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andrey2020 [161]
3 years ago
14

A firefighter of mass 81 kg slides down a vertical pole with an acceleration of 3 m/s 2 . The acceleration of gravity is 10 m/s

2 . What is the friction force that acts on him? Answer in units of N
Physics
1 answer:
natima [27]3 years ago
4 0

Answer:

The force of friction that acts on him is

F_k=567N

Explanation:

The firefighter with an acceleration of 3m/s^2 take the gravity acceleration as 10m/s^2 isn't necessary to know the coefficient of friction just to know the force of friction:

F=m*a

F=F_w-F_k

m*a=F_w-F_k

F_w=81kg*10m/s^2=810N

Sole to Fk

81kg*3m/s^2=810N-F_k

F_k=810N-243N

F_k=567N

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I do not know which substance you are referring to, but the freezing point of water is 32°F, or 0°C.

3 0
3 years ago
A 75kg hockey player is skating across the ice at a speed of 6.0m/s. What is the magnitude of the average force required to stop
liq [111]

Answer:

692.31 N

Explanation:

Applying,

F = ma............... Equation 1

Where F = Average force required to stop the player, m = mass of the player, a = acceleration of the player

But,

a = (v-u)/t............ Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t............ Equation 3

From the question,

Given: m = 75 kg, u = 6.0 m/s, v = 0 m/s (to stop), t = 0.65 s

Substitute these values into equation 3

F = 75(0-6)/0.65

F = -692.31 N

Hence the average force required to stop the player is 692.31 N

6 0
3 years ago
Which calculates the intensity of an electric field at a point where a 0.50 C charge experiences a force of 20. N?
tester [92]

Answer: 40\ N/C

Explanation:

Given

Magnitude of charge is q=0.5\ C

Force experienced is F=20\ N

Electric field intensity is the electrostatic force per unit charge

\therefore E=\dfrac{F}{q}\\\\\Rightarrow E=\dfrac{20}{0.50}\\\\\Rightarrow E=40\ N/C

Thus, the electric field intensity is 40\ N/C

6 0
3 years ago
The diagram does not represent a real electric field because the field lines, can someone help explain this for me
Papessa [141]

electric field lines are graphical presentation of electric field intensity

It is the graphical way to represent the electric field variation

If we draw the tangent to electric field line then it will give the direction of net electric field at that point

So whenever we draw the electric field lines of a charge distribution then it will always follow this basic properties

here we will always follow these basic properties of field lines

now as we can see that here two positive charges are placed nearby so the electric field must be like it can not intersect at any point because at intersection of two lines the direction of electric field not defined

As we have two directions of tangents at that point

So here the incorrect presentation is the intersection of two field lines which is not possible


4 0
4 years ago
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