PHYSICS COLLEGE

Suppose a boat moves at 16.4 m/s relative to the water. If the boat is in a river with the current directed east at 2.70 m/s, what is the boat's speed relative to the ground when it is heading east, with the current, and west, against the current? (Enter your answers in m/s.)

Answers

Answer 1

Answer:

Answer:

with the current: 19.1 m/s eastern

against the current: 13.7 m/s western

Explanation:

The boat speed relative to ground is the sum of the boat speed relative to water + the water speed relative to ground.

Suppose eastern is positive, water speed due east is 2.7m/s.

When the boat is heading east, its speed relative to water is 16.4m/s, its speed relative to ground is 16.4 + 2.7 = 19.1 m/s eastern

When the boat is heading west, its speed relative to water is -16.4m/s, its speed relative to ground is -16.4 + 2.7 = -13.7 m/s western

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A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 V. If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answers

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

The additional resistance is $R_z = 4.4 \Omega$

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

The rate at which internal energy increase in the battery is $Z_1 = 32 W$

The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

So this implies that

The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = (E)/(R_T)$

Substituting values

$4 = (28)/(R_z + 2.6)$

Making $R_z$ the subject of the formula

So $R_z = (28 - 10.4)/(4)$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

$Z_1 = 4^2 * 2$

$Z_1 = 32 W$

The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

The brakes of a car moving at 14m/s are applied, and the car comes to a stop in 4s. (a) What was the cars acceleration? (b) How long would the car take to come to a stop starting from 20m/s with the same acceleration? (c) How long would the car take to slow down from 20m/s to 10m/s with the same acceleration?

Answers

(1) The acceleration of the car will be $a=-3.5(m)/(s^2)$

(2) The time taken $t=5.7s$

(3) The time is taken by the car to slow down from 20m/s to 10m/s $t=2.9s$

What will be the acceleration and time of the car?

(1) The acceleration of the car will be calculated as

$a=(v-u)/(t)$

Here

u= 14 $(m)/(s)$

$a=(0-14)/(4) =-3.5(m)/(s^2)$

(2) The time is taken for the same acceleration to 20 $(m)/(s)$

$a=(v-u)/(t)$

$t=(v-u)/(a)$

u=20 $(m)/(s)$

$t=(0-20)/(-3.5) =5.7s$

(3) The time is taken to slow down from 20m/s to 10m/s with the same acceleration

From same formula

$t=(v-u)/(a)$

v=10 $(m)/(s)$

u=20 $(m)/(s)$

$t=(10-20)/(-3.5) =2.9s$

Thus

(1) The acceleration of the car will be $a=-3.5(m)/(s^2)$

(2) The time taken $t=5.7s$

(3) The time is taken by the car to slow down from 20m/s to 10m/s $t=2.9s$

To know more about the Equation of the motion follow

brainly.com/question/25951773

(a) $-3.5 m/s^2$

The car's acceleration is given by

$a=(v-u)/(t)$

where

v = 0 is the final velocity

u = 14 m/s is the initial velocity

t = 4 s is the time elapsed

Substituting,

$a=(0-14)/(4)=-3.5 m/s^2$

where the negative sign means the car is slowing down.

(b) 5.7 s

We can use again the same equation

$a=(v-u)/(t)$

where in this case we have

$a=-3.5 m/s^2$ is again the acceleration of the car

v = 0 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find the time the car takes to come to a stop:

$t=(v-u)/(a)=(0-20)/(-3.5)=5.7 s$

(c) $2.9 s$

As before, we can use the equation

$a=(v-u)/(t)$

Here we have

$a=-3.5 m/s^2$ is again the acceleration of the car

v = 10 is the final velocity

u = 20 m/s is the initial velocity

Re-arranging the equation and solving for t, we find

$t=(v-u)/(a)=(10-20)/(-3.5)=2.9 s$

Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?

Answers

Answer:

The two waves will add vectorially to produce a small amplitude wave in a valley phase.

Explanation:

The two waves will add vectorially to produce a small amplitude wave in a valley phase. This is because the amplitudes of the waves are slightly different and in opposite directions. When wave 1 cancels out all of wave 2, the resultant wave would be the slight difference between both waves, and it would be in the direction of wave 1 which is a valley phase.

Resonances of the ear canal lead to increased sensitivity of hearing, as we’ve seen. Dogs have a much longer ear canal—5.2 cm—than humans. What are the two lowest frequencies at which dogs have an increase in sensitivity? The speed of sound in the warm air of the ear is 350 m/s.A. 1700 Hz, 3400 Hz
B. 1700 Hz, 5100 Hz
C. 3400 Hz, 6800 Hz
D. 3400 Hz, 10,200 Hz

Answers

Answer:

B. 1700 Hz, 5100 Hz

Explanation:

Parameters given:

Length of ear canal = 5.2cm = 0.052 m

Speed of sound in warm air = 350 m/s

The ear canal is analogous to a tube that has one open end and one closed end. The frequency of standing wave modes in such a tube is given as:

f(m) = m * (v/4L)

Where m is an odd integer;

v = velocity

L = length of the tube

Hence, the two lowest frequencies at which a dog will have increased sensitivity are f(1) and f(3).

f(1) = 1 * [350/(4*0.052)]

f(1) = 1682.69 Hz

Approximately, f(1) = 1700 Hz

f(3) = 3 * [350/(4*0.052)]

f(3) = 5048 Hz

Approximately, f(3) = 5100 Hz

3. A particle of charge +7.5 µC is released from rest at the point x = 60 cm on an x-axis. The particle begins to move due to the presence of a charge ???? that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved 40 cm if a) ???? = +20 µC and b) ???? = −20 µC?

Answers

Answer:

HSBC keen vs kg get it yyyyyuuy

Explanation:

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Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits

Answers

Answer:

The maximum number of bright spot is $n_(max) =5001$

Explanation:

From the question we are told that

The slit distance is $d = 1 \ mm = 0.001 \ m$

The wavelength is $\lambda = 400 \ nm = 400*10^(-9 ) \ m$

Generally the condition for interference is

$n * \lambda = d * sin \theta$

Where n is the number of fringe(bright spots) for the number of bright spots to be maximum $\theta = 90$

=> $sin( 90 )= 1$

$n = (d )/(\lambda )$

substituting values

$n = ( 1 *10^(-3) )/( 400 *10^(-9) )$

$n = 2500$

given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as

$n_(max) = 2 * n + 1$

The 1 here represented the central bright spot

$n_(max) = 2 * 2500 + 1$

$n_(max) =5001$

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Please help really easy

Modern wind turbines are larger than they appear, and despite their apparently lazy motion, the speed of the blades tips can be quite high-many times higher than the wind speed. A turbine has blades 56 m long that spin at 13 rpm .A. At the tip of a blade, what is the speed?B. At the tip of a blade, what is the centripetal acceleration?C. A big dog has a torso that is approximately circular, with a radius of 16cm . At the midpoint of a shake, the dog's fur is moving at a remarkable 2.5m/s .D. What force is required to keep a 10 mg water droplet moving in this circular arc?E. What is the ratio of this force to the weight of a droplet?

One of two nonconducting spherical shells of radius a carries a charge Q uniformly distributed over its surface, the other carries a charge -Q, also uniformly distributed. The spheres are brought together until they touch.A.) What does the electric field look like, both outside and inside the shells?B.) How much work is needed to move them far apart?