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3 years ago

If you triple the speed from 20km/h to 60km/h, by what factor does the KE increase?

Physics

1 answer:

sertanlavr [38]3 years ago

5 0

Answer:

<em>The kinetic energy increases by a factor of 9 </em>

Explanation:

<u>Kinetic Energy </u>

It's the energy related to the speed of an object and is calculated by the formula

$\displaystyle K=\frac{mv^2}{2}$

Let's suppose the initial speed of the mass is $v_o$ , then the kinetic energy is

$\displaystyle K_o=\frac{mv_o^2}{2}$

Now the speed will triple, i.e. $v_1=3v_o$ . The new kinetic energy is

$\displaystyle K_1=\frac{mv_1^2}{2}$

$\displaystyle K_1=\frac{m(3v_o)^2}{2}$

$\displaystyle K_1=9\frac{mv_o^2}{2}$

Thus

$K_1=9K_o$

The kinetic energy increases by a factor of 9

Let's check with numbers: $v_o=20,\ v_1=60$

$\displaystyle K_o=\frac{m20^2}{2}=200m$

$\displaystyle K_1=\frac{m60^2}{2}=1800m$

The relation is

$\displaystyle \frac{K_1}{K_o}=9$

The new kinetic energy is nine times the initial kinetic energy.

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What is the voltage across each resistor?

kogti [31]

Answer:

see solution below

Explanation:

The given resistors are connected in series.

Equivalent resistance in series = 30 + 55 + 15

Equivalent resistance in series Rt = 100 ohms

Since the potential difference in the circuit = 36V

Get the current in the circuit first

I = V/Rt

I = 36/100

I = 0.36A

Get the voltage across 30ohms resistor;

V30 = 0.36 * 30

V30 = 10.8volts

Hence the voltage across the 30ohms resistor is 10.8volts

Get the voltage across 55ohms resistor;

V55 = 0.36 * 55

V55 = 19.8volts

Hence the voltage across the 55ohms resistor is 19.8volts

Get the voltage across 15ohms resistor;

V15 = 0.36 * 15

V15 = 5.4volts

Hence the voltage across the 15ohms resistor is 5.4volts

4 0

3 years ago

A photon detector captures a photon with an energy of 4.29 ✕ 10−19 J. What is the wavelength, in nanometers, of the photon?

serious [3.7K]

Answer : The wavelength of photon is, $4.63\times 10^{2}nm$

Explanation : Given,

Energy of photon = $4.29\times 10^{-19}J$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency of photon

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength of photon = ?

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$4.29\times 10^{-19}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}$

$\lambda=4.63\times 10^{-7}m=4.63\times 10^{-7}\times 10^9nm=4.63\times 10^{2}nm$

Conversion used : $1nm=10^{-9}m$

Therefore, the wavelength of photon is, $4.63\times 10^{2}nm$

6 0

3 years ago

On a trip to the Colorado Rockies, you notice that when the freeway goes steeply down a hill, there are emergency exits every fe

Zanzabum

Answer:

The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.

Explanation:

4 0

3 years ago

The center of gravity of an ax is on the centerline of the handle, close to the head. Assume you saw across the handle through t

andrew-mc [135]

Answer:

I believe it is they will weigh the same

Explanation:

Center of gravity is the axis on which the mass rotates evenly if I remember correctly from AP Physics

3 0

3 years ago

A man lifts a 120 kg barbell 2 m above the ground . What is the gain in gravitational PE of the barbell?

SpyIntel [72]

Answer:

2,352 Joules

Explanation:

At the ground, the barbell has a classical mechanical energy value of zero. There is no classical kinetic or potential energy for the barbell. The moment the man starts to lift the barbell, he does work on the barbell and transfers kinetic energy to it due to the motion. At its maximum height where the man lifts the barbell to a stop, the kinetic energy is zero because it transformed into gravitational potential energy stored in the gravitational field. Our reference point for potential was defined to be zero at the floor, therefore we can say that the gravitational potential energy at 2 meters is:

$U=mgh=(120kg)(9.8m/s^2)(2m)=2,352J$

7 0

2 years ago