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ICE Princess25 [194]
3 years ago
15

Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimu

m necessary. find the magnitude of the frictional force between the person and the wall.
Physics
2 answers:
Naddika [18.5K]3 years ago
7 0
In the magnitud is in good
Kamila [148]3 years ago
7 0

Answer:

friction force must be equal to the magnitude of weight of the object

Explanation:

When the object is stuck against the wall

then we will have

F_n = \frac{mv^2}{r}

also we know that

mg = F_f

so if the block is at rest and not sliding then in vertical direction the force must be balanced

so here we can say

F_g = F_f

so friction force must be equal to the magnitude of weight of the object

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A 90 kg painter is standing on a horizontal wooden scaffolding of length 10 m, which is supported on each end by a rope. The pai
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Explanation:

For equilibrium, \sum M = 0.

So,   8 m \times mg - (10 m) T_{1} = 0

             T_{1} = \frac{8 \times mg}{10}

                        = \frac{8 \times 90 \times 9.8}{10}

                        = 705.6 N

Also, for equilibrium \sum F_{y} = 0

              T_{1} + T_{2} - mg = 0

or,         T_{2} = mg - T_{1}

                        = 90 \times 9.8 - 705.6

                        = 176.4 N

Thus, we can conclude that the tension in the first rope is 176.4 N.

8 0
3 years ago
The distance between a charge and the source of an electric field changes from 3 mm to 6 mm. as a result of the change, the elec
STatiana [176]

The electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.

<h3>What is electric potential energy?</h3>

Electric potential energy can be defined as the energy needed to move a charge against an electric field.

It is calculated using the formula;

U = Kq1 q2 ÷ r

Where Q = electric potential energy

k = Coulombs constant

q1 and q2 = charges

r = distance of separation

Electric potential energy is inversely proportional to the distance of separation of the charges.

If the distance of the charges changes from 3mm to 6mm, then the electric potential energy of the charges is reduced because it decreases with increase in the distance of the charges.

Therefore, the electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.

Learn more about electric potential energy here:

brainly.com/question/14812976

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4 0
1 year ago
Distance meters Time (seconds) Trial Tria 2 Trials Average 3 053 059 0.52 055 6 106 719 104 Uning the values trom the data table
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5 0
2 years ago
Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da
valentinak56 [21]

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

k = 3906250 N/m

Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

3 0
2 years ago
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