Ask question

Ask question

All categories

3 years ago

11

A 9800-N automobile is pushed along a level road by four students who apply a total forward force of 600 N. Neglecting friction,

the acceleration of the automobile is: a) 0.0 m /s^2 b) 0.54 m/s^2 c) 0.6 m/s^2 d) 9.8 m/s^2

1 answer:

Alexandra [31]3 years ago

4 0

Answer:

c). $a = 0.60 m/s^2$

Explanation:

As we know that weight of the automobile is given here

so weight = mass times gravity

$W = mg$

$9800 = m(9.8)$

$m = 1000 kg$

now from Newton's law

$F = ma$

$600 = 1000 a$

$a = \frac{600}{1000}$

$a = 0.60 m/s^2$

You might be interested in

You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength

Ray Of Light [21]

Answer:

wavelength $\lambda$ = 437.27 nm

Explanation:

given data

first bright fringe = 2.96 mm

slit separation = 0.325 mm

distance D = 2.20 m

solution

we know that this is double slit experiment

so we apply here Fringe width formula that is

β = $\frac{D\lambda}{d}$ ....................1

$\lambda$ is Wavelength of light and D is Distance between screen and slit and d is slit width

so put here value and we get

$\lambda$ = $\frac{2.96*0.325*10^{-6}}{2.20}$

$\lambda$ = 437.27 × $10^{-9}$ m

wavelength $\lambda$ = 437.27 nm

4 0

3 years ago

You need a 450 microgram sample of gold, but you only have a mass balance that measures in decigrams. Convert the amount of gold

sveta [45]

The amount of gold in decigrams if 450 micrograms is needed is 4.5 × 10-³ decigrams.

<h3>How to convert micrograms to decigrams?</h3>

According to this question, 450 micrograms of a sample of gold is needed but we only have a mass balance that measures in decigrams.

This means that we are to convert the amount of gold you need to decigrams by comparing the exponents.

The conversion factor of micrograms to decigrams is as follows:

1 micrograms = 1 × 10-⁵ decigrams

This means 450 micrograms is equivalent to 450 × 1 × 10-⁵ = 4.5 × 10-³ decigrams

Therefore, the amount of gold in decigrams if 450 micrograms is needed is 4.5 × 10-³ decigrams.

Learn more about decigrams at: brainly.com/question/6869599

#SPJ1

7 0

1 year ago

Which of the following are ways the rising human population can lower their impact on the Earth's spheres and climate change? Se

Tju [1.3M]

a and c because its right im out of college

3 0

3 years ago

Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W

jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

$T^{2}=\frac{4\pi^{2}}{GM}r^{3}$ (1)

Where;

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24}kg$ is the mass of the Earth

$r$ is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period $T_{X}$ is:

$T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}$ (2)

Where $r_{X}=1.2(10)^{6}m$

$T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}$ (3)

$T_{X}=413.712 s$ (4)

For satellite Y, the orbital period $T_{Y}$ is:

$T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}$ (5)

Where $r_{Y}=1.9(10)^{5}m$

$T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}$ (6)

$T_{Y}=26.064 s$ (7)

This means $T_{X}>T_{Y}$

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

$V_{X}=\sqrt{\frac{GM}{r_{X}}}$ (8)

$V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}$

$V_{X}=18224.783 m/s$ (9)

<u>For Satellite Y:</u>

$V_{Y}=\sqrt{\frac{GM}{r_{Y}}}$ (10)

$V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}$

$V_{Y}= 45801.13 m/s$ (11)

This means $V_{Y}>V_{X}$

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

4 0

3 years ago

Read 2 more answers

Newtons second law lab report link

gulaghasi [49]

Answer:

ghittu iihg उह्स उउह्स उग्य्किव जिक्ह्ब

5 0

2 years ago