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dybincka [34]
3 years ago
11

A 65.0-kg woman steps off a 10.0-m diving platform and drops straight down into the water. 1) If she reaches a depth of 3.40 m,

what is the average resistance force exerted on her by the water
Physics
1 answer:
babymother [125]3 years ago
8 0

Answer:

19372.29 N

Explanation:

Kinetic Energy of the man = Work done by the average resistive force of water.

mg(H+d) = F×d............................ Equation 1

where, m = mass of the woman, H = Height of the platform, d = depth of water reached, F = average resistive force exerted by water.

make F the subject of the equation,

F = mg(H+d)/d...................... Equation 2

Given: m = 65.0 kg, H = 10 m, d = 3.4 m, g = 9.8 m/s²

Substitute into equation 2

F = 65(9.8)(100+3.4)/3.4

F = 19372.29 N

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A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl
Firdavs [7]

Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

m (Mass)=3.0 kg

α(incline) =34°

Spring Constant (force constant)=120 N/m

d (distance)=?

Solution

F=mg

F=(3.0)(9.8)

F=29.4 N

As we also know that

Force parallel to the incline=FSinα

F=29.4×Sin(34)

F=16.44 N

d(distance)=F/Spring Constant

d(distance)=16.44/120

d(distance)=0.137 m ⇒13.7 cm

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3 years ago
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State the success of j j Thompson model theory​
vovikov84 [41]

Answer:

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A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it
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The charge of the object must be 1.11 \times e^{-5} \text { coulomb }

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

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Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

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3 years ago
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Answer:

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