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Mademuasel [1]
3 years ago
13

If are spaced closely together on the map,there is a drastic temperature change over the distance

Physics
2 answers:
AlexFokin [52]3 years ago
6 0

If <em>the isotherms</em> are spaced closely together over some portion of the map, there is a drastic temperature change over that portion.

aev [14]3 years ago
6 0

Answer:

isotherms

Explanation:

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A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work
kifflom [539]

Answer: Option D: 5.5×10²Joules

Explanation:

Work done is the product of applied force and displacement of the object in the direction of force.

W = F.s = F s cosθ

It is given that the force applied is, F = 55 N

The displacement in the direction of force, s = 10 m

The angle between force and displacement, θ = 0°

Thus, work done on the object:

W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J

Hence, the correct option is D.

3 0
3 years ago
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
4 years ago
How much work is done when an engine generates 400 Watts of power in 25 seconds?
IceJOKER [234]

Answer:

10000 J or 10 KJ

Explanation:

power = workdone/time taken

400 = workdone/25

workdone = 400 * 25

=10000 J

8 0
3 years ago
Read 2 more answers
A person is riding a motorized tricycle. They weigh 180kg and are moving at 3 m/s over a distance of 300 m. How much work is don
agasfer [191]

If I am to understand this question correctly this is what asks you:

If a person is riding a motorized tricycle how much work do they do?

You may ask yourself, why did I only use part of the question. Simple, the rest is not relevant to what is being asked. The weight, speed, and distance wont affect the person riding any <em><u>motorized vehicle</u></em> other than the time it takes to get from one place to another.

So to answer this question I would say:

Not much, all they really have to do is to steer and set the motorized tricycle to cruise control. Just like any rode certified vehicle.

If you have any questions about my answer please let me know and I will be happy to clarify any misunderstandings. Thanks and have a great day!

3 0
3 years ago
In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 1.00 mm away. part a
mihalych1998 [28]
<span>We can use Coulomb's law to find the force F acting on the proton that is released. F = k x Q1 x Q2 / r^2 k = 9 x 10^9 Q1 is the charge on one proton which is 1.6 x 10^{-19} C Q2 is the same charge on the other proton r is the distance between the protons F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2 F = 2.304 x 10^{-22} N We can use the force to find the acceleration. F = ma a = F / m a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg) a = 1.38 x 10^5 m/s^2 The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
8 0
3 years ago
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