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o-na [289]
2 years ago
11

You have two beakers of different types of oil, which have different specific heat capacities. You put each beaker in a water ba

th at 50 °C. The water is heated and stays at 50 °C. Which statement is true?
A: If you put the same amount of oil in each beaker, the temperature of the oil in each beaker will be the same at all times.

B:If the mass of oil in the two beakers is the same, the oil with the lower specific heat capacity will heat up more slowly.

C:If the mass of oil in the two beakers is the same, the oil with the lower specific heat capacity will heat up more quickly.

D:The oil with the higher specific heat capacity will heat up more quickly regardless of the mass of oil.
Physics
1 answer:
eimsori [14]2 years ago
8 0

Answer:

C

Explanation:

C:If the mass of oil in the two beakers is the same, the oil with the lower specific heat capacity will heat up more quickly.

The answer is C

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When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6
Stella [2.4K]

Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 = +8.4\mu C

q_2 = +5.6 \mu C

force between two charges is given as

F = 0.66 N

now we have

F = \frac{kq_1q_2}{r^2}

0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}

r = 0.8 m

so it is separated by 80 cm distance

3 0
3 years ago
Read 2 more answers
How do resistors in parallel affect the total resistance?
4vir4ik [10]

Answer:

They're going to increase the total resistance as R_{T} = \sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}

Explanation:

If the resistors are in parallel, the potential difference is the same for each resistor. But the total current is the sum of the currents that pass through each of the resistors. Then

I = I_1 + I_2 + ... + I_N

where

I_i = \frac{V_i}{R_i}

but

V_i = V_j = V for i,j= 1, 2,..., N

so

I = \frac{V}{R_1}+ \frac{V}{R_2} + ... + \frac{V}{R_N} = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)V = \frac{V}{R_T}

where

R_T = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)^{-1} =\sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}

4 0
2 years ago
An 16.00-kg object is suspended by a rope at some height, h if the object has a potential energy of 900.01j what is the objects
qwelly [4]

Answer:

5.740 m

Explanation:

PE = mgh

900.0 J = (16.00 kg) (9.8 m/s²) h

h = 5.740 m

5 0
2 years ago
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The 88-lb force P is applied to the 210-lb crate, which is stationary before the force is applied. Determine the magnitude and d
Marina86 [1]

Answer:

F=-88Ib

Explanation:

From the question we are told that:

Force P=88Ib

Mass of crate M_c=210Ib

Generally the equation for Frictional force F is mathematically given by

Friction\ force (f) = friction\ coefficient\ (u) * Normal\ reaction (N)

F=u*N

with \mu =0.47

F=98.7Ib

Therefore since Static Friction supersedes applied force body remains at rest.

Frictional force =88Ib (negative)

F=-88Ib

5 0
2 years ago
What is the identity of a sample that has a volume of 92.5 cm3 and a mass of 249.8 g?
denpristay [2]
Here, We know, Density = Mass / Volume
Here, mass = 249.8 g
volume = 92.5 cm³

Substitute their values, 
d = 249.8 / 92.5
d = 2.7 g/cm³

In short, Your Answer would be "Aluminum" 

Hope this helps!
3 0
3 years ago
Read 2 more answers
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