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brilliants [131]
3 years ago
7

It is correct to say that impulse is equal toA) momentum.B) the change in momentum.C) the force multiplied by the distance the f

orce acts.D) all of the aboveE) none of the above
Physics
1 answer:
goldenfox [79]3 years ago
8 0

Answer:

B) the change in momentum

Explanation:

Impulse is defined as the product between the force exerted on an object (F) and the contact time (\Delta t)

I=F \Delta t

Using Newton's second law (F = ma), we can rewrite the force as product of mass (m) and acceleration (a):

I=(ma) \Delta t

However, the acceleration is the ratio between the change in velocity (\Delta v) and the contact time (\Delta t): a=\frac{\Delta v}{\Delta t}, so the previous equation becomes

I=m \frac{\Delta v}{\Delta t}\Delta t

And by simplifying \Delta t,

I=m \Delta v

which corresponds to the change in momentum of the object.

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4 years ago
g During the contraction of the heart, 65 cm3 blood is ejected from the left ventricle into the aorta with a velocity of approxi
Alisiya [41]

Answer:

The force is F = 0.1441 \  N

Explanation:

From the question we are told that

   The volume of blood ejected is  V_b  =  65cm^3 = 65*10^{-6} \  m^3

    The velocity of the ejected blood is  v  = 98 cm/ s = 0.98 \  m/s

    The mass density of blood is  \rho = 1060 \  kg/m^3

    The heart rate is  R  =  61 bpm(beats \  per \  miunite) = \frac{61}{60} = 1.0667\  bps

Generally the average force exerted is mathematically represented as

     F = 2 *  \rho *  V_b *  R *  v

=>   F = 2 * 1060  *  (65*10^{-6}) *1.0667   *   0.98

=>   F = 0.1441 \  N    

8 0
3 years ago
What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
What was the original mission of the voyager space probes?
devlian [24]
I'm pretty sure its C
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4 years ago
Read 2 more answers
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