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marshall27 [118]
3 years ago
8

The asteroid belt is located between Mars and Jupiter and contains thousands of asteroids.

Physics
2 answers:
Virty [35]3 years ago
6 0
Im not understanding the question but this is true
Wewaii [24]3 years ago
6 0
This is true but what's the question?
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Momentum is a measure of the<br> of an object.<br> Which term accurately completes the sentence?
RideAnS [48]

Answer:

Momentum is define as the product of the mass and velocity of a body. It is measured in Kgm/s.

Explanation:

Momentum is the product of mass and velocity of an object. When an object or a body of mass 'm' is moving with velocity 'v', then its momentum can be determined as;

momentum (P) = mass × velocity

i.e            P = m × v

                  = mv

It is measured in Kgm/s.

The change in momentum of a body is referred to as its impulse (Ft).

ΔP = m(v - u) = Ft

Where: P is the momentum of the object, m is its mass, v is its final velocity, u is the initial velocity, F is the force and t is the time in which the force acts.

8 0
3 years ago
Read 2 more answers
Seasons are caused by differences in daylight, temperature, and weather patterns due to :
Neko [114]

... the angular tilt of the Earth's position on its axis relative to the sun


4 0
4 years ago
Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin
Arte-miy333 [17]

Answer:

x=\dfrac{r}{\sqrt2}

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

E=K\dfrac{Q}{(r^2+x^2)^{3/2}}

For maximum condition

\dfrac{dE}{dx}=0

E=K{Q}{(r^2+x^2)^{-3/2}}

\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}

For maximum condition

\dfrac{dE}{dx}=0

K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0

r^2+x^2-3x^2=0

x=\dfrac{r}{\sqrt2}

At x=\dfrac{r}{\sqrt2} the electric field will be maximum.

3 0
3 years ago
How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C?
Svet_ta [14]

Answer:

229,098.96 J

Explanation:

mass of water (m) = 456 g = 0.456 kg

initial temperature (T) = 25 degrees

final temperature (t) = - 10 degrees

specific heat of ice = 2090 J/kg

latent heat of fusion =33.5 x 10^(4) J/kg

specific heat of water = 4186 J/kg

for the water to be converted to ice it must undergo three stages:

  • the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp

        Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J

  • the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp

         Q = 0.456 x 33.5 x 10^(4) = 152760 J

  • the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp

        Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J

The quantity of heat removed from all three stages would be added to get the total heat removed.

Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J

6 0
3 years ago
What profession is most likely to make use of amphoras and candelas
Monica [59]
Biologists probably
5 0
3 years ago
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