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3 years ago

Write short note on fulcrum

Physics

1 answer:

Semmy [17]3 years ago

6 0

Answer:

The definition of a fulcrum is a pivot point around which a lever turns, or something that plays a central role in or is in the center of a situation or activity.

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An object is dropped from rest from the top of a 400 m cliff on Earth. If air resistance is negligible, what is the distance the

IrinaVladis [17]

Answer:

176.58 m

Explanation:

t = Time taken = 6 seconds

u = Initial velocity = 0

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 6^2\\\Rightarrow s=176.58\ m$

The object travels 176.58 m from the cliff in 6 seconds.

8 0

3 years ago

Potential energy becomes kinetic energy when:

Sedbober [7]

Answer:

An object has potential energy (stored energy) when it is not in motion. Once a force has been applied or it begins to move the potential energy changes to kinetic energy (energy of motion).

EXAMPLE: A rock sitting on the edge of a cliff. If the rock falls, the potential energy will be converted to kinetic energy, as the rock will be moving. A stretched elastic string in a longbow.

5 0

2 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

3 years ago

If a hockey stick exerts a force of 40 N on a 0.5-kg puck, what will the

antoniya [11.8K]

Answer:

B. 80 m/s²

Explanation:

F = ma

a = F/m = (40 N)/(0.5 kg) = 80 m/s²

3 0

3 years ago

A 1.0-m-diameter vat of liquid is 2.0 m deep. The pressure at the bottom of the vat is 1.3 atm. What is the mass of the liquid i

gladu [14]

Answer:

The mass of the liquid = 10538 kg

Explanation:

The pressure in a liquid is

P = ρgh ......................... Equation 1

ρ = P/gh ...................... Equation 2

Where P = pressure, ρ = density, g = acceleration due to gravity, h = height.

Given: p = 1.3 atm, h = 2.0 m, g = 9.81 m/s²

If, 1 atm = 1.013×10⁵ N/m²

Then, P = 1.3×1.013×10⁵ N/m² = 1.3169×10⁵ N/m²

Substituting in equation 2,

ρ = 1.3169×10⁵/(9.81×2)

ρ = 1.3169×10⁵/19.62

ρ = 6712.03 kg/m³.

But Density,

ρ = m/v

m = ρ × v........................ Equation 3

Where m = mass of the liquid, v = volume of the liquid in the vat

v = πd²h/4, where d = diameter = 1.0 m, h = 2.0 m.

v = 3.14(1)²×2/4

v = 1.57 m³ also, ρ = 6712.03 kg/m³.

Substituting into equation 3

m = 1.57×6712.03

m = 10537.887 kg

m ≈ 10538 kg.

Thus the mass of the liquid = 10538 kg

3 0

3 years ago

Write short note on fulcrum​

Write short note on fulcrum