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iogann1982 [59]
1 year ago
7

In each part calculate the kinetic energy of the given objects in joules a) an automobile whose mass is 1260kg moving at a speed

of 66.3km/h b) a runner whose mass is 64.7 kg, moving at a speed of 8.52m/s c) an electron (mass 9.11x10^31kg) moving at a speed of 2.74x10^7m/s
Physics
1 answer:
Contact [7]1 year ago
5 0

a)

Kinetic energy (KE) = 1/2 m v^2

Where:

m = mass = 1260 kg

v = speed = 66.3 km/h = 18.42 m/s

Replacing:

KE = 1/2 (1260 kg ) (18.42 m/s )^2 = 213,756.7 J

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A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes
OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

y(t) = h +u_y t - \frac{1}{2}gt^2

where

h = 2.5 km = 2500 m is the initial height

u_y = 0 is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

y(t) = 0

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s

So the horizontal distance travelled is

d=v_x t

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

d=(333.3)(22.6)=7533 m = 7.5 km

7 0
3 years ago
A light wave travels through air at a speed of 3.0x108 m/s. Green light has a wavelength of about 5.76x1014Hz. What is the wavel
solong [7]

Answer:

521 nm

Explanation:

Given the values and units we are given, I'm assuming  5.76*10^14 Hz is frequency.

The formula to use here is λ * υ = c, where λ is wavelength, υ is frequency, and c is the speed of light.

λ = \frac{3*10^8\frac{m}{s} }{5.76*10^{14}Hz} = {5.20833*10^{-7} m}\approx{521 *10^{-9}m}={521 nm}

4 0
3 years ago
El peso de María es de 617,4 N. Si en dos meses pierde 2,5 Kg de masa, ¿cuál será su nuevo peso?
NeTakaya
Two hundred and thirsty years
5 0
3 years ago
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suppose that 273 g of one of the substances listed above displaces 26 mL of water. What is the substance?
guajiro [1.7K]
<span>The unknown substance is silver. I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so 273 g / 26 mL = 10.5 g/mL Looking up a list of elements sorted by density, I see the following: 10.07 Actinium 10.22 Molybdenum 10.5 Silver 11.35 Lead And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.</span>
6 0
4 years ago
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Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The long
nordsb [41]

Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

R = \dfrac{\rho L }{A}

where;

A = πr²

R = \dfrac{\rho L }{\pi r ^2}

For the shorter cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{ 4 \rho L }{\pi \ D   ^2}

For the longer cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{32\rho L }{\pi \ (4 D)   ^2}

R = \dfrac{2\rho L }{\pi \ (D)   ^2}

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D   ^2}}{ \dfrac{2\rho L }{\pi \ (D)   ^2}}

\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D   ^2}}* { \dfrac  {\pi \ (D)   ^2} {2\rho L}}

\dfrac{R_s}{R_L} =2

{R_s}=2{R_L}

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

7 0
3 years ago
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