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3 years ago

15

Answers: 64% 36% 61% 39% Please explain reason!

1 answer:

NeTakaya3 years ago

3 0

It is 61 percent in nasisidinsnskscc

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Please help. Questions are in the image.

yuradex [85]

Speed: measure of how fast an object travels
Velocity: measure of how fast, and in what direction, an object travels
Distance: how far an object has traveled on some path
Position: where an object is located in some reference system
Displacement: the difference between an objects starting position and it’s ending position

8 0

3 years ago

The rate at which energy is transferred is called a. joules. b. power. c. work. d. time.

Marrrta [24]

B. Power
I hope this helps you, have a great day!

7 0

3 years ago

Read 2 more answers

kakasveta [241]

Answer:

Wavelength!

Explanation:

At least I think? Or wavelength might be crest to crest! Sorry if I'm incorrect. Let me know how I did!

8 0

3 years ago

g A spherical container of inner diameter 0.9 meters contains nuclear waste that generates heat at the rate of 872 W/m3. Estimat

zhannawk [14.2K]

Answer: The total rate of heat transfer from the container to its surroundings ignoring radiation is 332.67 W.

Explanation:

Given: Inner diameter = 0.9 m

q = 872 $W/m^{3}$

Now, radii is calculated as follows.

$r = \frac{diameter}{2}\\= \frac{0.9}{2}\\= 0.45 m$

Hence, the rate of heat transfer is as follows.

$Q = q \times V$

where,

V = volume of sphere = $\frac{4}{3} \pi r^{3}$

Substitute the values into above formula as follows.

$Q = q \times \frac{4}{3} \pi r^{3}\\= 872 W/m^{3} \times \frac{4}{3} \times 3.14 \times (0.45 m)^{3}\\= 332.67 W$

Thus, we can conclude that the total rate of heat transfer from the container to its surroundings ignoring radiation is 332.67 W.

3 0

3 years ago

A U-tube is open to the atmosphere at both ends. Water is poured into the tube until the water column on the vertical sides of t

zmey [24]

Answer:

The value is $\Delta h = 0.003 \ m$

Explanation:

From the question we are told that

The height of the water is $h_1 = 10 \ cm = 0.10 \ m$

The density of oil is $\rho_o = 950 \ kg/m^3$

The height of oil is $h_2 = 6 \ cm = 0.06 \ m$

Given that both arms of the tube are open then the pressure on both side is the same

So

$P_a = P_b$

=> Here

$P_a = P_z + \rho_w * g * h$

where $\rho_w$ is the density of water with value $\rho_w = 1000 \ kg/m^3$

and $P_z$ is the atmospheric pressure

and

$P_b = P_z + \rho_o * g * h_2$

=> $P_z + \rho_w * g * h = P_z + \rho_o * g * h_2$

=> $\rho_w * h = \rho_o * h_2$

=> $h = \frac{950 * 0.06 }{1000}$

=> $h = 0.057 \ m$

The difference in height is evaluated as

$\Delta h = 0.06 - 0.057$

$\Delta h = 0.003 \ m$

6 0

4 years ago