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maks197457 [2]
3 years ago
7

A block of wood mass 0.60kg is balanced on top of a vertical port 2.0m high. A 10gm bullet is fired horizontally into the block

and the embedded bullet land at a 4.0m from the base of the port. Find the initial velocity of the bullet.
Physics
1 answer:
anzhelika [568]3 years ago
7 0

Answer:

Mass of bullet is m=0.01kg

Mass of the block is M=4kg

Coefficient=0.25,distance=20m

So, let the speed of the block just after the bullet embedded in it be V and v be the speed of bullet before striking the block,

By applying conservation of momentum,

mv=(m+M)V

V=

M+m

mv

Explanation:

please mark me as the brainliest answer and please follow me for more answers to your questions..

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Find the noise level of a sound having an intensity of 1.5x10^-14W/m^2 given I0=10^12 W/m2
Llana [10]

Answer:

Noise level will be -18.2 watt

So option (b) will be correct answer

Explanation:

We have given sound intensity I=1.5\times 10^{-14}w/m^2

And threshold intensity I_0=\times 10^{-12}w/m^2 ( in question it is given as 10^{12}w/m^2 but its standard value is 10^{-12}w/m^2 )

Now noise level  =10log\frac{I}{I_0}=10log\frac{1.5\times 10^{-14}}{10^{-12}}=10log0.015-18.23

So the noise level will be -18.2

So option (b) will be correct answer

5 0
3 years ago
Scientific laws explain_____.
arsen [322]

Howdy! I have your answer:

Scientific laws explain repeated observations that are experimental.

It also describes aspects of the universe.

~sofia

3 0
3 years ago
Block 1, with mass m1 and speed 3.6 m/s, slides along an x axis on a frictionless floor and then undergoes a one-dimensional ela
irina1246 [14]

Answer:

a) The block 1 slides 0.24 m into the rough region.

b) The block 2 slides 2.7 m

Explanation:

Hi there!

First, let´s find the final velocity of each block. With that velocities, we can calculate the kinetic energy of each block. The kinetic energy of the blocks will be equal to the work done by friction to stop them. From the equation of work, we can calculate the distance traveled by the blocks.

Since the collision is elastic, the momentum and kinetic energy of the system composed of the two blocks is constant.

The momentum of the system is calculated as the sum of the momenta of each block:

m1 · v1 + m2 · v2 = m1 · v1´ + m2 · v2´

Where:

m1 and m2 = mass of blocks 1 and 2 respectively.

v1 and v2 = velocity of blocks 1 and 2 respectively.

v1´ and v2´ = final velocity of blocks 1 and 2 respectively.

Using the data we have, we can solve the eqaution for v1´:

m1 · 3.6 m/s + 0.40 m1 · 0 = m1 · v1´ + 0.40 m1 · v2´

3.6 m/s · m1 = m1 · v1´ + 0.40 m1 · v2´

3.6 m/s = v1´ + 0.40 v2´

v1´ = 3.6 m/s - 0.40 v2´

The kinetic energy of the system also remains constant:

1/2 m1 · (v1)² + 1/2 m2 · (v2)² = 1/2 m1 · (v1´)² + 1/2 m2 · (v2´)²

Multiply by 2 both sides of the equation:

m1 · (v1)² + m2 · (v2)² = m1 · (v1´)² + m2 · (v2´)²

Let´s replace with the data:

m1 · (3.6 m/s)² + 0.40 m1 · 0 = m1 · (v1´)² + 0.40 m1 (v2´)²

divide by m1:

(3.6 m/s)² = (v1´)² + 0.40 (v2´)²

Replace v1´ = 3.6 m/s - 0.40 v2´

(3.6 m/s)² = (3.6 m/s - 0.40 v2´)² + 0.40 (v2´)²

Let´s solve for v2´:

(3.6 m/s)² = (3.6 m/s)² - 2.88 v2´ + 0.16 (v2´)² + 0.40 (v2´)²

0 = 0.56 (v2´)² - 2.88 v2´

0 = v2´(0.56 v2´ - 2.88)   v2´ = 0 (the initial velocity)

0 = 0.56 v2´ - 2.88

2.88/0.56 = v2´

v2´ = 5.1 m/s

Now let´s calculate v1´:

v1´ = 3.6 m/s - 0.40 v2´

v1´ = 3.6 m/s - 0.40 (5.1 m/s)

v1´ = 1.56 m/s

Now, let´s calculate the final kinetic energy (KE) of each block:

a) Block 1:

KE = 1/2 · m1 · (1.56 m/s)² = m1 · 1.2 m²/s²

The work done by friction is calculated as follows:

W = Fr · s

Where:

Fr = friction force.

s = traveled distance.

The friction force is calculated as follows:

Fr = N · μ

Where:

N = normal force.

μ = coefficient of friction.

And the normal force is calculated in this case as:

N = m1 · g

Where g is the acceleration due to gravity.

Then, the work done by friction will be:

W = m1 · g · μ · s

The kinetic energy of an object is the negative work that must be done on that object to bring it to stop. Then:

m1 · 1.2 m²/s² = m1 · g · μ · s

Solving for s:

s = m1 · 1.2 m²/s²  / m1 · g · μ

s = 1.2 m²/s²/ 9.8 m/s² · 0.50

s = 0.24 m

The block 1 slides 0.24 m into the rough region.

b) For block 2 the kinetic energy will be the following:

KE = 1/2 · 0.4 · m1 · (5.1 m/s)² = m1 · 5.2 m²/s²

The friction force will be:

Fr = 0.4 m1 · g · μ

And the work done will be:

W = 0.4 m1 · g · μ · s

Since W = ΔKE,

Then:

m1 · 5.2 m²/s² = 0.4 m1 · g · μ · s

Solving for s:

5.2 m²/s²/(0.4 · g · μ) = s

s =  5.2 m²/s²/(0.4 · 9.8 m/s² · 0.50)

s = 2.7 m

The block 2 slides 2.7 m

3 0
3 years ago
Which example shows energy?
olasank [31]

D because u r using your energy to do all the things

4 0
2 years ago
Read 2 more answers
Se lanza una bala con una velocidad inicial de 200 m/s y con un ángulo de inclinación de 30º respecto a la horizontal. Si se con
puteri [66]

Answer:

El alcance de la bala es 3464,1 m.

Explanation:

El alcance de la bala se puede calcular como sigue:

y = y_{0} + tan(\theta)*x - \frac{g}{2}*\frac{x^{2}}{(v_{0}cos(\theta))^{2}}

En donde:

y: es la altura final = 0

y_{0}: es la altura inicial = 0

x: es el alcance

θ: es el angulo respecto a la horizontal = 30°

v_{0}: es la velocidad inicial = 200 m/s

g: es la gravedad = 10 m/s²

Entonces, tenemos:

y = y_{0} + tan(\theta)*x - \frac{g}{2}*\frac{x^{2}}{(v_{0}cos(\theta))^{2}}

x = \frac{2tan(\theta)*(v_{0}cos(\theta))^{2}}{g} = \frac{2tan(30)*(200 m/s*cos(30))^{2}}{10 m/s^{2}} = 3464,1 m

Por lo tanto, el alcance de la bala es 3464,1 m.

Espero que se te sea de utilidad!

8 0
3 years ago
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