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EleoNora [17]
3 years ago
13

A steel cable lying flat on the floor drags a 20 kg block across a horizontal, frictionless floor. A 110 N force applied to the

cable causes the block to reach a speed of 3.9 m/s at a distance of 1.9 m. What is the mass of the cable?
Physics
1 answer:
sesenic [268]3 years ago
8 0

Answer:

The answer is;

The mass of the cable is 7.428 kg

Explanation:

Force = 110 N

Mass of block = 20 kg

velocity = 3.9 m/s

We apply the law of motion which is

v² = u² + 2·a·s

Where a = acceleration

s = distance = 1.9 m and

u = initial velocity = 0 m/s

v = velocity = 3.9 m/s

∴ v² = 2·a·s

3.9² = 2×1.9×a which gives

a = 4.003 m/s²

F= m×a

Where F= Force = 110 N

m = mass

a = acceleration = 4.003 m/s²

∴ m = F/a = 110/4.003 = 27.482 kg

Since the mass of the block = 20 kg

Mass of cable + block = 27.482 kg

Mass of cable = 27.428 kg - mass of block = 27.428 kg - 20 kg = 7.428 kg

Mass of cable = 7.428 kg

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What is the speed of a beam of electrons when under the simultaneous influence of E = 1.64×104 V/m B = 4.60×10−3 T Both fields a
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Explanation:

Magnetic force(B) = 4.60×10^-3 T

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Both forces having equal magnitude ;

Magnetic force = electric force

qvB = qE

vB = E

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v = 3.57×10^6 m/s

2.) Assume no electric field

qvB = ma

Where a = v^2 ÷ r

R = radius

a = acceleration

v = velocity

qvB = m(v^2 ÷ R)

R = (m×v) ÷ (|q|×B)

q=1.6×10^-19C

m = 9.11×10^-31kg

R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)

R = 32.5227×10^-25 ÷ 7.36×10^-22

R = 4.42×10^-3m

3.) period(T)

T = (2*pi*R) ÷ v

T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)

T = (27.775×10^-3) ÷(3.57×10^6)

T = 7.78×10^-9 s

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What adaptation of a cheetah is shown in the image?
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These are the physical and behavioral features which ensures that they survive in their changing environment.

Cheetahs are known to have long and strong legs which is the reason why it runs at a very high speed.

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When can we be certain that the average velocity of an object is always equal to its instantaneous velocity?
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3 years ago
In the diagram, 91, 92, and q3 are in a straight line. Each of these particles has a charge of -2.35 x 10-6 C. Particles q₁ and
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The net force on particle q₃ is  6.2128125 N.

<h3>What is electrostatic force?</h3>

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F = kq₁q₂/d²

where k = 9 x 10⁹ N.m²/C²

Given is the diagram in which each of the particles has a charge of -2.35 x 10⁻⁶ C. Particles q₁ and q₂ are separated by 0.100 m and particles q₂ and q₃ are separated by 0.100 m.

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Net force acting on particle q₃ is

F₃ = F₃₁ +F₃₂

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