Ask question

Ask question

All categories

3 years ago

10

A 2.1 times 103 - kg car starts from rest at the top of a 5.0 - m - long driveway that is inclined at 20 deg with the horizontal

. If an average friction force of 4.0 times 103 N impedes the motion, find the speed of the car at the bottom of the driveway.

1 answer:

VLD [36.1K]3 years ago

4 0

Answer:

speed of the car at the bottom of the driveway is 3.8 m/s

Explanation:

given data

mass = 2.1× 10³ kg

distance = 5 m

angle = 20 degree

average friction force = 4 × 10³ N

to find out

find the speed of the car at the bottom of the driveway

solution

we find acceleration a by force equation that is

force = mg×sin20 - friction force

ma = mg×sin20 - friction force

put here value

2100a = 2100 ( 9.8)×sin20 - 4000

a = 1.447 m/s²

so from motion of equation

v²-u² = 2as

here u is 0 by initial speed and v is velocity and a is acceleration and s is distance

v²-0 = 2(1.447)(5)

v = 3.8

speed of the car at the bottom of the driveway is 3.8 m/s

You might be interested in

About how much of the United States' electricity is produced by nuclear reactors?

Firdavs [7]

Answer:

19% total electrical output

Explanation:

3 0

2 years ago

Read 2 more answers

This is my question. I need the answer stat.

Kipish [7]

Answer:

Explanatioyour answers look right, but if there has , has to be another answer its a , but your answers are right

5 0

2 years ago

Read 2 more answers

Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti

Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

$c = \lambda \cdot \nu$ (1)

Where c is the speed of light, $\lambda$ is the wavelength and $\nu$ is the frequency.

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, $\lambda$ can be isolated from equation 1:

$\lambda = \frac{c}{\nu}$ (2)

since the value of c is $3x10^{8}m/s$ . It is necessary to express the frequency in units of hertz.

$\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz}$ ⇒ $100200000Hz$

But $1Hz = s^{-1}$

$\nu = 100200000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}$

$\lambda = 2.99 m$

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

$\nu = 1070kHz . \frac{1000Hz}{1kHz}$ ⇒ $1070000Hz$

But $1Hz = s^{-1}$

$\nu = 1070000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}$

$\lambda = 280.3 m$

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

$\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz}$ ⇒ $835600000Hz$

But $1Hz = s^{-1}$

$\nu = 835600000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}$

$\lambda = 0.35 m$

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

6 0

3 years ago

Light is emitted by electrons when they drop from one energy level to a lower level. Which transition results in the emission of

natulia [17]

Answer:

Level 4 to level 2

Explanation:

Electrons in an atom are contained in specific energy levels (1, 2, 3, and so on) having different distances from the nucleus. When light is emitted by electrons from one energy level to a lower level, level 4 to level 2 has the greatest energy.

Hence, the correct option is "Level 4 to level 2".

3 0

3 years ago

A ball of mass m, attached to the end of a horizontal cord, is rotated in a circle of radius r on a frictionless horizontal surf

kirza4 [7]

Answer: $v=\sqrt{\frac{FL}{m}}$

Explanation:

Given

Ball of mass m

maximum Bearable Tension in string is F

Let length of the cord be L m and moving at a speed of v m/s

Here Tension will Provide Centripetal Force

T=Centripetal Force

$F=T=\frac{mv^2}{L}$

$v=\sqrt{\frac{FL}{m}}$

8 0

3 years ago