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slavikrds [6]
2 years ago
7

Do the data for the first part of the experiment support or refute the first hypothesis? Be sure to explain your answer and incl

ude how the variables changed in the first part of the experiment.
Physics
2 answers:
Sphinxa [80]2 years ago
5 0

The data for the first part of the experiment support the first hypothesis. As the force applied to the cart increased, the acceleration of the cart increased. Since the increase in the applied force caused the increase in the cart's acceleration, force and acceleration are directly proportional to each other, which is in accordance with Newton's second law.

klio [65]1 year ago
3 0

no experiment shown here ?

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As the external magnetic field decreases, an induced current flows in the coil. what is the direction of the induced magnetic fi
GrogVix [38]
As the external magnetic field decreases, an induced current flows in the coil. The direction of the induced magnetic field would be pointing to the screen. The flux through the coil is said to decrease. In order to counter this change, the coil would generate or produce a magnetic field that is induced that would be pointing to the same direction as the external field that is flowing which is into the the screen. This is according to Lenz's law or the right hand rule. It states that an induced current in a circuit that is due to the change or motion in   magnetic field should be directed opposing to the change in the flux.
4 0
1 year ago
8) A flat circular loop having one turn and radius 5.0 cm is positioned with its plane perpendicular to a uniform 0.60-T magneti
Marrrta [24]

Answer:

EMF induced in the loop is 9.4 V

Explanation:

As we know that initial magnetic flux of the loop is given as

\phi_1 = B.A

\phi_1 = (0.60)(\pi (0.05)^2)

\phi_1 = 4.7 \times 10^{-3} Wb

As soon as the area of the loop becomes zero the final magnetic flux of the loop is ZERO

Now as per faraday's law of electromagnetic induction the EMF is induced due to rate of change in magnetic flux

so we will have

EMF = \frac{\Delta \phi}{\Delta t}

so we will have

EMF = \frac{4.7 \times 10^{-3} - 0}{0.50 \times 10^{-3}}

EMF = 9.4 V

7 0
2 years ago
You then measure Polly's internal temperature to be 13oC, which is quite a drop from the normal human body temperature of 37oC.
ankoles [38]

Answer:

The specific heat is 3.47222 J/kg°C.

Explanation:

Given that,

Temperature = 13°C

Temperature = 37°C

Mass = 60 Kg

Energy = 5000 J

We need to calculate the specific heat

Using formula of energy

Q= mc\Delta T

c =\dfrac{Q}{m\Delta T}

Put the value into the formula

c=\dfrac{5000}{60\times(37-13)}

c=3.47222\ J/kg^{\circ}C

Hence, The specific heat is 3.47222 J/kg°C.

5 0
1 year ago
Any person who opens the door he applies​
miskamm [114]

Answer:

any person who opens the door he applies pulling force

5 0
1 year ago
Two objects are maintained at constant temperatures, one hot and one cold. Two identical bars can be attached end to end, as in
irina1246 [14]

Answer:

Qa/Qb = k/2×2k = 1/4

Explanation:

a situation is series and b situation is parallel.

let conductance of each plate = k

so net conductance in series = k/2

net conductance in parallel = 2k

so Qa/Qb = k/2×2k = 1/4

7 0
2 years ago
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