All categories

3 years ago

a 2-kg rifle that is suspended by strings fires a 0.01-kg bullet at 200 m/s. the recoil velocity of the rifle is about

Physics

1 answer:

Trava [24]3 years ago

3 0

Explanation:

from Newton's third law of motion we know every action has equal and opposite reaction.

So, the velocity of going out of the rifle = 200 m/s

therefore the recoil is 200 m/s on the backward direction.

You might be interested in

Chemical testing of water

NNADVOKAT [17]

Tap water isn't pure. it just isn't H₂O.

4 0

3 years ago

Which shows the correct order of processes leading up to the formation of a main sequence star?

Sphinxa [80]

C. Should be the answer it worked for me

8 0

3 years ago

Read 2 more answers

A wagon has a mass of 100 grams and an

olga2289 [7]

Answer:

Net force on the wagon is 200 N

Explanation:

As we know by Newton's II law that net force on the system of mass is given as product of mass and acceleration

Here we know that

mass = 100 kg

a = 2 m/s/s

now we have

$F = ma$

$F = (100)(2)$

$F = 200 N$

3 0

3 years ago

What is the de Broglie wavelength of an object with a mass of 2.50 kg moving at a speed of 2.70 m/s? (Useful constant: h = 6.63×

xxMikexx [17]

Answer:

9.82 × $10^{-35}$ Hz

Explanation:

De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:

λ = $\frac{h}{mv}$

where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.

Given that: h = 6.63 × $10^{-34}$ Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;

λ = $\frac{h}{mv}$

= $\frac{6.63*10^{-34} }{2.5*2.7}$

= $\frac{6.63 * 10^{-34} }{6.75}$

= 9.8222 × $10^{-35}$

The wavelength of the object is 9.82 × $10^{-35}$ Hz.

4 0

3 years ago

You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30

12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

$\beta = 10 log(\dfrac{I_1}{I_0})$

I₀ = 10⁻¹² W/m²

now,

$30 = 10 log(\dfrac{I_1}{10^{-12}})$

$\dfrac{I_1}{10^{-12}}= 10^3$

$I_1= 10^{-8}\ W/m^2$

to hear the whisper sound = 80 dB

$80 = 10 log(\dfrac{I_2}{10^{-12}})$

$\dfrac{I_2}{10^{-12}}= 10^8$

$I_2= 10^{-4}\ W/m^2$

we know intensity of sound is inversely proportional to square of distances

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}$

$\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}$

$10^{-4}=\dfrac{r_2^2}{20^2}$

r₂ = 0.2 m

6 0

3 years ago