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ratelena [41]
3 years ago
15

What do you mean by momentum?

Physics
1 answer:
Shtirlitz [24]3 years ago
7 0
The answer to this question is that --- The momentum possessed by a body is generally defined as the product of its mass and velocity.

Momentum is a vector and it also has magnitude as it is the product of the multiplication of the mass and velocity. 
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How do I solve for time if I have initial velocity and final speed?
xeze [42]
Final speed = initial speed + (acceleration x time)

(final speed - initial speed) = acceleration x time

Time = (final speed - initial speed) / acceleration
4 0
3 years ago
How must a fuse be connected in a circuit to prevent current from flowing when the circuit becomes ""overloaded""?
Lady bird [3.3K]

Answer:

The fuse must be connected between the device and the power intake source.

Explanation:

A fuse is a protective component of electrical appliances that is designed to be sensitive to a particular range of electric current

The fuse is made of a thing metal strip with a known melting point. Once current abive its carrying capacity flows through it, large heat is generated in the metal strip which melts it and causes the metal strip to cut int two protecting the device from the power spike.

3 0
2 years ago
Anger is to angry as fire is to what
finlep [7]
<span>Anger is to angry as fire is to blazing. </span>
4 0
3 years ago
What experimental evidence led to the development of this atomic model from the one before it?
Marina86 [1]

Answer:

A few of the positive particles aimed at a gold foil seemed to bounce back

Explanation:

7 0
3 years ago
Read 2 more answers
the jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at t
givi [52]

Explanation:

Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

→ The tangential component of acceleration is rate of increase in the speed of plane so,

a_{t} = v = 0.8 m/s^{2}

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).

dy/dx = d(0.4x²)/dx

         = 0.8x

Take the derivative again,

d²y/dx² = d(0.8x)/dx

          = 0.8

at x= 5 Km

dy/dx = 0.8(5)

         = 4

p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} }   }{\frac{d^{2y} }{dx^{2} } }

now insert the values,

p = \frac{[1+(4)^{2}]^{\frac{3}{2} }  }{0.8}  = 87.62 km

→ Now the normal component of acceleration is given by

a_{n} = \frac{v^{2} }{p}

    = (200)²/(87.6×10³)

aₙ = 0.457 m/s²

→ Now the total acceleration is,

a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}

a = [(0.8)^{2} + (0.457)^{2}]^{0.5}

a = 0.921 m/s²

4 0
3 years ago
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