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Reptile [31]
2 years ago
8

The specific heat capacity of aluminum is 0.22 cal/g°C. How much energy needs to flow into 20.0 grams of aluminum to change its

temperature by 15°C?
0.017 cal
0.29 cal
66 cal
1363 cal
Physics
2 answers:
Leokris [45]2 years ago
3 0
Its C. 66 cal , that's your answer to your question.
stellarik [79]2 years ago
3 0

Answer: 66 cal

Explanation:

The Heat energy required to flow into a unit mass object to raise its temperature by 1 degree is known as specific heat capacity.

It is given by:

Q = m c ΔT

where, m is the mass, c is the specific heat capacity and ΔT is the change in temperature.

Given:

specific heat capacity of aluminum, c = 0.22 cal/g°C

mass of aluminium, m = 20.0 g  

Change in temperature, ΔT = 15°C

⇒Q = 20.0 g  × 0.22 cal/g°C × 15°C =  66 cal

The amount of energy that needs to flow into 20.0 grams of aluminum to change its temperature by 15°C is 66 cal.

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The answer is -0.5m/s2

Explanation:

I did the test and got it right.

6 0
2 years ago
A laser beam is incident at an angle of 30.2° to the vertical onto a solution of corn syrup in water. (a) If the beam is refract
mote1985 [20]

Answer:

a) n2 = 1.55

b) 408.25 nm

c) 4.74*10^14 Hz

d) 1.93*10^8 m/s

Explanation:

a) To find the index of refraction of the syrup solution you use the Snell's law:

n_1sin\theta_1=n_2sin\theta_2   (1)

n1: index of refraction of air

n2: index of syrup solution

angle1: incidence angle

angle2: refraction angle

You replace the values of the parameter in (1) and calculate n2:

n_2=\frac{n_1sin\theta_1}{sin\theta_2}=\frac{(1)(sin30.2\°)}{sin18.82\°}=1.55

b) To fond the wavelength in the solution you use:

\frac{\lambda_2}{\lambda_1}=\frac{n_1}{n_2}\\\\\lambda_2=\lambda_1\frac{n_1}{n_2}=(632.8nm)\frac{1.00}{1.55}=408.25nm

c) The frequency of the wave in the solution is:

v=\lambda_2 f_2\\\\f_2=\frac{v}{\lambda_2}=\frac{c}{n_2\lambda_2}=\frac{3*10^8m/s}{(1.55)(408.25*10^{-9}m)}=4.74*10^{14}\ Hz

d) The speed in the solution is given by:

v=\frac{c}{n_2}=\frac{3*10^8m/s}{1.55}=1.93*10^8m/s

8 0
3 years ago
If you lift two loads up one story, how much work do you do compared to lifting just one load up one story?
Daniel [21]

Answer:a

Explanation:

We have to lift the load two loads up one story, so energy required is

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Energy E_1=2\times m\times g\times \h

E_1=2mgh

Here energy gained is the potential energy which depends upon the datum(floor).

For lifting one load up one story

Energy required

E_2=mgh

thus E_2 is half of E_1

So option a is correct                                

6 0
3 years ago
A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.79 rad/s . Its total moment of inertia is 1790
Nezavi [6.7K]

Answer:

w_2=0.467rad/s

Explanation:

Four people standing on the ground each of mass and usually this questions have to find the final angular velocity

m_t=4*70kg=280kg

The radius r=4.2/2=2.1m

Angular velocity  w_1=0.79rad/s

The moment of inertia total is I_t=1790 kg/m^2

Momento if inertia

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Angular momentum

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Solve to w2

w_2=\frac{I_1*w_1}{I_t}

w_2=\frac{1790kg*m^2*0.79rad/s}{3024.8kg*m^2}

w_2=0.467rad/s

8 0
3 years ago
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