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Yuliya22 [10]
3 years ago
8

Sue and Jenny kick a soccer ball at exactly the same time. Sue’s foot exerts a force of 75.9 N to the north. Jenny’s foot exerts

a force o 105.8 N to the east. What is the magnitude of the resultant force on the ball?
Physics
1 answer:
Lady_Fox [76]3 years ago
6 0

Answer:

Fr^2 = 75.9N+105.8N=181.7

<u><em>Fr = </em></u><u><em>181.7N.</em></u>

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What is the mechanical advantage of the wheel and axle shown below?
goldenfox [79]

Answer:

B. 120

Explanation:

24/.2 = 120

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Which of the following is not an example of an engine?
SashulF [63]
The answer is A. waterfall
To be considered as an engine , it should be a Man-made objects that could be used to produce power that creates motions.
From all the options above could be used to produce such power, but the waterfall is not made by mandkind
4 0
3 years ago
A water tank is filled with water. What will be the pressure of the water when the level of the water is 6m?
Luden [163]

Answer:

pressure = density x g x height

= 1000 x 10 x 6 Pascal

=60000 Pascal

OR 60 kP

3 0
3 years ago
Read 2 more answers
A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffici
Ivenika [448]

Answer:

Coefficient of friction will be 0.296

Explanation:

We have given initial speed of the stone u = 8 m /sec

It comes to rest so final speed v = 0 m /sec

Distance traveled before coming to rest s = 11 m

According to third equation of motion

v^2=u^2+2as

So 0^2=8^2+2\times a\times 11

a=\frac{-64}{22}=-2.90m/sec^2

Acceleration due to gravity g=9.8m/sec^2

We know that acceleration is given by

a=\mu g

So 2.90=9.8\times \mu \\

\mu =\frac{2.9}{9.8}=0.296

So coefficient of friction will be 0.296

3 0
3 years ago
In Fig. P24.59, each capacitance C1 is 6.9 mF, and each capacitance C2 is 4.6 mF. (a) Compute the equivalent capacitance of the
mrs_skeptik [129]

(a) Equivalent capacitance of network between points a and b is 2.3μF.

(b) Charge on each of the three capacitors nearest a and b is 920 μC.

A) Let's consider the right side that is farthest from the point 'a'.

Three C1 are connected in series

C= 1/C1 + 1/C1 +1/C1 = C1/3

C = 6.9/3 = 2.3μF

Now this C is connected in parallel with C2

So, C= 2.3 + 4.6 = 6.9μF

Again we get three capacitors of 6.9μF each, connected in series.

C = C1/3 = 2.3μF

It again combines with 4.6μF in parallel

C = 4.6 + 2.3 =6.9μF

Now, this final reduction is the same as that of the first.

In the end, we have 3 capacitors of 6.9μF each connected in series.

Equivalent Capacitance C(eq) = C1/3 = 6.9/3 = 2.3μF

         = 2.3 × 10^{-6} F

B) C(eq) = 2.3 × 10^{-6} F

Vab = 400 V (Given)

Charge on each of the three capacitors nearest a and b (Q) =?

Q = C(eq) × Vab

   = 2.3 × 10^{-6} × 400 = 9.2 × 10^{-4} C = 920μC

Hence, the equivalent capacitance of the network between points a and b is 2.3μF.

And Charge on each of the three capacitors nearest a and b is 920 μC.

Learn more about Capacitance here brainly.com/question/13578522

#SPJ1

8 0
2 years ago
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