All categories

3 years ago

HELP PLEASE !!

1 answer:

3 0

If a car crashes into another car like this, the wreck should go nowhere. Besides this being an unrealistic question, the physics of it would look like this:

Momentum before and after the collision is conserved.

Momentum before the collision:
p = m * v = 50000kg * 24m/s + 55000kg * 0m/s = 50000kg * 24m/s
Momentum after the collision:
p = m * v = (50000kg + 55000kg) * v

Setting both momenta equal:
50000kg * 24m/s = (50000kg + 55000kg) * v

Solving for the velocity v:
v = 50000kg * 24m/s/(50000kg + 55000kg) = 11,43m/s

You might be interested in

How much would you have to shrink a baseball to a black hole?

Vesna [10]

The Schwarzchild radius is . . . R = 2GM/c² .

Just like you, I'm not completely sure what that means. But it DOES use the mass of a black hole to calculate a radius associated with it, so with 15 Brainly points at stake, it seems like a good-enough formula to use for an answer.

Before I proceed, I really should ask you whether you're talking a softball or a hardball, but again . . . . .

So R = 2GM/c²

G = the gravitational constant = 6.67 x 10⁻¹¹ N-m²/kg²

M = mass of the baseball = 145 grams = 0.145 kg

c = speed of light = 3 x 10⁸ m/s

R = 2 (6.67 x 10⁻¹¹ m³/kg-s²) (0.145 kg) / (3 x 10⁸ m/s)²

R = (2 x 6.67 x 10⁻¹¹ x 0.145 / 9 x 10¹⁶) (m³-kg-s² / kg-s²-m²)

R = ( 1.9343 x 10⁻¹¹ / 9 x 10¹⁶) (m³-kg-s²/m²-kg-s²)

R = (0.2149 x 10⁻²⁷) meter

R = 2.149 x 10⁻²⁸ meter

For reference: The radius of a Hydrogen atom is 1.2 x 10⁻¹⁰ meter.

So in order to make a black hole out of a baseball, you have to crunch the baseball down to around 0.000000000000000001791 the size of a Hydrogen atom.

Would that be a problem for you ?

3 0

3 years ago

Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plate

vichka [17]

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

CB = 4.45 x 10⁻⁹ F = 4.45 nF

5 0

3 years ago

Alligators and other reptiles don't use enough metabolic energy to keep their body temperatures constant. they cool off at night

drek231 [11]

To compute for the heat, $Q$ needed to be absorbed or released, we need

$Q = mc\Delta T$

where $m$ is the mass of the alligator, $\Delta T$ is the change in temperature, and $c$ is the specific heat of the alligator's body. Plugging in all the information we have,

$Q = (300 kg)(3400 kJ/K)(30-25) = 5100000 J$

Recall that 1 Watt = 1 J/s, thus time needed to absorb radiation from the sun is

$t = \frac{Q}{W}\\ t = 4250 seconds$

That means it takes 4250 seconds for the alligator to warm up to be able to absorb the radiation from the sun.

Answer: 4250 seconds

4 0

4 years ago

Why do electrons rather than protons make up the flow of charge in a metal wire?

andrew-mc [135]

Protons are tightly bound to the nucleus of atoms, and the nucleus is
'shielded' from the outside world by the 'cloud' of electrons around it.
It takes a lot of energy to separate a proton from the nucleus. When
that happens, it's a 'radioactive' or a 'nuclear' event, and the atom has
changed into an atom of a different substance. This isn't something
that's happening around us very often.

On the other hand, the electrons are on the outside of the atom, and
it's rather easy to convince an electron to leave its atom and flow off
to somewhere else.

6 0

3 years ago

Which type of wave can only be transmitted through matter?

Doss [256]

I only doing this for points, but i think its c

8 0

3 years ago

Read 2 more answers