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3 years ago

8

A cylindrical coil of radius 6 cm is placed in magnetic field, which is changing in time with the rate 0.5 T/s. The magnetic fie

ld direction is parallel to the cylinder axis. What should be the number of turns in the coil to induce the emf

1 answer:

Jlenok [28]3 years ago

3 0

induced emf value is missing..

please correct the question

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Which statement is true?

Fynjy0 [20]

If it's Kepler's law of equal areas you're talking about,
then the first of the four statements is true.

4 0

3 years ago

Read 2 more answers

Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke

umka2103 [35]

Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

Thrust = $v_{e} \frac{dM}{dt}$

$v_{f}$ -v₀ = v_{e} $ln ( \frac{M_{o} }{M_{f}} )$

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

The final mass is the mass of the engines + the mass of the rocket

M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

thrust = 5.26 N

t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

thrust = v_{e} $\frac{M_{f} - M_{o} }{t_{f} - t_{o} }$

v_{e} = thrust $\frac{\Delta t}{\Delta M}$

v_{e} = 5.26 $\frac{1.90}{0.080 -0.0927}$

v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

v_{f}-0 = v_{e} $ln ( \frac{M_{o} }{M_{f} } )$

we calculate

v_{f} = 786.93 ln (0.0927 / 0.080)

v_{f} = 115.95 m / s

5 0

3 years ago

1. What are the units of area under the line/curve?<br> 2. Does the area have any meaning?

TiliK225 [7]

The area under the velocity time graph is 125 m and the meaning of the area is displacement.

<h3>What is area under velocity - time graph?</h3>

The area under a velocity time graph represents the displacement of the object.

total area of the graph = A1 + A2

total area of the graph = ¹/₂ (base₁)(height₁) + ¹/₂ (base₂)(height₂)

total area of the graph = ¹/₂(4)(40) + ¹/₂(3)(30)

total area of the graph = 125 m

Thus, the area under the velocity time graph is 125 m and the meaning of the area is displacement.

Learn more about velocity time graph here: brainly.com/question/4710544

#SPJ1

6 0

1 year ago

1<br> Verify the identity. Show your work.<br><br> cot θ ∙ sec θ = csc θ

Fittoniya [83]

To verify the identity, we can make use of the basic trigonometric identities:
cot θ = cos θ / sin θ
sec θ = 1 / cos <span>θ
csc </span>θ = 1 / sin θ<span>

Using these identities:
</span>cot θ ∙ sec θ = (cos θ / sin θ ) (<span> 1 / cos </span><span>θ)
</span>
We can cancel out cos <span>θ, leaving us with
</span>cot θ ∙ sec θ = 1 / sin θ
cot θ ∙ sec θ = = csc <span>θ</span>

6 0

3 years ago

(3) What is the weight of a 50-kg astronaut (a) on Earth (b) On the Moon ,(g=1.7m/s2), (c) on Mars (g=3.7m/s2) (d)in outer space

artcher [175]

Answer:

a) On Earth

490N

b) On the Moon

85N

c) On Mars

185N

d)in outer space traveling with constant velocity.

0

Explanation:

The weight is defined as:

$W = mg$ (1)

Where m is the mass and g is the gravity

a) On Earth $g = 9.8m/s^{2}$

Then, equation 1 can be used:

$W = (50Kg)(9.8m/s^{2})$

$W = 490Kg.m/s^{2}$

but 1N = Kg.m/s^{2}

$W = 490N$

Hence, the weight of the astronaut on Earth is $490N$

b) On the Moon $g = 1.7m/s^{2}$

$W = (50Kg)(1.7m/s^{2})$

$W = 85N$

Hence, the weight of the astronaut on the Moon is $85N$

c) On Mars $g = 3.7m/s^{2}$

$W = (50Kg)(3.7m/s^{2})$

$W = 185N$

Hence, the weight of the astronaut on Mars is $185N$

(d) in outer space traveling with constant velocity.

Tanking into consideration that the astronaut is traveling in outer space at a constant velocity, it can be concluded that the acceleration will be zero.

Remember that the acceleration is defined as:

$a = \frac{v_{f} - v_{i}}{t}$

Since the acceleration is the variation of the velocity in a unit of time.

Therefore, from equation 1 is gotten.

$W = (50kg)(0)$

Remember that g is the acceleration that a body experience as a consequence of the gravitational field.

$W = 0$

5 0

2 years ago