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3 years ago

The acceleration vector of a particle in projectile motion ________.

Physics

1 answer:

Alex73 [517]3 years ago

4 0

Answer:

Points downward, and its magnitude is 9.8 m/s^2

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform horizontal motion, with constant velocity and zero acceleration. In fact, there are no forces acting on the projectile along the horizontal direction (if we neglect air resistance), so the acceleration along this direction is zero.

- A vertical motion, with constant acceleration g = 9.8 m/s^2 towards the ground (downward), due to the presence of gravity wich "pulls" the projectile downward.

The total acceleration of the projectile is given by the resultant of the horizontal and vertical components of the acceleration. But we said that the horizontal component is zero, therefore the total acceleration corresponds just to its vertical component, therefore it is a vector with magnitude 9.8 m/s^2 which points downward.

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Katelyn (55 kg) is practicing a drop jump in the biomechanics lab. She steps off a plyometrics box, lands on the force plate, an

suter [353]

Answer:

J = 357.5 kg*m/s

Explanation:

The impulse exerted on Katelyn when she was on the force plate, is equal to the change in her momentum, according to Newton's 2nd Law.
Assuming as the positive direction the upward direction (coincident with the positive y-axis) we can express the initial momentum as follows:

$p_{o} = m*v_{o} = 55 kg * (-3.0 m/s) (1)$

By the same token, the final momentum is as follows:

$p_{f} = m*v_{f} = 55 kg * (3.5 m/s) (2)$

As we have already said, the impulse J is just equal to the change in momentum, i.e., the difference between (2) and (1):

$J = p_{f} - p_{o} = m* (v_{f} -v_{o}) = 55 kg* (3.5m/s- (-3.0m/s)) = 357.5 kg*m/s (3)$

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3 years ago

An exoplanet is in an elliptical orbit around a distant star. At its closest approach, the exoplanet is 0.540 AU from the star a

atroni [7]

Answer:

0.71121 km/s

Explanation:

$v_1$ = Velocity of planet initially = 54 km/s

$r_1$ = Distance from star = 0.54 AU

$v_2$ = Final velocity of planet

$r_2$ = Final distance from star = 41 AU

As the angular momentum of the system is conserved

$mv_1r_1=mv_2r_2\\\Rightarrow v_1r_1=v_2r_2\\\Rightarrow v_2=\frac{v_1r_1}{r_2}\\\Rightarrow v_2=\frac{54\times 0.54}{41}\\\Rightarrow v_2=0.71121\ km/s$

When the exoplanet is at its farthest distance from the star the speed is 0.71121 km/s.

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3 years ago

A Tale f Two Elephants

Murljashka [212]

Answer:

u need it now?

Explanation:

I'll answer it

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4 years ago

Define volume charge density <img src="https://tex.z-dn.net/?f=%20%28%20%5C%3A%20%5Crho%20%5C%3A%20%29%20" id="TexFormula1" titl

azamat

Answer:

Volume charge density: When the charge density is throughout the volume of a body, then the charge per unit volume is called volume charge density and it is represented by ρ.

∴ρ=

$\frac{v}{p}$

Hence the unit of ρ is Cm

−3

Example: If a charge q is uniformly distributed in the whole volume of a

sphere of radius R, then p =

$\frac{q}{ \frac{4}{3} } \pi {r}^{3}$ =3q/4πR-3

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2 years ago

Read 2 more answers

An electric motor exerts a constant torque of 10 Nm on a grindstone mounted on its shaft. The moment of inertia of the grindston

Dmitry_Shevchenko [17]

Answer:

Angular acceleration = 5 rad /s ^2

Kinetic energy = 0.391 J

Work done = 0.391 J

P =6.25 W

Explanation:

The torque is given as moment of inertia × angular acceleration

angular acceleration = torque/ moment of inertia

= 10/2= 5 rad/ s^2

The kinetic energy is = 1/2 Iw^2

w = angular acceleration/time

=5/8= 0.625 rad /s

1/2 × 2× 0.625^2

=0.391 J

The work done is equal to the kinetic energy of the motor at this time

W= 0.391 J

The average power is = torque × angular speed

= 10× 0.625

P = 6.25 W

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4 years ago