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2 years ago

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A bird flying at a height of 12 m doubles its speed as it descends to a height of 6.0 m. The kinetic energy has changed by a fac

tor of : a) 2 b) 4 c) 1 d) 0.25

1 answer:

Alexandra [31]2 years ago

7 0

Answer:

Option d)

Explanation:

As the Kinetic energy of the body ids due to the speed of the body and depends on it, it does not depend on the height of the body.

The kinetic energy is given by:

KE = $\frac{1}{2}mv^{2}$

where

m = mass of the body

v = kinetic energy of the body

Now, as per the question:

Initial velocity is u

and final velocity, v = 2u

Thus

Initial KE = $\frac{1}{2}mu^{2}$ (1)

FInal KE = $\frac{1}{2}mv^{2}$ = $\frac{1}{2}m(2u)^{2}$

FInal KE = 4 $\frac{1}{2}mu^{2}$ (2)

Thus from eqn (1) and (2):

Final KE = 4(Initial KE)

The Kinetic Energy has changed by a factor of 4

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A solenoid that is 35 cm long and contains 450 circular coils 2.0 cm in diameter carries a 1.75-A current. (a) What is the magne

Taya2010 [7]

Answer:

Explanation:

a ) No of turns per metre

n = 450 / .35

= 1285.71

Magnetic field inside the solenoid

B = μ₀ n I

Where I is current

B = 4π x 10⁻⁷ x 1285.71 x 1.75

= 28.26 x 10⁻⁴ T

This is the uniform magnetic field inside the solenoid.

b )

Magnetic field around a very long wire at a distance d is given by the expression

B = ( μ₀ /4π ) X 2I / d

= 10⁻⁷ x 2 x ( 1.75 / .01 )

= .35 x 10⁻⁴ T

In the second case magnetic field is much less. It is due to the fact that in the solenoid magnetic field gets multiplied due to increase in the number of turns. In straight coil this does not happen .

6 0

2 years ago

Read 2 more answers

Chris and Jamie are carrying Wayne on a horizontal stretcher. The uniform stretcher is 2.00 m long and weighs 100 N. Wayne weigh

PIT_PIT [208]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The value is $F_j = 550\ N$

Explanation:

From the question we are told that

The length of the stretcher is $d = 2.0 \ m$

The weight of the stretcher is $W = 100 \ N$

The weight for Wayne is $W_w = 800 \ N$

The distance of center of gravity for Wayne from Chris is $c_w = 75 cm = 0.75 \ m$

Generally taking moment about the first end where Chris is

$F_j * d$ => upward moment

Here $F_j$ is the force applied by Jamie

Generally taking moment about the second end where Jamie is

$W * ( \frac{d}{2} ) + W_w * (d - c_w)$ => downward moment

Generally at equilibrium , the upward moment is equal to the downward moment

$F_j * d = W * ( \frac{d}{2} ) + W_w * (d - c_w)$

=> $F_j * 2 = 100 * ( \frac{ 2}{2} ) + 800 * (2 - 0.75)$

=> $F_j = 550\ N$

3 0

2 years ago

A rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical accel

Vikki [24]

acceleration of rocket is given here as

$a_y = 2.60* t$

now we know that

$\frac{dv}{dt} = 2.60t$

now integrating both sides

$\int dv = \int 2.60t dt$

$v = 2.60\frac{t^2}{2}$

$v = 1.30 t^2$

here since its given that rocket will accelerate for t = 10 s

so here we have

$v = 1.30 * 10^2$

$v = 130 m/s$

so after t = 10 s the speed of rocket will be 130 m/s upwards

5 0

1 year ago

Two small aluminum spheres, each of mass 0.0250 kilograms, areseparated by 80.0 centimeters.

sineoko [7]

Answer:

Total number of electrons

$N = 7.25 \times 10^{24}$

electrons removed from each sphere

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = 7.27 \times 10^{-10}$

Explanation:

As we know that moles is defined as

$n =\frac{mass}{molar mass}$

$n = \frac{0.0250}{0.026982}$

$n = 0.93$

so number of atoms of Al in each sphere is given as

$N = 0.93(6.02 \times 10^{23})$

$N = 5.58 \times 10^{23}$

Now number of electrons in each atom is given as

atomic number = number of electrons in each atom = 13

total number of electrons in each sphere is

$N = 13 \times (5.58 \times 10^{23})$

$N = 7.25 \times 10^{24}$

Also we know that force of attraction between them is given as

$F= \frac{kq_1q_2}{r^2}$

$1.00 \times 10^4 = \frac{(9\times 10^9)q^2}{0.80^2}$

$q = 8.4 \times 10^{-4} C$

now we have

$q = Ne$

$8.4 \times 10^{-4} = N(1.6 \times 10^{-19}$

$N = \frac{(8.4 \times 10^{-4})}{1.6 \times 10^{-19}}$

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = \frac{5.27 \times 10^{15}}{7.25 \times 10^{24}}$

$f = 7.27 \times 10^{-10}$

6 0

1 year ago

the millersburg ferry (m=13000.0 kg loaded) puts its engines in full reverse and stops in 65 seconds. if the speed before brakin

kenny6666 [7]

The braking force is -400 N

Explanation:

We can solve this problem by using the impulse theorem, which states that the impulse applied on the ferry (the product of force and time) is equal to its change in momentum:

$F \Delta t = m(v-u)$

where in this problem, we have:

F is the force applied by the brakes

$\Delta t = 65 s$ is the time interval

m = 13,000 kg is the mass of the ferry

u = 2.0 m/s is the initial velocity

v = 0 is the final velocity

And solving for F, we find the force applied by the brakes:

$F=\frac{m(v-u)}{\Delta t}=\frac{(13000)(0-2.0)}{65}=-400 N$

where the negative sign indicates that the direction is backward.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

4 0

2 years ago