Ask question

Ask question

All categories

3 years ago

13

A bullet has a mass of 8 grams and a muzzle velocity of 340m/sec. A baseball has a mass of 0.2kg and is thrown by the pitcher at

40m/sec. What is the momentum of the baseball? What is the momentum of the bullet?

1 answer:

Thepotemich [5.8K]3 years ago

3 0

Answer:

Momentum of bullet

$P = 2.72 kg m/s$

momentum of baseball

$P = 8 kg m/s$

Explanation:

As we know that momentum is defined as the product of mass and velocity

here we know that

mass of the bullet = 8 gram

velocity of bullet = 340 m/s

momentum of the bullet is given as

$P = mv$

$P = (\frac{8}{1000})(340)$

$P = 2.72 kg m/s$

Now we have

mass of baseball = 0.2 kg

velocity of baseball = 40 m/s[/tex]

momentum of baseball is given as

$P = (0.2)(40)$

$P = 8 kg m/s$

You might be interested in

The mass of a triton is

Margarita [4]

2.14x 1022 kg

the 22 is a xponet of ten

7 0

3 years ago

Read 2 more answers

Select the correct answer. Eli and Andy want to find out which of the two is stronger. Eli pushes a table with a force of 120 ne

sergiy2304 [10]

Acceleration of the table: B. 0.50 meters/second2

Explanation:

The problem can be solved by using Newton's second law of motion, which states that the net force acting on an object is the product of its mass and its acceleration. Mathematically:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

For the table in this problem, we have:

$\sum F = 125 N - 120 N = 5 N$ is the net force on the table, because there are two forces of 125 N and 120 N acting in opposite directions

m = 10.0 kg is the mass of the table

Solving for a, we find the acceleration:

$a=\frac{\sum F}{m}=\frac{5}{10.0}=0.50 m/s^2$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

7 0

2 years ago

Read 2 more answers

What is the Nearest Galaxy from the Milky Way

Vsevolod [243]

Answer:

the andromeda galaxy

Explanation:

hope it helps , pls mark me as brainliest

6 0

2 years ago

Read 2 more answers

A plastic rod has been bent into a circle of radius R = 8.00 cm. It has a charge Q1 = +2.70 pC uniformly distributed along one-q

vovikov84 [41]

Answer:

Part a)

$V = -1.52 V$

Part b)

$V = -1.16 V$

Explanation:

Part a)

Electric potential is a scalar quantity

so here we can say that total potential due to a ring on its center is given as

$V = \frac{kQ}{R}$

here we know that

$Q = Q_1 - 6Q_1$

$Q = - 5 Q_1$

$Q_1 = 2.70 pC$

now we have

$V = \frac{(9\times 10^9)(-5\times 2.70 \times 10^{-12})}{0.08}$

$V = -1.52 V$

Part b)

Potential on the axis of the ring is given as

$V = \frac{kQ}{\sqrt{r^2 + R^2}}$

$V = \frac{(9\times 10^9)(-5\times 2.70\times 10^{-12})}{\sqrt{0.08^2 + 0.0671^2}}$

$V = -1.16 V$

8 0

2 years ago

The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu

ivolga24 [154]

Answer: $0.10233nm$

Explanation:

The mean free path $\lambda$ of an atom is given by the following formula:

$\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}$ (1)

Where:

$\lambda=0.2\mu m=0.2(10)^{-6}m$

$R=8.3145J/mol.K$ is the Universal gas constant

$T=0\°C=273.115K$ is the absolute standard temperature

$d$ is the diameter of the helium atom

$N_{A}=6.0221(10)^{23}/mol$ is the Avogadro's number

$P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3}$ absolute standard pressure

Knowing this, let's find $d$ from (1), in order to find the radius $r$ of the helium atom:

$d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}$ (2)

$d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}$ (3)

$d=2.0467(10)^{-10}m$ (4)

If the radius is half the diameter:

$r=\frac{d}{2}$ (5)

Then:

$r=\frac{2.0467(10)^{-10}m}{2}$ (6)

$r=1.0233(10)^{-10}m$ (7)

However, we were asked to find this radius in nanometers. Knowing $1nm=(10)^{-9}m$ :

$r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm$ (8)

Finally:

$r=0.10233nm$ This is the radius of the helium atom in nanometers.

5 0

2 years ago