Ask question

Ask question

All categories

2 years ago

12

Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless sur

face, so it moves freely when the bullet hits it). The wood block is initially at rest. The bullet is traveling 300 m/s when it hits the wood block and sticks inside it. Now the bullet and the wood block move together as one object. How fast are they traveling?

1 answer:

d1i1m1o1n [39]2 years ago

8 0

Answer:

The speed of bullet and wooden bock coupled together, V = 22.22 m/s

Explanation:

Given that,

Mass of the bullet, m = 0.04 Kg

Mass of the wooden block, M = 0.5 Kg

The initial velocity of the bullet, u = 300 m/s

The initial velocity of the wooden block, U = 0 m/s

The final velocity of the bullet and wooden bock coupled together, V = 0 m/s

According to the conservation of linear momentum, the total momentum of the body after impact is equal to the total momentum before impact.

Therefore,

mV + MV = mu + MU

V(m+M) = mu

V = mu/(m+M)

Substituting the values in the above equation,

V = 0.04 Kg x 300 m/s / (0.04 Kg+ 0.5 Kg)

= 22.22 m/s

Hence, the speed of bullet and wooden bock coupled together, V = 22.22 m/s

You might be interested in

What is the torque τb about axis b due to the force f⃗ ? (b is the point at cartesian coordinates (0,b), located a distance b fr

yKpoI14uk [10]

Check the attached file for the solution.

8 0

2 years ago

A 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Afterwards, the 355 kg car moves right at 5.80

Sidana [21]

Explanation:

It is given that,

Mass of bumper car, m₁ = 202 kg

Initial speed of the bumper car, u₁ = 8.5 m/s

Mass of the other car, m₂ = 355 kg

Initial velocity of the other car is 0 as it at rest, u₂ = 0

Final velocity of the other car after collision, v₂ = 5.8 m/s

Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁

Using the conservation of linear momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$202\ kg\times 8.5\ m/s+355\ kg\times 0=m_1v_1+355\ kg\times 5.8\ m/s$

p₁ = m₁v₁ = -342 kg-m/s

So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.

7 0

2 years ago

A ray of light passes from air into carbon disulfide (n = 1.63) at an angle of 28.0 degrees to the normal. what is the refracted

Snezhnost [94]

We can solve the problem by using Snell's law, which states
$n_i \sin \theta_i = n_r \sin \theta_r$
where
$n_i$ is the refractive index of the first medium
$\theta_i$ is the angle of incidence
$n_r$ is the refractive index of the second medium
$\theta_r$ is the angle of refraction

In our problem, $n_i=1.00$ (refractive index of air), $\theta_i = 28.0^{\circ}$ and $n_r=1.63$ (refractive index of carbon disulfide), therefore we can re-arrange the previous equation to calculate the angle of refraction:
$\sin \theta_r = \frac{n_i}{n_r} \sin \theta_r = \frac{1.00}{1.63} \sin 28.0^{\circ} = 0.288$
From which we find
$\theta_r = \arcsin (0.288)=16.7^{\circ}$

6 0

2 years ago

A rectangular loop of wire with length a=2.2 cm, width b=0.80 cm,and resistance R=0.40m ohms is placed near an infinitely long w

ser-zykov [4K]

Answer:

magnetic flux ΦB = 0.450324 × $10^{-7}$ weber

current I = 1.02484 $10^{-8}$ A

Explanation:

Given data

length a = 2.2 cm = 0.022 m

width b = 0.80 cm = 0.008 m

Resistance R = 0.40 ohms

current I = 4.7 A

speed v = 3.2 mm/s = 0.0032 m/s

distance r = 1.5 b = 1.5 (0.008) = 0.012

to find out

magnitude of magnetic flux and the current induced

solution

we will find magnitude of magnetic flux thorough this formula that is

ΦB = ( μ I(a) /2 π ) ln [(r + b/2 ) /( r -b/2)]

here μ is 4π × $10^{-7}$ put all value

ΦB = (4π × $10^{-7}$ 4.7 (0.022) /2 π ) ln [(0.012+ 0.008/2 ) /( 0.012 -0.008/2)]

ΦB = 0.450324 × $10^{-7}$ weber

and

current induced is

current = ε / R

current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]

put all value

current = μ I(a) bv / 2πR [(r² ) - (b/2 )² ]

current = 4π × $10^{-7}$ (4.7) (0.022) (0.008) (0.0032) / 2π(0.40) [(0.012² ) - (0.008/2 )² ]

current = 1.02484 $10^{-8}$ A

5 0

2 years ago

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Juli2301 [7.4K]

Answer:

0.0984

Explanation:

From the first diagram attached below; a free flow diagram shows the interpretation of this question which will be used to solve this question.

From the diagram, the horizontal component of the force is:

$F_X = F_{cos \ \theta}$

Replacing 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

$F_X =64.65 \ N$

On the other hand, the vertical component is ;

$F_Y = Fsin \ \theta$

Replacing 42° for θ and 87.0° for $F$

$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

However, resolving the vector, let A be the be the component of the mutually perpendicular directions.

The magnitude of the two components is shown in the second attached diagram below and is now be written as A cos θ and A sin θ

The expression for the frictional force is expressed as follows:

$f = \mu \ N$

Where;

$\mu$ is said to be the coefficient of the friction

N = the normal force

Similarly the normal reaction (N) = mg - F sin θ

Replacing $F_Y \ for \ F_{sin \ \theta}$ . The normal reaction can now be:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals to frictional force.

The horizontal component of the force is given as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Making $\mu$ the subject of the formular in the above equation; we have the following:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Replacing the following values: i.e

$F_X \ = \ 64.65 \ N$

m = 73 Kh

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Then:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

Thus, the coefficient of friction is = 0.0984

5 0

2 years ago