Light travels at 300,000,000 m/s. This is an example

If the rifle is stopped by the hunter's shoulder in a distance of 3.23 cm, what is the magnitude of the average force exerted on

never [62]

By conservation of momentum, the initial recoil speed of the rifle is:
u = 904kg * 0.0106 kg / 3.53 kg = 2.7146 m/s

The final recoil speed v is zero in a distance of d = 0.0346 m

So the rifle's acceleration is a = (0 - u^2) / 2*d = -2.7146^2 / 2*.0323 m/s^2
a = -114.07m/s^2

6 0

3 years ago

Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the a

Naily [24]

Answer:

The angular acceleration of the pencil α = 17 rad·s⁻²

Explanation:

Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:

τ = I α (1)

W r = I α (2)

The weight is that the pencil has is,

sin 10 = r / (L/2)

r = L/2(sin(10))

The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:

I = 1/3 M L²

Thus,

mg(L / 2)sin(10) = (1/3 m L²)(α)

α(f) = 3/2(g) / Lsin(10)

α = 3/2(9.8) / 0.150sin(10)

α = 17 rad·s⁻²

Therefore, the angular acceleration of the pencil is 17 rad·s⁻²

3 0

3 years ago

What is the internal resistance of a 12.0 V car battery whose terminal voltage drops to 8.2 V when the starter draws 75 A

jonny [76]

Answer:

The internal resistance of the cell is 0.051 ohm.

Explanation:

Given;

emf of the battery, E = 12 V

terminal voltage of the cell, V = 8.2 V

current in the circuit, I = 75 A

let the potential drop of the cell due to internal resistance (r) = Ir

The internal resistance of the cell is calculated from the equation below;

E = V + Ir

where;

r is the internal resistance of the cell

$Ir = E - V\\\\r = \frac{E-V}{I} \\\\r = \frac{12-8.2}{75} \\\\r = 0.051 \ ohm$

Therefore, the internal resistance of the cell is 0.051 ohm.

6 0

3 years ago

Sequence eye light enters the eye

Burka [1]

Answer:

Well it is the courtnes

Explanation:

8 0

3 years ago

Integrated Concepts A lightning bolt strikes a tree, moving 20.0 C of charge through a potential difference of 1.00×102 MV . (a)

vodka [1.7K]

Answer:

a) 2*10^9 J

b) 764.8 kg

Explanation:

Given that

Energy of charge, q = 20 C

Potential difference, ΔV = 1*10^2 MV = 1*10^2 * 10^6 V = 1*10^8 V

a)

To find the energy dissipated, we use the formula

ΔU = qΔV

ΔU = 20 * 1*10^8

ΔU = 2*10^9 J

b)

Change in temperature, ΔT = 100 - 15°

ΔT = 85° C

Change in energy, ΔU = 2*10^9 J

Specific heat of water, C = 4180 j./Kg.K

Latent heat of vaporization, L(v) = 2.26*10^6 J/Kg

Q1 = mcΔT

Q2 = mL(v)

Net energy needed, U = Q1 + Q2

U = mcΔT + mL(v)

U = m (cΔT + L(v))

m = U /[cΔT + L(v)]

Being that we have all the values, we then substitute

m = 2*10^9 / [(4180 * 85) + 2.26*10^6]

m = 2*10^9 / (3.553*10^5 + 2.26*10^6]

m = 2*10^9 / 2.615*10^6

m = 764.8 kg

c)

Having 765 kg of steam at the temperature would have extreme effect on the tree, damaging it permanently. Possibly even blowing it to pieces

6 0

3 years ago