What energy is directly dependent upon velocity and mass?

Which term describes the increase in a material's volume due to an increase in temperature?

AlladinOne [14]

Thermal Expansion. best describes the increase in a materials volume

7 0

2 years ago

A submarine is 2.84 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the

xxMikexx [17]

Answer:

468 m

Explanation:

So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.

The length CE is length AE - length AC = 284 - 234 = 50 m

We can calculate the angle ECD:

$tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2$

$\hat{ECD} = tan^{-1} 2 = 63.43^o$

This is also the angle ACB, so we can find the length AB:

$tan(\hat{ACB}) = \frac{AB}{AC} = \frac{AB}{234}$

$2 = \frac{AB}{234}$

$AB = 2*234 = 468 m$

So the height of the building is 468m

5 0

3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

1 year ago

Fear me or i am glory i give you free p oints \ / _| |_

8_murik_8 [283]

Answer:

kayyy

Explanation:

3 0

2 years ago

Read 2 more answers

If m represent mass in kg, v represents speed in m/s and r represents radius in m show F in the formula F= (mv^2)/r can be expre

Dmitrij [34]

M represent mass in kg
v represents speed in m/s
r represents radius in m

Now, just substitute these into the formula:
 $F = \frac{m* v^{2} }{r} =\frac{kg* ( \frac{m}{s} )^{2} }{m} =\frac{kg* \frac{m^{2}}{s^{2}} }{m} = \frac{kg*m^{2}}{s^{2}*m } =\frac{kg*m}{s^{2} }$

3 0

3 years ago