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zepelin [54]
2 years ago
11

How do you determine the acceleration of an object?

Physics
2 answers:
Gelneren [198K]2 years ago
8 0

Are you familiar with any basic calculus? If so, we can just look at this derivative and see what's happening with our units here..

a = \frac{dv}{dt}

Here it shows that acceleration is the derivative of velocity with respect to time. In other words, we can say that:

a=\frac{\Delta v}{\Delta t}

We can read that equation as: "acceleration is the change of velocity divided by the change in time (aka the time interval)."

If you're not familiar with calculus, we can use a simple equation of motion:

v_f = v_i+at

where:

vf = final velocity

vi = initial velocity

a = acceleration

t = observed time interval

We can rearrange this equation to find:

a = \frac{v_f - v_i}{t}

This is the same exact thing we wrote before!

Shtirlitz [24]2 years ago
6 0

Answer:the correct option is A

Explanation:

To determine the acceleration of an object, you divide the change in velocity by the time interval

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Near the surface of Earth an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (
pentagon [3]

Answer:

q=2.997\times 10^{-4}C

Sign-Negative

Explanation:

We are given that

Electric field =E=100NC^{-1} (Radially downward)

Acceleration=0.19 ms^{-2}(Upward)

Mass of charge=3 g=3\times 10^{-3}kg

1kg=1000g

We have to find the magnitude and sign of  charge would have to be placed on a penny .

By newton's second law

\sum F_y=ma

\sum F_y=qE-mg

Substitute the values then we get

qE-mg=ma

Substitute the values then we get

q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)

100q-29.4\times 10^{-3}=0.57\times 10^{-3}

100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}

q=\frac{29.97\times 10^{-3}}{100}

q=2.997\times 10^{-4}C

Sign of charge =Negative

Because electric force acting  in opposite direction of electric field therefore,charge on penny will be negative.

4 0
3 years ago
What's the Genotype for these? If your answer is something off-topic I ShAlL rEpOrT iT u.U
RideAnS [48]

Pick any free letter for each trait. For example, "A" for the dominant allele and "a" for the recessive allele for the number of arms. Assuming complete dominance, the genotype for the dominant trait will always have at least one "capital" allele, so a creature with 2 arms would have genotype either AA (homzygous dominant) or Aa (heterozygous), while a creature with the recessive trait of having 4 arms will always be homzygous recessive with genotype aa.

4 0
2 years ago
The moment of a force is calculated from the product of the ———— and the———— distance from the line of action of the force to th
bixtya [17]

Answer: It is the product of the (force)multiplied by the (perpendicular) distance from the line of action of the force to the pivot

Explanation:

7 0
2 years ago
A car mass of 1.2 x 10 kilograms starts from rest and attains a speed of 20 meters/seconds in 5 seconds. What net force acted on
lorasvet [3.4K]

Net force on the car=F=4.8 x 10³ N

Explanation:

mass of car= 1.2 x 10³ Kg

initial velocity= Vi=0

Final velocity= Vf= 20 m/s

time = t= 5 s

Using kinematic equation,

Vf= Vi + at

20= 0 + a (5)

5 a=20

a= 20/5

a= 4 m/s²

Now force is given by F = ma

F= 1.2 x 10³ (4)

F=4.8 x 10³ N

7 0
2 years ago
If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa
shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

r_{P}=2r_{E}

If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

\frac{g_{P}}{g_{E}}=\frac{1}{4}

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0
3 years ago
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