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3 years ago

A force of 348.0 N is used to push a car 33.4 m horizontally in 3.21 s. Calculate the power developed.

Physics

1 answer:

Alekssandra [29.7K]3 years ago

7 0

Answer:

3620.9 Watts

Explanation:

So first you want to find the work (J) and to find the work you need to multiply force (N) times distance (m).

348.0*33.4=11623.2 J

After that you want to find the power (Watt) and to find the power you need to divide the work (J) by the time (s).

11623.2/3.21=3620.9 Watts

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Falling objects drop with an average acceleration of 9.8 m/s2. An arrow is shot with a velocity of 11.76 m/s straight down from

In-s [12.5K]

Answer:

3.8 secs

Explanation:

Parameters given:

Acceleration due to gravity, g = 9.8 $m/s^2$

Initial velocity, u = 11.76 m/s

Final velocity, v = 49 m/s

Using one of Newton's equations of linear motion, we have that:

$v = u + gt$

where t = time of flight of arrow

The sign is positive because the arrow is moving downward, in the same direction as gravitational force.

Therefore:

$49 = 11.76 + 9.8*t\\\\\\\49 - 11.76 = 9.8t\\\\=> 9.8t = 37.24\\\\\\t = \frac{37.24}{9.8} \\\\\\t = 3.8 secs$

The arrow was in flight for 3.8 secs

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3 years ago

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3 years ago

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Suppose a car manufacturer tested its cars for front-en4 collisions by hauling them up on a crane and dropping then; from a cert

Brrunno [24]

Answer:

Generally from third equation of motion we have that

$v^2 = u^2 + 2a[s_i - s_f]$

Here v is the final speed of the car

u is the initial speed of the car which is zero

$s_i$ is the initial position of the car which is certain height H

$s_i$ is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

$v^2 = 0 + 2g[H - 0]$

=> $v = \sqrt{ 2 g H}$

$H = 9.86 \ m$

Explanation:

Generally from third equation of motion we have that

$v^2 = u^2 + 2a[s_i - s_f]$

Here v is the final speed of the car

u is the initial speed of the car which is zero

$s_i$ is the initial position of the car which is certain height H

$s_i$ is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

$v^2 = 0 + 2g[H - 0]$

=> $v = \sqrt{ 2 g H}$

When $v = 50 \ km/h = \frac{50 *1000}{3600} = 13.9 \ m/s$ we have that

$13.9 = \sqrt{ 2 g H}$

=> $H = \frac{13.9^2}{2 * 9.8}$

=> $H = 9.86 \ m$

6 0

3 years ago

A solenoid 25.0 cmcm long and with a cross-sectional area of 0.550 cm^2 contains 460 turns of wire and carries a current of 90.0

ankoles [38]

Answer:

a. B = 0.20T

b. u = 17230.6 J/m³

c. E = 0.236J

d. L = 5.84*10^-5 H

Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

$B=\frac{\mu_o n i}{L}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

n: turns of the solenoid = 460

L: length of the solenoid = 25.0cm = 0.25m

i: current = 90.0A

You replace the values of the parameters in the equation (1):

$B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T$

The magnetic field in the solenoid is 0.20T

b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:

$u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}$

The energy density is 17230.6 J/m³

c. The total energy contained in the solenoid is:

$E=uV$ (2)

V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

$V=AL$

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

$V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3$

Then, the energy contained in the solenoid is:

$E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J$

The energy contained is 0.236J

d. The inductance of the solenoid is calculated as follow:

$L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H$

The inductance of the solenoid is 5.84*10^-5 H

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3 years ago

Why won’t a magnet help you separate a mixture of salt and water

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Answer: nor are magnetic

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3 years ago

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