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katen-ka-za [31]
4 years ago
9

How many joules of heat are needed to raise the temperature of 50.0 g of aluminum from 10°C to 110°C, if the specific heat of al

uminum is 0.95 J/g°C?
Physics
1 answer:
dusya [7]4 years ago
5 0

Answer:

Heat required to raise the temperature of the aluminium is 4750 J

Explanation:

As we know that the heat energy required to raise the temperature of the aluminium is given as

Q = ms\Delta T

here we know that

m = 50 g

\Delta T = 110 - 10

\Delta T = 100 ^oC

so we have

Q = 50(0.95)(100)

Q = 4750 J

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Explain why science writers show connections between subjects and ideas within a science text.
vova2212 [387]

Answer:

They show connections because actual life problems help the students understand the topic better!! Hope this helps

Explanation:

Brainliest would be appreciated

5 0
3 years ago
Read 2 more answers
A student on a skateboard is moving at a speed of 1.40 m/s at the start of a 2.15 m high and 12.4 m long incline. The total mass
saw5 [17]

Answer:

W = 609.97J

Explanation:

In order to calculate the work done by the student, you take into account that the total work is equal to the change of the kinetic energy of the student, as follow:

W_T=\Delta K\\\\W+W_g-W_f=\frac{1}{2}m(v^2-v_o^2)\\\\W+Mgsin\alpha d-F_fd=\frac{1}{2}m(v^2-v_o^2)         (1)

The work done by the friction force is negative because it is against the motion of the student.

W: work done by the student = ?

Wf: work done by the friction force

Wg: work done by the gravitational force

Ff: total friction force = 41.0N

m: mass of the skateboard = 53.0kg

d: distance traveled by the student

v: final speed of the student = 6.90m/s

vo: initial speed of the student = 1.40m/s

α: angle of the incline

You first calculate the distance d, with the Pythagoras' theorem

d=\sqrt{(2.15m)^2+(12.4m)^2}=12.58m

Furthermore, the angle α is:

\alpha=tan^{-1}(\frac{2.15m}{12.4m})=9.83\°

Then, you solve the equation (1) for W and replace the values of all parameters:

W=\frac{1}{2}m(v^2-v_o^2)+F_fd-Mgsin\alpha d\\\\W=\frac{1}{2}(53.0kg)((6.90m/s)^2-(1.40m/s)^2)+(41.0N)(12.58m)\\-(53.0kg)(9.8m/s^2)sin(9.83\°)(12.58m)\\\\W=609.97J

The work done by the student is 609.97J

6 0
4 years ago
Friction between two flat surfaces is distinguished as being either static friction or ________ friction
cricket20 [7]
Friction between two flat surfaces is distinguished as being kinetic friction.
3 0
3 years ago
Read 2 more answers
Speed<br> =<br> distance = 400m; time = 20s
Natalka [10]
?.............................
6 0
3 years ago
Prof. Marcia Grail, supervillain and superscientist, wishes to spy on her foes (they'll all pay!) from orbit. She wishes to be a
Sloan [31]

Answer:

563.64 m

Explanation:

Given that as per the question

x = 5 cm = 0.05 m

D = 4.2 × 107 m

d = smallest aperture size

As per the situation the solution of the smallest aperture telescope that she can get away with is below :-

We will use Rayleigh's diffraction limit which is

d\frac{x}{D} = 1.22\lambda

The equation will be

d\frac{0.05}{4.2\times 10^7} = 1.22[550\times 10^{-9}]

d = 563.64 m

So, the answer is d = 563.64 m

6 0
3 years ago
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