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Goryan [66]
3 years ago
11

Difference between constant speed and average speed

Physics
1 answer:
Tresset [83]3 years ago
8 0
If a body is traveling with constant speed , it means that it's distance is constantly increasing with time.

If it goes 5m in 3min , it will go 5m in next 3 min.

Average velocity is the total displacement divided by total time.

When the body travels 5m in 3 min and 25 m in next 3min , average velocity
=(25+5)/(3+3)
=30/6
=5m/min
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HELP ME PLEASE!!!!!!!!!
Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

F = \frac{kq_1q_3}{d_{13}^2}

now from the above equation

F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}

F = 0.288 N

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

F_{net} = 2Fcos\theta

here we know that angle is

\theta = 37 degree

now we have

F_{net} = 2\times 0.288 cos37

F_{net} = 0.46 N

so net force on q3 is 0.46 N vertically upwards along +Y axis

6 0
3 years ago
Jaime lifts a package weighing 75N. if she lifts it 1.2 m, what work has she done
9966 [12]
Answer:
90 J
Explanation:
W=fd
W=(75)(1.2)
W= 90 J
6 0
3 years ago
In regions where few species existed before or where species were wiped out what occurs?
marysya [2.9K]
I cant really say but i believe it is called extinction
5 0
3 years ago
The image formed by a plane mirror is _____.
Thepotemich [5.8K]
The image formed by a plane mirror is virtual, upright and the same size with the actual object. The upright image of an object in a plane mirror is can be found on the other side of the mirror which is why it is also virtual. 
4 0
3 years ago
The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a
mario62 [17]

Answer:

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a)Given I = 55A−(0.65A/s2)t²

I = dQ/dt

dQ = I×dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

Q = [55t – 0.65/3×t³]

Q between t=0 and t= 7.5s

Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.

3 0
3 years ago
Read 2 more answers
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