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kkurt [141]
3 years ago
5

What wa can friction do to things like brake pads

Physics
2 answers:
aleksandr82 [10.1K]3 years ago
5 0
Friction wears things like brake pads down over time. it just keeps rubbing and rubbing til their is nothing left.
andrey2020 [161]3 years ago
4 0
Ware them down, its like rubbing two pieces of chalk together. 
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A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary
Archy [21]

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

  • m = 10 kg
  • g = 9.8 m/s²
  • T = ?

Use formula:

  • \boxed{\bold{T=m*g}}

Replace and solve:

  • \boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}
  • \boxed{\boxed{\bold{T=98\ N}}}

The tension in the rope is <u>98 Newtons.</u>

Greetings.

5 0
3 years ago
2. A 15 kg mass fastened to the end of a steel wire of un-
Flauer [41]

Explanation:

Elongation of the wire is:

ΔL = F L₀ / (E A)

where F is the force,

L₀ is the initial length,

E is Young's modulus,

and A is the cross sectional area.

ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)

ΔL = T (1.25×10⁻⁶ m/N)

T = (80,000 N/m) ΔL

Draw a free body diagram of the mass at the bottom of the circle.  There are two forces: tension force T pulling up and weight force mg pulling down.

Sum of forces in the centripetal direction:

∑F = ma

T − mg = mv²/r

T − mg = mω²r

T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)

T − 147 N = (2368.7 N/m) (0.5 m + ΔL)

Substitute:

(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)

(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL

(797631.3 N/m) ΔL = 1331.35 N

ΔL = 0.00167 m

ΔL = 1.67 mm

6 0
3 years ago
CYCLONE SAFETY MEASURES
Slav-nsk [51]

Answer:

Below are NDMA's recommended safety measures:

Ignore rumours, Stay calm, Don't panic.

Keep your mobile phones charged for emergency communication; use SMS.

Listen to the radio, watch TV, read newspapers for weather updates.

Keep cattle/animals untied to ensure their safety.

5 0
3 years ago
Read 2 more answers
Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyon
Nutka1998 [239]

Answer:

16.63min

Explanation:

The question is about the period of the comet in its orbit.

To find the period you can use one of the Kepler's law:

T^2=\frac{4\pi}{GM}r^3

T: period

G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2

r: average distance = 1UA = 1.5*10^11m

M: mass of the sun = 1.99*10^30 kg

By replacing you obtain:

T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min

the comet takes around 16.63min

8 0
3 years ago
Air resistance is a type of<br> a. motion.<br> b. acceleration.<br> c. velocity.<br> d. friction.
Alex73 [517]
Air resistance is a type of friction.

Hope this helps! :D
3 0
3 years ago
Read 2 more answers
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