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2 years ago

What is 1.3 for PE=mgh and for KE=1(m)v^2 and E=PE+KE and V

Physics

1 answer:

emmainna [20.7K]2 years ago

8 0

Answer:

1.3 is equal to m times v to the second times PK, meaning the answer to your question is 3.3 K

Explanation:

Have a gday!

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An airplane is moving at 350 km/hr. If a bomb is

bogdanovich [222]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final height

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb's initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's final velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity </h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

3 years ago

What is the speed shown in the graph? which number do i start on?

den301095 [7]

Answer:

4 m/s

Explanation:

speed = distance/time

speed= 20/5 = 4

similarly for all no. the answer is constant,i.e. 4

3 0

3 years ago

Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the

Assoli18 [71]

Answer:

Part a)

$\rho = 3\epsilon_0 k r^2$

Part b)

$Q = 4\pi \epsilon_0kR^5$

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

$E = kr^3 \hat r$

now electric flux through a spherical surface of radius r is given as

$\phi = E. A$

$\phi = kr^3(4\pi r^2)$

now by Guass law we know that

$E.A = \frac{q}{\epsilon_0}$

$q = 4\pi \epsilon_0kr^5$

now volume charge density is given as

$\rho = \frac{q}{\frac{4}{3}\pi r^3}$

$\rho = 3\epsilon_0 k r^2$

Part b)

Total charge inside the radius R is given as

$Q = 4\pi \epsilon_0kR^5$

7 0

3 years ago

A friend of yours who has not taken an astronomy class looks at your textbook and really likes the picture of the Pleiades, a cl

Andrei [34K]

Answer:

<em>C. the blue colour of the Earth's sky</em>

Explanation:

The Pleiades is a cluster of sister stars that are among the closest star cluster to earth.

The reflection nebula of the Pleiades is due to the scattering of the blue light from the hot blue luminous stars that dominate the star cluster. Th blue light is scattered from dust molecules, thought to be predominantly carbon compound like diamond dusts, and other compounds like iron.

The blue colour of the Earth's sky is the closest terrestrial phenomenon to the reflection nebula. On a clear cloudless day, molecules in the air scatter the blue component of light more than the other component colours of white light, giving the sky its characteristic blue coluor.

The common characteristics of the luminous nebula and the Earth's blue sky is that they both have their light scattered by the presence of small particles.

8 0

3 years ago

What percent of the world lumber harvest goes to paper production<br> 10%<br> 20%<br> 25%<br> 50%

sergejj [24]

The answer is 50% of the paper production

8 0

3 years ago

What is 1.3 for PE=mgh and for KE=1(m)v^2 and E=PE+KE and V​

What is 1.3 for PE=mgh and for KE=1(m)v^2 and E=PE+KE and V