A physics book slides off a table at 1.25ms and hits the ground after 0.4s

Which of these properties of an object best quantifies its inertia: velocity, acceleration, volume, mass, or temperature?

LuckyWell [14K]

Answer:

Mass

Explanation:

Inertia is essentially an object's tendency to stay in motion or at rest unless it is forced to do otherwise (pun intended). It only makes sense to me that mass would best quantify an object's inertia, because an object with more mass would be harder to move and/or stop from moving.

3 0

2 years ago

This is a small rocky body that orbits the sun, and when close to the sun exhibit a small tail of ice and gas.

Over [174]

This is a comet. It is also called a dirty snowballs.

3 0

2 years ago

There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

$q_1=q_2=2.06\mu C=2.06\times 10^{-6} C$

$q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C$

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force, $F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2$

$F_1=F_3=F$

$F_{net}=\sqrt{F^2+F^2+0}-F_2$

$F_{net}=\sqrt 2F-F_2$

$F=\frac{kq^2}{d^2}$

$F_2=\frac{Kq^2}{2d^2}$

$F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})$

Where $k=9\times 10^9$

$F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})$

$F_{net}=0.208 N$

3 0

2 years ago

A golf club hits a stationary 0.05kg golf ball with and average force of 5.0 x 10^3 newtons accelerating the ball at 44 meters p

maxonik [38]

Answer: The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

Explanation:

Force applied on the golf ball = $5.0\times 10^3 N$

Mass of the ball = 0.05 kg

Velocity with which ball is accelerating = 44 m/s

Time period over which forece applied = t

$f=ma=\frac{m\times v}{t}$

$t=\frac{0.05 kg\times 44m/s}{5.0\times 10^3 N}=4.4\times 10^{-4} seconds$

$Impulse=(force)\times (time)=f\times t = 5.0\times 10^3\times 4.4\times 10^{-4} seconds=2.2$ Newton seconds

The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

7 0

2 years ago

At what altitude above the earth's surface would the acceleration due to gravity be 4.9 m/s^2? Assume the mean radius of the ear

Mandarinka [93]

We apply the gravity calculation expressed in the formula: g=GM/r2
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1) Radius = √6.674e-11*5.972e24/8 = 7058 kms Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2 Radius=√6.674e-11*5.972e24/4.9 = 9019 kms Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.

4 0

2 years ago