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Naddika [18.5K]
3 years ago
8

Objects falling through air experience a type Of friction called ____

Physics
2 answers:
DaniilM [7]3 years ago
6 0

Answer: That's air resistance.

Explanation: Well, air resistance is an upward force exerted on falling objects.

( I hope this helped <3 )

77julia77 [94]3 years ago
4 0

Answer:air resistance

Explanation:

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A boiling liquid absorbs thermal energy (heat) at a rate of 450 W. The specific latent heat of vaporisation is 2.7 × 106 J / kg
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3 years ago
Students have a small steel ball and a large steel ball, and they have a short ramp and a tall ramp.
lozanna [386]
I would say it’s “B” for the reason that the student did not experiment with a short ramp and large ball as well as a large ramp and a short ball. They have to do those experiments as well before drawing their conclusion.
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1 year ago
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The bolts on a car wheel require tightening to a torque of 4 N*m. If a 9 cm long wrench is used, what is the magnitude of the fo
notsponge [240]

Answer:

320N

Explanation:

The magnitude of the torque required is expressed using the formula;

T = Fr sin theta where;

F is the force

r is the radius = 9cm = 0.09m

theta is the angle of inclination = 8 degrees

Torque T = 4Nm

Substitute the given values and get F

4 = F(0.09)sin8

4 = 0.0125F

F = 4/0.0125

F = 320N

Hence the magnitude of the force required when the force is applied at 8 degrees to the wrench is 320N

7 0
2 years ago
Your best friend weighs 81.5 kg and is a rugby player. In one of his games, he slides to a stop in a phenomenal manner. The coef
Alex777 [14]

A. The acceleration during the slide is 6.86 m/s²

B. The time taken to slide until he stops is 1.2 s

<h3>How to determine the force of friction</h3>
  • Mass (m) = 81.5 Kg
  • Coefficient of friction (μ) = 0.7
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Normal reaction (N) = mg = 81.5 × 9.8 = 798.7 N
  • Frictional force (F) =?

F = μN

F = 0.7 × 798.7

F = 559.09 N

<h3>A. How to determine the acceleration</h3>
  • Mass (m) = 81.5 Kg
  • Frictional force (F) = 559.09 N
  • Acceleration (a) =?

a = F / m

a = 559.09 / 81.5

a = 6.86 m/s²

<h3>B. How to determine the time </h3>
  • Initial velocity (u) = 8.23 m/s
  • Final velocity (v) = 0 m/s
  • Decceleration (a) = -6.86 m/s²
  • Time (t) =?

a = (v – u) / t

t = (v – u) / a

t = (0 – 8.23) / -6.86

t = 1.2 s

Learn more about acceleration:

brainly.com/question/491732

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4 0
1 year ago
A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k
Luba_88 [7]

Answer:

The frictional force  F_{fri} = 6.446 N

The acceleration of the block a = 6.04 \frac{m}{s^{2} }

Explanation:

Mass of the block = 3.9 kg

\theta = 40°

\mu = 0.22

(a). The frictional force is given by

F_{fri} = \mu R_{N}

R_{N} = mg \cos \theta

R_{N} = 3.9 × 9.81 × \cos 40

R_{N} = 29.3 N

Therefore the frictional force

F_{fri} = 0.22 × 29.3

F_{fri} = 6.446 N

(b). Block acceleration is given by

F_{net} = F - F_{fri}

F = 30 N

F_{fri} = 6.446 N

F_{net} = 30 - 6.446

F_{net} = 23.554 N

The net force acting on the block is given by

F_{net}  = ma

23.554 = 3.9 × a

a = 6.04 \frac{m}{s^{2} }

This is the acceleration of the block.

8 0
2 years ago
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