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1 year ago

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QC Suppose you hear a clap of thunder 16.2s after seeing the associated lightning strike. The speed of light in air is 3.00× 10⁸

m/s . (a) How far are you from the lightning strike?

1 answer:

MAVERICK [17]1 year ago

7 0

If you hear a clap of thunder in a time of 16.2s after seeing the associated lightning strike, you are: 5508 m far from the lightning strike

To solve this problem we must consider that the speed of light is greater than the speed of sound, therefore to calculate the distance we must use the speed of sound (340 m/s).

The formula and procedure we will use to solve this exercise is:

x = v * t

Where:

x = distance
t = time
v = velocity

Information about the problem:

v(sound) = 340 m/s
t = 16.2 s
x=?

Applying the distance formula we have that:

x = v * t

x= 340 m/s * 16.2 s

x = 5508 m

<h3>What is velocity?</h3>

It is a physical quantity that indicates the displacement of a mobile per unit of time, it is expressed in units of distance per time, for example (miles/h, km/h).

Learn more about velocity at: brainly.com/question/80295?source=archive

#SPJ4

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A moving object has a kinetic energy of 150 J and a momentum of 25.1 kg·m/s.

PolarNik [594]

Answer:

11.95m/s

Explanation:

A moving object has a kinetic energy of 150 J and a momentum of 25.1 kg·m/s.

a) Find the speed of the object. Answer in units of m

K. E =½mv²

150= ½mv²

Multiply both sides by 2

mv² = 300

Divide both sides by v²

m = 300/v² .................. Equation 1

Momentum is the product of mass and velocity

Momentum = mv

25.1 = mv

Divide both sides by v

m = 25.1/v ................ Equation 2

Equate equations 1 and 2

300/v² = 25.1/v

Cross multiply

25.1v² = 300v

Multiply v with both sides

25.1v = 300

Divide both sides by 25.1

v = 300/25.1

V = 11.95m/s

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3 years ago

Define the term dimension<br>

Vladimir79 [104]

Answer:

Q1. A measurable extent of a particular kind, such as length, breadth, depth, or height.

Q2. A dimensional constant is a physical quantity that has dimensions and has a fixed value. Some of the examples of the dimensional constant are Planck's constant, gravitational constant, and so on.

Q3. Physical quantities which posses dimensions and have variable are called dimensional variables. Examples are length, velocity, and acceleration etc.

Q4. Dimensionless variables are the quantities which doesn't have any dimensions the the value is a variable. Eg: angle = arc/ radius. Dimensions = L/L. = 1. So angle does not have any dimensions and the value can vary.

Q5. Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. This principle is helpful because it helps us convert the units from one form to another.

Q6. Dimensional analysis has been around a long time, Newton called it the "Great principle of Similitude", but the modern form can be traced back to James Clerk Maxwell. It was Maxwell who distinguished mass [A/], length [£], and time [7"] as the independent dimensions from which others could be derived.

Q7. Mass, length, time, temperature, electric current, amount of light, and amount of matter.

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2 years ago

Cuales son los factores para que un sobre conductor se genere y aumente la energía termíca

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2 years ago

You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe agai

QveST [7]

Answer:

Explanation:

Given mass of grindstone $m=90\ kg$

radius of stone $r=0.34\ m$

angular speed of disc $\omega =90\ rpm$

Steel Axle applying a force of $F=20\ N$

coefficient of kinetic friction $\mu =0.2$

Frictional Torque applied by steel is given by

$\tau=r\times f_r$

$\tau =r\times \mu F$

where $f_r$ =frictional force

$\tau =r\times \mu \times F$

$\tau =0.34\times 0.2\times 20$

$\tau =1.36\ N-m$

Torque is also given by

$\tau =I\cdot \alpha$

where $\alpha$ =angular acceleration

I=moment of Inertia

$\tau =0.5Mr^2\times \alpha$

$0.5Mr^2\times \alpha =1.36\ N-m$

$0.5\times 90\times 0.34^2\times \alpha =1.36$

$\alpha =0.261\ rad/s^2$

3 0

3 years ago

A 11.3-kg object oscillates at the end of a vertical spring that has a spring constant of 2.20 ✕ 104 N/m. The effect of air resi

3241004551 [841]

Answer:

(a) the frequency of the dampened oscillation is 7.02 Hz

(b) percentage decrease in amplitude of the oscillation in each cycle is 2%

Explanation:

Given;

mass of the object = 11.3 kg

the spring constant = 2.2 X 10⁴ N/m.

damping coefficient b = 3.00 N · s/m

Part (a) the frequency of the dampened oscillation

The oscillation frequency is calculated as follows;

$\omega _D = \sqrt{\omega_o^2 -(\frac{b}{2m})^2}\\\\\omega_o^2 = \frac{k}{m} =\frac{2.2X10^{4}}{11.3} = 1946.903rad/s\\\\thus, \omega _D = \sqrt{1946.903-(\frac{3}{2*11.3})^2} =44.12 rad/s$

The damped frequency $= \frac{\omega _D}{2\pi } = \frac{44.12}{2\pi } = 7.02 Hz$

Part (b) percentage decrease in amplitude of the oscillation in each cycle

The amplitude of the oscillation depends on the damping coefficient (b) and period (T), and it is given as;

$A(t) = e^{-\frac{b}{2m}(t)}$

After one cycle, the amplitude changes from A(t) to A(t+T), where T is the period of the oscillation.

$A(t +T) = e^{-\frac{b}{2m}(t+T)}$

Percentage decrease in amplitude is gotten by dividing A(t) by A(t+T)

$= \frac{e^{-\frac{b}{2m}(t)}}{e^{-\frac{b}{2m}(t+T)}} =e^{-\frac{b}{2m}(T)}$

But T = 1/f

Substituting the values of the parameters in the above equation, we will have;

$=e^{-\frac{b}{2m}(T)} = e^{-\frac{3}{2X11.3}(\frac{1}{7.02})} = 0.98$

Percentage decrease = 1 - 0.98 = 0.02 = 2%

4 0

3 years ago