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Ber [7]
3 years ago
6

The set of points in a plane that are equidistant from a given point form a ____.

Physics
2 answers:
il63 [147K]3 years ago
8 0

Answer:

Circle

Explanation:

gtnhenbr [62]3 years ago
6 0
Circle

.......................
You might be interested in
Help me please i dont understand
solniwko [45]

Answer:

It is b

Explanation:

The question tells you that the distance was increased 3 times the original distance, which means it was moved 3 times

5 0
3 years ago
Calculate the final velocity right after a 117 kg rugby player who is initially running at 7.45 m/s collides head‑on with a padd
Free_Kalibri [48]

Answer:

v_f = 0.87 m/s

Explanation:

We are given;

F_avg = -17700 N (negative because it's backward)

m = 117 kg

Δt = 5.50 × 10^(−2) s

v_i = 7.45 m/s

Now, formula for impulse is given by;

I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s

From impulse momentum theory, we know that;

Change in momentum of particle is equal to impulse.

Thus,

Δp = I = m•v_f - m•v_i

Thus,

-973.5= 117(v_f - 7.45)

Thus,

-973.5/117 = (v_f - 7.45)

-8.3205 + 7.45 = v_f

v_f = - 0.87 m/s

We'll take absolute value as;

v_f = 0.87 m/s

5 0
3 years ago
A 12kg cheetah accelerates 24 m/s". What is the force the cheetah needed to run?
Kobotan [32]

Answer:

288N

Explanation:

Given parameters:

Mass of Cheetah = 12kg

Acceleration  = 24m/s²

Unknown:

Force needed by the cheetah to run  = ?

Solution:

The force needed by the Cheetah to run is the net force.

According to Newton's law;

    Force  = mass x acceleration

Insert the given parameters and solve;

   Force  = 12 x 24  = 288N

7 0
3 years ago
How old is the universe
Virty [35]
Roughly 13.8 billion years old according to science
5 0
3 years ago
Read 2 more answers
#1 Not sure where to start. This is for AP Physics!
yaroslaw [1]

First,

\rho=\dfrac mV

where \rho is density, m is mass, and V is volume. We can compute the volume of the roll:

2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V

\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness x. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

V=Ax

where A is the given area, so

4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x

\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}

If we're taking significant digits into account, the volume we found would have been V=470\,\mathrm m^3, in turn making the thickness x=250\,\mathrm{mm}.

8 0
3 years ago
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