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igor_vitrenko [27]
3 years ago
7

Why do temperatures differ between air, land, and water?

Physics
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer:the ocean water isn't moving into or through the atmosphere as a result of the sun's heating of the water, convection isn't the process influencing air and water temperature difference. The heat of the ocean is being given off radiated into the air, making the air substantially warmer

Explanation:

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What color do blind people see lol
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I think most see black and white.

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Where are you likely to find a cooling plant located in the ductwork? A. A medium-sized office B. A large factory C. A single fa
Rufina [12.5K]

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i think it would be B, a large factory

Explanation:

8 0
3 years ago
If =12a andthe distance from each wire to point p is 0.12m, then what is the magnitude of the magnetic force per unit length on
vaieri [72.5K]

The magnitude of the magnetic force per unit length on the top wire is

2×10⁻⁵  N/m

<h3>How can we calculate the magnitude of the magnetic force per unit length on the top wire ?</h3>

To calculate the magnitude of the magnetic force per unit length on the top wire, we are using the formula

F= \frac{\mu_0 I_f}{2\pi d}

Here we are given,

\mu_0= magnetic permeability

= 4\pi×10⁻⁷ H m⁻¹

If= 12 A

d= distance from each wire to point.

=0.12m

Now we put the known values in the above equation, we get

F= \frac{\mu_0 I_f}{2\pi d}

Or, F = \frac{4\pi \times 10^{-7}\times  12}{2\pi \times 0.12}

Or, F= 2×10⁻⁵ N/m.

From the above calculation, we can conclude that the magnitude of the magnetic force per unit length on the top wire is 2×10⁻⁵ N/m.

Learn more about magnetic force:

brainly.com/question/2279150

#SPJ4

7 0
2 years ago
A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
3 years ago
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