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cupoosta [38]
3 years ago
8

Calculate the magnitude of the effort in the given​

Physics
1 answer:
Otrada [13]3 years ago
8 0
Moment = Force x Distance from pívot
500x0.5=2.5 x effort
250= 2.5 x effort
effort = 250/2.5= 100N
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A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar
Ne4ueva [31]

Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

Coefficients of static = 0.8

Kinetic friction = 0.4

Tension = 24 N

We need to calculate the maximum friction force

Using formula of friction force

f=\mu mg

Put the value into the formula

f=0.8\times37

f=29.6\ N

So, the tension must exceeds 29.6 N for the block to move

We need to calculate the frictional force acting on the block

Using formula of frictional force

f = \mu N

Put the value in to the formula

f=0.4\times37

f=14.8\ N

Hence, The frictional force acting on the block is 14.8 N.

6 0
3 years ago
NEED HELP PLEASE 15 POINTS PLEASEE NEED HELP
Ostrovityanka [42]

Answer:

I wish I knew

Explanation:

5 0
3 years ago
Read 2 more answers
A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0
3 years ago
How far will a train with a speed of 20 m/s travel in: 10s
Citrus2011 [14]

Answer:

The correct answer would be 200m

Explanation:

20x10=200

the previous answer is incorect from the other user.

8 0
3 years ago
How is the lifetime of a star related to its mass?
Flauer [41]
Stars having less mass collapses early than those with more mass. This can be explained by Einstein's equation E=mc².
According to this equation, mass of stars is converted into light due to thermonuclear reactions occuring in the core of star which acts as engine of the stars. This thermonuclear reactions keeps star alive. Thermonuclear reactions occurs slowly in massive stars hence massive stars live more than light stars.
6 0
3 years ago
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