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blagie [28]
3 years ago
15

What would be a good reason to increase friction between surfaces? to allow objects to slide past each other easily to reduce we

ar and tear to provide traction so that slipping does not occur to avoid contact that results in increased temperature and overheating
Physics
2 answers:
4vir4ik [10]3 years ago
6 0

Answer:

B to provide traction so that slipping does not occur

Explanation:

Lelu [443]3 years ago
5 0
So basically the objects would be sandpaper and smooth metal, the sandpaper can indirectly touch the metal since it’s so smooth and it won’t cause any temp change either
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A block of ice with mass 5.50 kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal
Yuki888 [10]

Answer:

A) 3.13 m/s

B) 5.34 N

C) W = 26.9 J

Explanation:

We are told that the position as a function of time is given by;

x(t) = αt² + βt³

Where;

α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³

Thus;

x(t) = 0.21t² + 0.0204t³

A) Velocity is gotten from the derivative of the displacement.

Thus;

v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)

v(t) = 0.42t + 0.0612t²

v(4.5) = 0.42(4.5) + 0.0612(4.5)²

v(4.5) = 3.1293 m/s ≈ 3.13 m/s

B) acceleration is gotten from the derivative of the velocity

a(t) = v'(t) = 0.42 + 2(0.0612t)

a(4.5) = 0.42 + 2(0.0612 × 4.5)

a(4.5) = 0.9708 m/s²

Force = ma = 5.5 × 0.9708

F = 5.3394 N ≈ 5.34 N

C) Since no friction, work done is kinetic energy.

Thus;

W = ½mv²

W = ½ × 5.5 × 3.1293²

W = 26.9 J

6 0
4 years ago
A 0.15-kg ball is thrown into the air and rises to a height of 20.0 m. How much kinetic energy did the ball initially have?
zzz [600]
IF the toss was straight upward, then the kinetic energy it got
from the toss is the gravitational potential energy it has at the top,
where it stops rising and starts falling.

Potential energy =  (mass)  x   (gravity) x (height)

                           = (0.15 kg) x (9.8 m/s²) x (20 m)

                           =      29.4 kg-m²/s²  =  29.4 joules .
7 0
3 years ago
Read 2 more answers
A hockey puck slides 55.0 m along the length of the rink in just 1.25 s. The slight friction between the puck and the ice provid
topjm [15]

Answer:

44.8 m/s

Explanation:

Use the Initial Speed Formula:

InS = 2(d/t) - Final Speed

InS = 2(55/1,25) - 43.2

InS = 2.44 - 43,2

InS = 88 - 43,2

InS = 44.8 m/s

3 0
3 years ago
An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 × 10^6 m and a mass of 7.35 × 10^22 k
ExtremeBDS [4]

Answer:

2.87 km/s

Explanation:

radius of planet, R = 1.74 x 10^6 m

Mass of planet, M = 7.35 x 10^22 kg

height, h = 2.55 x 10^6 m

G = 6.67 x 106-11 Nm^2/kg^2

Use teh formula for acceleration due to gravity

g=\frac{GM}{R^{2}}

g=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{1.74^{2}\times 10^{12}}

g = 1.62 m/s^2

initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0

Use third equation of motion

v^{2}=u^{2}-2gh

0 = v² - 2 x 1.62 x 2.55 x 10^6

v² = 8262000

v = 2874.37 m/s

v = 2.87 km/s

Thus, the initial speed should be 2.87 km/s.

6 0
3 years ago
An emf source of 6.0V is connected to a purely resistive lamp and a current of 2.0 amperes flows. All the wires are resistance-f
IgorC [24]

Resistance = (voltage) / (current)

Resistance = (6.0 v) / (2.0 A)

Resistance = 3.0 ohms 
7 0
3 years ago
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