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3 years ago

Pls do question 1 part d). tysm

Physics

1 answer:

Aneli [31]3 years ago

5 0

Answer:

Explanation:

cuz its informing the length of 5 and weight on 20N

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A 40.0 turn coil of wire of radius 3.0 cm is placed between the poles of an electromagnet. The field increases from 0 to 0.75 T

nadezda [96]

Answer:

Induced emf, $V=3.76\times 10^{-4}\ V$

Explanation:

We have,

Number of turns in the coil, N = 40

Radius of coil, r = 3 cm = 0.03 m

The field increases from 0 to 0.75 T at a constant rate in a time interval of 225 s.

It is required to find the magnitude of the induced emf in the coil if the field is perpendicular to the plane of the coil. The induced emf is given by :

$\epsilon=\dfrac{d\phi}{dt}$

$\phi=NBA\cos\theta$ , is magnetic flux

$\epsilon=NA\dfrac{dB}{dt}\\\\\epsilon=\dfrac{40\times \pi (0.03)^2\times (0.75-0)}{225}\\\\\epsilon=3.76\times 10^{-4}\ V$

So, the magnitude of induced emf is $3.76\times 10^{-4}\ V$ .

7 0

3 years ago

Can you plz make me brainliast?

bekas [8.4K]

Answer:

I dont know what to answer so im just gonna say ok

8 0

3 years ago

Which type of coolant(s) usually is (are) used to remove heat from a nuclear reactor core?

Ilya [14]

Answer:

(B). Liquid sodium or water

Explanation:

I don't know how to explain this But hope it right!

If I'm wrong I'm sorry

If I'm right Thank you and (brainliest plz)

6 0

3 years ago

A 9.0 kg test rocket is fired vertically from Cape Canaveral. Its fuel gives it a kinetic energy of 1905 J by the time the rocke

mrs_skeptik [129]

Answer:

21.6m

Explanation:

Since the rocket engine burns all the fuel hence the kinetic energy will be converted to potential energy

Potential Energy = mass × acceleration due to gravity × height

Given

PE = 1905J

Mass = 9.0kg

Acceleration due to gravity =9.8m/s²

Required

Height h

Substitute into the formula

1905 = 9(9.8)h

1905 = 88.2h

h =1905/88.3

h = 21.6m

Hence the required height is 21.6m

5 0

3 years ago

A wye-connected load has a voltage of 480v applied to it. What is the voltage drop across each phase

rodikova [14]

Answer:

$Y_A=277.128 \angle 30v$

$Y_B=277.128 \angle (-150)v$

$Y_C=277.128 \angle (90)v$

Explanation:

From the question we are told that

Voltage $V_L_L =480v$

Generally in a case of Y_connection $V_p_ h$ is mathematical represented as

$V_p_h=\frac{V_l_l}{\sqrt{3}} \angle (\phi-30)v$

Generally voltage drop across phase A

$Y_A=\frac{408}{\sqrt{3}} \angle -(0-30)$

$Y_A=277.128 \angle 30v$

Generally voltage drop across phase B

$Y_B=277.128 \angle (-30-120)$

$Y_B=277.128 \angle (-150)v$

Generally voltage drop across phase C

$Y_C=277.128 \angle (-30+120)$

$Y_C=277.128 \angle (90)v$

3 0

3 years ago