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max2010maxim [7]
4 years ago
10

A group of students decides to set up an experiment in which they will measure the specific heat of a small amount of metal. The

metal has a mass of about 5 grams.
They hang the metal in a beaker of boiling water for a long time (10 minutes or so). Then, they very quickly (within a few seconds) remove the metal from the boiling water and transfer it to a styrofoam cup of 150 mL of water at room temperature. There is a thermometer in the styrofoam cup. They know that the rise in temperature will tell them what they need to know in order to determine the specific heat of the metal, so they watch the thermometer closely... but nothing happens. The temperature does not appear to change at all.

Each student has a different suggestion for how to improve the experiment. Which of the suggestions is least likely to help?

A. Use less room temperature water in the styrofoam cup
B. Use more metal (50 grams instead of 5 grams)
C. Use more boiling water in the first beaker
D. Use a more sensitive thermometer.
Physics
1 answer:
lesya692 [45]4 years ago
7 0
I think the answer is C
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A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500
defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

M = \frac{\mu_0 N_1 N_2A_1}{l_1}

Where,

\mu =permeability of free space

N_1 = Number of turns in solenoid 1

N_2 = Number of turns in solenoid 2

A_1= Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

\mu_0 = 4\pi *10^{-7}H/m

N_1 = 500

N_2 = 40

A = 7.5*10^{-4}m^2

l = 0.5m

Substituting,

M = \frac{\mu_0 N_1 N_2A_2}{l_1}

M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}

M = 3.77*10^{-4}H

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0
4 years ago
Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the bat
RSB [31]

Answer:

a. Capacitance

b. Charge on the plates  

e. Energy stored in the capacitor

Explanation:

Let A be the area of the capacitor plate

The capacitance of a capacitor is given as;

C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\

where;

V is the potential difference between the plates

The charge on the plates is given as;

Q = \frac{V\epsilon _0 A}{d}

The energy stored in the capacitor is given as;

E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2

Thus, the physical variables listed that will change include;

a. Capacitance

b. Charge on the plates  

e. Energy stored in the capacitor

3 0
3 years ago
Brainliest to first to answer. What is the maximum stress which a material can withstand when it is pulled apart?
USPshnik [31]

Answer:

stress tension tensile strength

Explanation:

The maximum stress which a material can withstand when it is pulled apart is its: stress tension tensile strength.

3 0
4 years ago
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How do you know that forces are balanced when static friction acts on an object?
lyudmila [28]
By looking at the acceleration of the object.
In fact, Netwon's second law states that the resultant of the forces acting on an object is equal to the product between the mass m of the object and its acceleration:
\sum F = ma

So, when static friction is acting on the object, if the object is still not moving we know that all the forces are balanced: in fact, since the object is stationary, its acceleration is zero, and so the resultant of the forces (left term in the formula) must be zero as well (i.e. the forces are balanced).
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3 years ago
A solid, uniform sphere with a mass of 2.5 kg rolls without slipping down an incline plane starting from rest at a vertical heig
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Answer:

1/2 m v^2 + 1/2 I ω^2 = m g h       conservation of energy

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1/2 m v^2 + 1/5 m ω^2 R^2 = m g h

1/2 v^2 + 1/5 v^2 = g h

v^2 = 10 g h / 7 = 1.43 * 9.80 * 19 m^2/s^2 = 266 m^2/s^2

v = 16.3 m/s

v = R ω

ω = 16.3 / .6 = 27.2 / sec

8 0
2 years ago
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