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1 year ago

A 22-g bullet traveling 265 m/s penetrates a 1.9 kg block of wood and emerges going 125 m/s .

Physics

1 answer:

Usimov [2.4K]1 year ago

6 0

The body moves at a velocity of 1.62m/s after the bullet emerges.

<h3>Given:</h3>

Mass of bullet, $m_1$ = 22g

= 0.022 kg

Mass of the block, $m_2$ = 1.9 kg

Velocity of bullet , $v_1$ = 265 m/s

$v_2 = 0$

According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.

After penetration;

$v^{'}_1 =125 m/s$

$v^{'}_2=?$

The formula for calculating the collision of a body is expressed as:

p = mv

m is the mass of the body

v is the velocity of the body

∴ Momentum before = Momentum after

Substitute the given parameters into the formula as shown:

$m_1v_1+ m_2v_2 = m_1v^{'}_1+ m_2v^{'}_2\\0.022* 265 + 0 = 0.022*125+1.9*v^{'}_2\\5.83 = 1.9 v^{'}_2\\v^{'}_2 = 1.62 m/s$

Therefore, It moves with a velocity of 1.62 m/s.

Learn more about momentum here:

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What is the speed that is measured in speedometer to track speed violation?

BartSMP [9]

Answer:

The officer's unit detects this 135-mile-per-hour speed and should subtract the patrol car's 70-mile -per-hour ground speed to get your true speed of 65 miles per hour. Instead, the officer's ground-speed beam fixes on the truck ahead and measures a false 50-mile-per-hour ground speed.

Explanation:

A speedometer or speed meter is a gauge that measures and displays the instantaneous speed of a vehicle. Now universally fitted to motor vehicles, they started to be available as options in the early 20th century, and as standard equipment from about 1910 onwards.

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3 years ago

two point charges of magnitude 4.0 μc and -4.0 μc are situated along the x-axis at x1 = 2.0 m and x2 = -2.0 m, respectively. wha

user100 [1]

The electric potential at the origin of the xy coordinate system is negative infinity

<h3>What is the electric field due to the 4.0 μC charge?</h3>

The electric field due to the 4.0 μC charge is E = kq/r² where

k = electric constant = 9.0 × 10 Nm²/C²,
q = 4.0 μC = 4.0 × 10 C and
r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m

<h3>What is the electric field due to the -4.0 μC charge?</h3>

The electric field due to the -4.0 μC charge is E = kq'/r² where

k = electric constant = 9.0 × 10 Nm²/C²,
q' = -4.0 μC = -4.0 × 10 C and
r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m

Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is

E" = E + E'

= -2E

= -2kq/r²

<h3>What is the electric potential at the origin?</h3>

So, the electric potential at the origin is V = -∫₂⁰E".dr

= -∫₂⁰-2kq/r².dr

Since E and dr = dx are parallel and r = x, we have

= -∫₂⁰-2kqdxcos0/x²

= 2kq∫₂⁰dx/x²

= 2kq[-1/x]₂⁰

= -2kq[1/x]₂⁰

= -2kq[1/0 - 1/2]

= -2kq[∞ - 1/2]

= -2kq[∞]

= -∞

So, the electric potential at the origin of the xy coordinate system is negative infinity

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brainly.com/question/26978411

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3 0

1 year ago

The temperature of air changes from 0 to 10°C while its velocity changes from zero to a final velocity, and its elevation change

aliya0001 [1]

Answer:

Final velocity = 119.83 m/s

Final elevation = 731.9 m

Explanation:

We are told that temperature of air changes from 0 to 10°C

Thus;

Change in temperature; ΔT = 10 - 0 = 10°C

Also, its velocity changes from zero to a final velocity. Thus;

v1 = 0 m/s

v2 is unknown

Also, its elevation changes from zero to a final elevation.

So, z1 = 0 and z2 is unknown

Now, we want to find v2 and z2 when the internal, kinetic and potential energy are equal.

Thus Equating the formula for both kinetic and internal energy gives;

½m(v2² - v1²) = mc_v•ΔT

m will cancel out and v1 is zero to give;

v2² = 2c_v•ΔT

v2 = √(2c_v•ΔT)

Where c_v is specific heat of constant volume of air with a constant value of 718 J/Kg.K

Thus;

v2 = √(2 × 718 × 10)

v2 = √14360

v2 = 119.83 m/s

To find z2, we will equate potential energy formula to that of the internal energy.

Thus;

mg(z2 - z1) = mc_v•ΔT

m will cancel out and since z1 is zero, then we have;

z2 = (c_v•ΔT)/g

z2 = 718 × 10/9.81

z2 = 731.9 m

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3 years ago

J. J. Thomson s experiments provided evidence that an atom

gulaghasi [49]

J.J. Thomson's experiments provided evidence that an atom B.contains negatively charged particles.

<span>♡♡Hope I helped!!! :)♡♡
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3 years ago

During lightning strikes from a cloud to the ground, currents as high as 2.50×104 A can occur and last for about 40.0 μs . How m

Nezavi [6.7K]

Answer:

<h3>The charge transferred from the cloud to earth is 1 Coulomb.</h3>

Explanation:

Given :

Current $I = 2.5 \times 10^{4}$ A

Time $t = 40 \times 10^{-6}$ sec

We know that the current is the rate of flow of charge.

From the formula of current,

Where $Q =$ charge transfer between cloud and earth.

$Q =I t$

$Q = 2.5 \times 10^{4} \times 40 \times 10^{-6}$

$Q = 1$ C

Hence, the charge transferred from the cloud to earth is 1 Coulomb.

3 0

3 years ago