=w/t so P= 300/6= 50 W

6 0
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Imagine you are standing in the antarctic during the southern hemisphere summer. the sky is clear and blue. there are no clouds
faltersainse [42]

Answer:

The shadow will appear dark

Explanation:

Your shadow is caused by your body blocking incident light rays from reach the ground under the shadow area. The absence of light means there will be no light reflected from the ground making it appear dark.

6 0
3 years ago
The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,
kondaur [170]

Answer:

a= -2\ m/s^2

a=-12.5\ m/s^2

x=2.17 m

x=8.4 m

Explanation:

Given that

v=A+Bt^{-1}

v=2+0.25t^{-1}

To find acceleration :

we know that

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=0-0.5t^{-2}

a=-0.5t^{-2}

Acceleration at t= 2 s

a=-0.5\times 2^{-2}

a= -2\ m/s^2

Acceleration at t= 5 s

a=-0.5\times 5^{-2}

a=-12.5\ m/s^2

We know that

v=\dfrac{dx}{dt}

dx=\left(2+\dfrac{1}{4t}\right)dt

Position at t= 2 s:

\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{2}

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt

x=\left [2t+0.25\ lnt \right ]_{1}^{5}

x=8+0.25 ln5

x=8.4 m

4 0
3 years ago
Suppose that a teacher driving a 1972 LeMans zooms out of a darkened tunnel at 34.5 m/s. He is momentarily blinded by the sunshi
valkas [14]

Answer:

489.19m

Explanation:

To find the minimum distance you first calculate the time in which the teacher stops:

v=v_o-at\\\\t=\frac{v_o-v}{a}=\frac{34.5m/s-0m/s}{2.5m/s^2}=13.8s

however, the reaction of the teacher is 0.31s later, then you use

t=13.8-0.31s=13.49s

during this time the camper has traveled a distance of:

x=vt=(15.1m/s)(13.49s)=203.69m   (1)

Next you calculate the distance that teacher has traveled for 13.6s:

x=x_o+v_ot+\frac{1}{2}at^2\\\\x=0m+34.5m/s(13.49s)+\frac{1}{2}(2.5m/s^2)(13.49s)^2=692.88m  (2)

The minimum distance between the driver and the camper will be the difference between (2) and (1):

x_{min}=692.88m-203.69m=489.19m

5 0
3 years ago
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