A copper wire has a mass of 29.33 mg/cm and has a length of 2.5 cm.

A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver n

satela [25.4K]

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

c) v0^2/a0

Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

t2 = v0/a0

C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

now distance covered by car before coming to full stop

Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

Now the separation between train and car

= (2v0^2)/a0 - v0^2/a0

= v0^2/a0

8 0

3 years ago

A basketball with a mass of 0.5 kilograms is accelerated at 2

Paul [167]

Answer: 1N

Explanation: its not 0N.

5 0

2 years ago

Sound with a frequency of 1250 Hz leaves a room through a doorway with a width of 1.05 m.At what minimum angle relative to the c

kiruha [24]

Answer:

15.19°, 31.61°, 51.84°

Explanation:

We need to fin the angle for m=1,2,3

We know that the expression for wavelenght is,

$\lambda = \frac{c}{f}$

Substituting,

$\lambda = \frac{344}{1250}$

$\lambda = 0.2752m$

Once we have the wavelenght we can find the angle by the equation of the single slit difraction,

$sin\theta = \frac{m \lambda}{W}$

Where,

W is the width

m is the integer

$\lambda$ the wavelenght

Re-arrange the expression,

$\theta = sin^{-1} \frac{m\lambda}{W}$

For m=1,

$\theta = sin^{-1} \frac{1 (0.2752)}{1.05}= 15.19\°$

For m=2,

$\theta = sin^{-1} \frac{2 (0.2752)}{1.05}= 31.61\°$

For m=3,

$\theta = sin^{-1} \frac{3 (0.2752)}{1.05}= 51.84\°$

<em>The angle of diffraction is directly proportional to the size of the wavelength.</em>

5 0

2 years ago

You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla

lesantik [10]

Answer:

When they are connected in series

The 50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

The power rating of the first bulb is $P_1 = 100 \ W$

The power rating of the second bulb is $P_2 = 50 \ W$

Generally the power rating of the first bulb is mathematically represented as

$P_1 = V^2 R$

Where $V$ is the normal household voltage which is constant for both bulbs

So

$R_1 = \frac{V^2}{P_1 }$

substituting values

$R_1 = \frac{V^2}{100}$

Thus the resistance of the second bulb would be evaluated as

$R_2 = \frac{V^2}{50}$

From the above calculation we see that

$R_2 > R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 1 = I^2_1 R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 2 = I^2_2 R_2$

Now given that they are connected in series which implies that the same current flow through them so

$I_1^2 = I_2^2$

This means that

$P \ \alpha \ R$

So when they are connected in series

$P_2 > P_1$

This means that the 50 W bulb glows more than the 100 \ W bulb

3 0

3 years ago

An alert driver can apply the brakes fully in about 0.5 seconds. How far would the car travel if it

dybincka [34]

Answer:

The car would travel after applying brakes is, d = 14.53 m

Explanation:

Given that,

The time taken to apply brakes fully is, t = 0.5 s

The velocity of the car, v = 29.06 m/s

The distance traveled by the car in 0.5 s, d = ?

The relation between the velocity, displacement, and time is given by the formula

d = v x t m

Substituting the values in the above equation,

d = 29.06 m/s x 0.5 s

= 14.53 m

Therefore, the car would travel after applying brakes is, d = 14.53 m

8 0

3 years ago