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OLga [1]
3 years ago
9

A long solenoid has a diameter of 12.0 cm. When a current i exists in its windings, a uniform magnetic field of magnitude B = 30

.0 mT is produced in its interior. By decreasing i, the field is caused to decrease at the rate of 6.50 mT/s. Calculate the magnitude of the induced electric field
(a) 4.20 cm and
(b) 10.3 cm from the axis of the solenoid.

Physics
1 answer:
goldfiish [28.3K]3 years ago
4 0

Answer:

(A) 5.8×10-⁴V/m

(B) 1.13×10-⁴V/m

Explanation:

Please see attachment below.

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10. According to Newton's First Lawy , if a box is pushed with no external resistance, what will the action of the box be ?
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Explanation:

According to Newton's First Law of motion, if a box is pushed with no external resistance, the box will keep on moving due to the absence of external force. It might gets stopped due to frictional force that is acting between the surface and the ball. The first law of motion is also known as law of inertia. the magnitude of force acting on the object is given by second law of motion.

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A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill. there is no friction, what is
HACTEHA [7]

Answer:

the maximum vertical height the person in the cart can reach is 18.42 m

Explanation:

Given;

mass of the person in cart, m₁ = 45 kg

mass of the cart, m₂ = 43 kg

acceleration due to gravity, g = 9.8 m/s²

final speed of the cart before it goes up the hill, v = 19 m/s

Apply the principle of conservation of energy;

mgh_{max} = \frac{1}{2}mv^2_{max}\\\\ gh_{max} = \frac{1}{2}v^2_{max}\\\\h_{max} = \frac{v^2_{max}}{2g} \\\\h_{max} =\frac{(19)^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m

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5 0
3 years ago
When the spring, with the attached 275.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the
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Answer:

The period of oscillation is 1.33 sec.

Explanation:

Given that,

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Using formula of angular frequency

\omega=\sqrt{\dfrac{k}{m}}

Where, m = mass

k = spring constant

Put the value into the formula

\omega=\sqrt{\dfrac{6.2}{275.0\times10^{-3}}}

\omega=4.74\ rad/s

We need to calculate the period of oscillation,

Using formula of time period

T=\dfrac{2\pi}{\omega}

Put the value into the formula

T=\dfrac{2\pi}{4.74}

T=1.33\ sec

Hence, The period of oscillation is 1.33 sec.

4 0
3 years ago
A very long train is rolling at 4 m/s along a straight track. An observer is standing on the ground very dangerously close to th
hichkok12 [17]

Answer:

A. \vec{r}=(6\frac{m}{s})t\ \ \hat{i}

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Explanation:

A. The vectorial equation of the person who is getting closer to the other person is:

\vec{r}=\vec{v}t

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Then, you replace and obtain:

\vec{r}=(6\frac{m}{s})t\ \ \hat{i}

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t=\frac{300m}{6m/s}=50s

4 0
3 years ago
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