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2 years ago

What is the density of a rock with a mass of .235kg

Physics

1 answer:

Andre45 [30]2 years ago

4 0

Answer:

where is volume? formula of density is: mass/volume so volume must be there

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Which country had the largest population in 1997

nignag [31]

China i hope this helped

6 0

2 years ago

During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.146 kg baseball crashing through the pane

Nonamiya [84]

Answer:

Impulse, |J| = 0.6716 kg-m/s

Force, F = 63.35 N

Explanation:

It is given that,

Mass of the baseball, m = 0.146 kg

Initial speed of the ball, u = 15.3 m/s

Final speed of the ball, v = 10.7 m/s

To find,

(a) The magnitude of this impulse.

(b) The magnitude of the average force of the glass on the ball.

Solution,

(a) Impulse of an object is equal to the change in its momentum. It is given by :

$J=m(v-u)$

$J=0.146\ kg(10.7-15.3)\ m/s$

J = -0.6716 kg-m/s

|J| = 0.6716 kg-m/s

(b) Another definition of impulse is given by the product of force and time of contact.

t = 0.0106 s

$J=F\times \Delta t$

$F=\dfrac{J}{\Delta t}$

$F=\dfrac{0.6716\ kg-m/s}{0.0106\ s}$

F = 63.35 N

Hence, this is the required solution.

6 0

2 years ago

a 0.15 kg baseball has a momentum of 0.78 kg*m/s just before it lands on the ground. What was the ball's speed just before landi

vovangra [49]

5.2m/s

Explanation:

Given parameters:

Mass of baseball = 0.15kg

Momentum of baseball = 0.78kgm/s

Unknown:

Speed of baseball = ?

Solution:

The momentum of the baseball is a function of the product of the mass and velocity. It is a vector quantity:

Momentum = mass x velocity

Since the speed of the ball is unknown:

Velocity = $\frac{momentum }{mass}$

= $\frac{0.78}{0.15}$

= 5.2m/s

The speed of the baseball before it lands is 5.2m/s

Learn more:

Momentum brainly.com/question/9484203

#learnwithBrainly

4 0

2 years ago

A uniform bridge span of weight 1200 kN and of

Scorpion4ik [409]

45-4753that is your answer

3 0

2 years ago

An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 17601760 × 103 seconds (ab

Yuliya22 [10]

Answer:

The value is $a_r = 3.81 *10^{-3} m/s^2$

Explanation:

Generally the moon's radial acceleration is mathematically represented as

$a_r = r * w^2$

Here $w$ is the angular velocity which is mathematically represented as

$w =\frac{2 \pi }{ T}$

substituting $1760 * 10^3 \ seconds$ for T(i.e the period of the moon ) we have

$w =\frac{2 * 3.142 }{ 1760 * 10^3}$

=> $w = 3.57 *10^{-6} \ rad/s$

From the question r(which is the radius of the orbit ) is evaluated as

$r = R + H$

substitute $3.60 * 10^6 m$ for R and $295.0 * 10^6 \ m$ H

$r = 295.0 * 10^6 +3.60 * 10^6$

=> $r = 2.986 *10^{8} \ m$

$a_r = 2.986 *10^{8} * ( 3.57 *10^{-6} )^2$

$a_r = 3.81 *10^{-3} m/s^2$

4 0

2 years ago