De ce cand bate puternic vantul, cad fructele din pom?

A very elastic rubber ball is dropped from a certain height and hits the floor with a (2pts) downward speed v. Since it is so el

artcher [175]

Answer:

3.True. The magnitude of momentum is the same

Explanation:

Let's propose the solution of the problem

The initial moment is

p₀ = m v

The final moment

$p_{f}$ = m (-v)

p₀ = - $p_{f}$

Now we can review the claims

1. False. We see that the moment module is the same, but its direction changes

2. False. The impulse is a vector

3.True. The magnitude of momentum is the same

7 0

3 years ago

Will give brainliest to correct answer!

ad-work [718]

The international system of units is the designated system of units used by scientist in every part of the world to keep data in the same form and measurements, this is to avoid confusion and the need to convert data when being shared. typically described in meters or kilometer over a time form usually seconds or hours.

6 0

3 years ago

Read 2 more answers

Unlike other states of matter, what expand to fill their containers

padilas [110]

Answer:

gas

Explanation:

8 0

3 years ago

Read 2 more answers

When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do

Semenov [28]

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

$F=\frac{\Delta p}{\Delta t}$

where

F is the force

$\Delta p$ is the change in momentum

$\Delta t=0.030 s$ is the time interval

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m = 0.04593 kg is the mass of the ball

u = 0 is the initial velocity of the ball

$v=281 km/h =78.1 m/s$ is the final velocity of the ball

Substituting into the original equation, we find the force exerted on the golf ball:

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.04593)(78.1-0)}{0.030}=128.9 N$

7 0

3 years ago

A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber

Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

$V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}$

Now calculate the bouyant force acting on the lead

$F_L = Vpg$

$F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N$

This force will act in upward direction

Gravitational force on the lead due to its mass will act in downward direction

Hence the difference of this two force

$T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N$

If V is the volume submerged in the water then bouyant force on the bobber is

$F_B=V'pg$

Equate bouyant force with the tension and gravitational force

$F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}$

Now Total volume of bobble is

$\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\$

= $\large\boxed{4.52 \%}$

7 0

3 years ago