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iVinArrow [24]
3 years ago
7

A uniform rod of length L rests on a frictionless horizontalsurface. The rod pivots about a fixed, frictionless axis at oneend.

the rod is initially at rest. A bullet travelling parallel tothe horizontal surface and perpendicular to the rod with speed vstrikes the rod at its center and becomes embedded in it. The massof the bullet is one-fourth the mass of the rod. a) What is thefinal angular speed of the rod? b) What is the ratio of the kineticenergy of teh system after the collision to the kinetic energy ofthe bullet before the collision?

Physics
1 answer:
coldgirl [10]3 years ago
5 0

Answer: a ) 6/19 V/L

b ) 3/19

Explanation:

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(ASAP) would it be 125 m/s2 to calculate for her speeding up?
serg [7]

Answer:

0\:\mathrm{ m/s^2}

Explanation:

Recall the formula for acceleration:

\displaystyle\\a=\frac{v_f-v_i}{\Delta t}, where v_f is final velocity, v_i is initial velocity, and \Delta t is elapsed time (change in velocity over this amount of time).

Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.

We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).

We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.

Substituting values in our formula, we have:

\displaystyle a=\frac{25-25}{5}=\frac{0}{5}=\boxed{0\:\mathrm{m/s^2}}

Alternative:

Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!

8 0
3 years ago
2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t
serious [3.7K]

Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v (1)

Where:

m_{S}, m_{C} - Masses of the semi truck and the car, measured in kilograms.

v_{S}, v_{C} - Initial velocities of the semi truck and the car, measured in meters per second.

v - Final speed of the system after collision, measured in meters per second.

If we know that m_{S} = 2200\,kg, m_{C} = 2000\,kg, v_{C} = 45\,\frac{m}{s} and v = -15\,\frac{m}{s}, then the initial velocity of the semi truck is:

m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}

v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}

v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}

v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})

v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s}  \right)

v_{S} = -69.545\,\frac{m}{s}  

The semi truck travels at an initial speed of 69.545 meters per second downwards.

3 0
3 years ago
What happens to a radioactive isotope as it decays?
kupik [55]
It becomes a different element
8 0
3 years ago
HALP me!! This is a physics question.
emmasim [6.3K]
<h2>Answer:</h2>

<u>Distance covered is 6.9 meters</u>


<h2>Explanation:</h2>

Data given:

Work Done = 345 kJ = 345000 J

Force = 5 x 10 ^ 4 =  50000 N

Distance = ?


Solution:

As we know that

Work Done = Force applied x Distance covered

By arranging the equation we get

Work / Force = Distance covered

By putting the values

345000 / 50000 = 6.9

So distance covered is 6.9 meters

4 0
3 years ago
Read 2 more answers
An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is W/m². What is the rms
inn [45]

I think your question should be:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is

S = 1.23*10^9 W/m^2

What is the rms value of (a) the electric field and

(b) the magnetic field in the electromagnetic wave emitted by the laser

Answer:

a) 6.81*10^5 N/c

b) 2.27*10^3 T

Explanation:

To find the RMS value of the electric field, let's use the formula:

E_r_m_s = sqrt*(S / CE_o)

Where

C = 3.00 * 10^-^8 m/s;

E_o = 8.85*10^-^1^2 C^2/N.m^2;

S = 1.23*10^9 W/m^2

Therefore

E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]}

E_r_m_s= 6.81 *10^5N/c

b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:

B_r_m_s = E_r_m_s / C;

= 6.81*10^5 N/c / 3*10^8m/s;

B_r_m_s = 2.27*10^3 T

8 0
4 years ago
Read 2 more answers
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