Calculate the weight of a 4.5 kg rabbit.

A 96-mH solenoid inductor is wound on a form 0.80 m in length and 0.10 m in diameter. A coil is tightly wound around the solenoi

Sholpan [36]

Answer:

$i = 7.83 \mu A$

Explanation:

Induced EMF in the coil is given by the equation

$EMF = M\frac{di}{dt}$

so we have

$M = 31 \mu H$

also we know that rate of change in current in solenoid is given as

$\frac{di}{dt} = 2.5 A/s$

so induced EMF of coil is given as

$EMF = (31 \times 10^{-6})(2.5)$

$EMF = 77.5 \times 10^{-6} A/s$

now induced current in the coil will be given as

$i = \frac{EMF}{R}$

$i = \frac{77.5 \times 10^{-6}}{9.9}$

$i = 7.83 \mu A$

4 0

2 years ago

The drawing shows three identical springs hanging from the ceiling. Nothing is attached to the first spring, whereas a 4.50-N bl

Pavlova-9 [17]

Answer:

a. 30 N / m

b. 9.0 N

Explanation:

Given that

Unstretched length of the spring, $L_o$ = 20.0cm = 0.2m

a) When the mass of 4.5N is hanging from the second spring, then extended length Is

$L_1$ = 35.0cm = 0.35m

So, the change in spring length when mass hangs is

$x = L_1 - L_o$

= (0.35 - 0.20) m

= 0.15m

As spring are identical

Let us assume that the spring constant be "k", so at equilibrium

Restoring Force on spring = Block weightage

kx = W = 4.50

$k= \frac{4.50}{x} = \frac{4.50}{0.15}$

= 30 N / m

b) Now for the third spring, stretched the length of spring is

$L_2$ = 50cm = 0.5m

So, the change in spring length is

$x'= L_2 - L_o$

= (0.5-0.20)m

= 0.30m

At equilibrium,

Restoring Force on spring = Block weightage

Now using all mentioned and computed values in above,

$W'= kx'$

= 30(0.3)

= 9.0 N

4 0

2 years ago

Nc-1 has the same dimension as

lisabon 2012 [21]

Answer:

Explanation:

The answer is electric field intensity. Electric field intensity is the force per unit positive charge which the charge exerts at any point.

8 0

2 years ago

Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hourmiles/hour

slavikrds [6]

Answer:

The acceleration of the cheetahs is 10.1 m/s²

Explanation:

Hi there!

The equation of velocity of an object moving along a straight line with constant acceleration is the following:

v = v0 + a · t

Where:

v = velocity of the object at time t.

v0 = initial velocity.

a = acceleration.

t = time

We know that at t = 2.22 s, v = 50.0 mi/h. The initial velocity, v0, is zero.

Let's convert mi/h into m/s:

50.0 mi/h · (1609.3 m / 1 mi) · (1 h / 3600 s) = 22.4 m/s

Then, using the equation:

v = v0 + a · t

22.4 m/s = 0 m/s + a · 2.22 s

Solving for a:

22.4 m/s / 2.22 s = a

a = 10.1 m/s²

The acceleration of the cheetahs is 10.1 m/s²

5 0

2 years ago

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of

adoni [48]

Answer:

3311N

Explanation:

r = radius = 600m

V = speed = 150m/s

Mass = weight = 70kg

The weight of pilot when calculated due to circular motion

W = tv

Fv = mv²/r

Fv = 70x150²/600

Fv = 79x22500/600

= 15750000/600

= 2625N

Real Weight of the pilot = m x g

= 70 x 9.8

= 686N

The apparent Weight is calculated by

Mv²/r + mg

= 2625N + 686N

= 3311 N

Therefore the apparent Weight is 3311N

6 0

1 year ago