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Maru [420]
2 years ago
6

1. What are the two factors affecting friction

Physics
1 answer:
bija089 [108]2 years ago
5 0
Answer: a.) Roughness of the surfaces in contact with each other .
Higher the roughness of surfaces in contact with each other, greater is the friction between bodies. Force of friction will be less between smooth surfaces.

b.) Weight of the sliding/rolling body: greater the weight of the moving body on the surface, more is the force of friction on the body by the surface.

I hope this helps

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Carbon dioxide being released from a fire extinguisher a good example of how volume __________.
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"<span>increases when pressure decreases". Pressure and volume of gasses are related from Boyle's law, which states that Pressure is proportional to 1/V, so as pressure decreases, volume increases. </span>
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3 years ago
During meiosis maternal and paternal chromosomes that occur in a gamete line up and exchange segments in a process called crossi
sergiy2304 [10]

Explanation:

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3 years ago
Read 2 more answers
A box at rest is in a state of equilibrium half way up on a ramp. The ramp has an incline of 42° . What is the force of static f
olchik [2.2K]

Answer:

Explanation:

The force of static friction acting on the box is the frictional force;

Frictional force Ff = Wsin theta (force acting along the ramp)

W is the gravitational force known as the weight

Ff = 112.1sin42°

Ff = 112.1(0.6691)

Ff = 75.00N

Hence he force of static friction acting on the box  if box has a gravitational force of 112.1 N id 75.00N

4 0
3 years ago
The legend that Benjamin Franklin flew a kite as a storm approached is only a legend; he was neither stupid nor suicidal. Suppos
UNO [17]

Answer:

The current on the water layer = 1.64×10^-3A

Explanation:

Let's assume that the radius given for the string originates from the centre of the string. The equation for determining the current in the water layer is given by:

I = V × pi[(Rwater + Rstring)^2 - (Rstring)^2/ ( Resitivity × L)

I =[ 166×10^6 ×3.142[(0.519×10^-4) + (2.15×10^-3])^2 - ( 2.15×10^-3)^2] / ( 183 × 831)

I =[ 521572000(4.848×10^6)- 4.623×10^-6]/ 154566

I = 252.83 -(4.623×10^-6)/ 154566

I = 252.83/154566

I = 1.64× 10^-3A

3 0
3 years ago
A Cessna 150 aircraft has a lift-off speed of approximately 125 km/h. What minimum constant acceleration does this require if th
aivan3 [116]

Answer:

Part a)

a = 3.65 m/s^2

Part b)

t = 9.5 s

Part c)

v_f = 55.1 m/s

Explanation:

Part a)

As we know that it starts from rest and moves on runway by total distance 165 m

so we will have

v_f^2 - v_i^2 = 2ad

v_f^2 - 0 = 2(a)(165)

v_f = 125 km/h = 34.7 m/s

now we have

a = 3.65 m/s^2

Part b)

Now for take off time we will have

v_f - v_i = at

34.7 - 0 = 3.65 t

t = 9.5 s

Part c)

v_f = v_i + at

v_f = 0 + (3.65)(15.1)

v_f = 55.1 m/s

7 0
3 years ago
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