PLEASE HELP ANSWER ASAP. 10 points. I will give brainliest.

A ball is tossed straight up from the surface of a small, spherical asteroid with no atmosphere. The ball rises to a height equa

alukav5142 [94]

Answer:

d. equal to one-fourth the acceleration at the surface of the asteroid.

Explanation:

The explanation is attached as a picture with this answer

Newton's law of universal gravitation is being used to compare the accelerations at the surface and at the top of the ball's path.

as it can be seen in the explanation that the proportional form of the equation is used because we do not need to necessarily use to final form with "G" for comparison calculations.

As per the given scenario only difference between the two points in the gravitational field is the distance from center of the spherical asteroid, i.e. r.

It is taken 2r for the top is the path. hence we obtain (1/4)g as our answer.

4 0

3 years ago

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Is (kr) an element a compound or a mixture?

Xelga [282]

Krypton ... symbol Kr ... is an element. It's element number 36, because
there are 36 protons in the nucleus of every atom of Krypton.

Krypton has nothing to do with Superman, except that about 75 years ago,
the creators of Superman thought it was a cool-sounding scientific word,
so they chose it for the name of his home planet.

8 0

3 years ago

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How far apart (in mm) must two point charges of 70.0 nC (typical of static electricity) be to have a force of 1.30 N between the

Sever21 [200]

Answer:r=5.824 mm

Explanation:

Given

Charge $q_1=70 nC$

$q_2=70 nC$

Force between them $F=1.30 N$

Electrostatic Force is given by

$F=\frac{kq_1q_2}{r^2}$

where $q_1$ and $q_2$ are the charge on the Particles

r=distance between them

$k=Coulomb\ constant =9\times 10^9 N-m^2/C^2$

$1.3=\frac{9\times 10^9\times 70\times 70\times 10^{-18}}{r^2}$

$r^2=\frac{9\times 10^9\times 70\times 70\times 10^{-18}}{1.3}$

$r=\sqrt{33.923\times 10^{-6}}$

$r=5.824\times 10^{-3}$

$r=5.824 mm$

5 0

3 years ago

A rocket takes off from Earth's surface, accelerating straight up at 63.2 m/s2. Calculate the normal force (in N) acting on an a

jarptica [38.1K]

From the calculation, the normal force is 6161.2 N.

<h3>What is the normal force?</h3>

The normal force is given by the expression;

N - mg = ma

Then;

N = mg + ma

m = 84.4 kg

g = 9.8 m/s^2

a = 63.2 m/s2

Now we have;

N = m(g + a)

N = 84.4 (9.8 + 63.2)

N = 6161.2 N

Learn more about normal force:brainly.com/question/15520313

#SPJ1

4 0

1 year ago

A 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in 0.50 s. What net force

Julli [10]

Answer:

The net force acting on the otter along the incline is 13.96 N.

Explanation:

It is given that,

Mass of the otter, m = 2 kg

Distance covered by otter, d = 85 cm = 0.85 m

It takes 0.5 seconds.

We need to find the net force acts on the otter along the incline. If a is the acceleration of the otter. It can be calculated using second equation of motion as :

$d=ut+\dfrac{1}{2}at^2$