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2 years ago

How can a large force result in a relatively small power?

Physics

1 answer:

Sav [38]2 years ago

8 0

Answer:

hey mate here is your answer

So if an object has a very small velocity (not moving very far over time, even though a large force may be applied to it, the Power will remain small. ... Stepping on the gas, or "speeding up" the car, is applying a force which will increase velocity and increase power.

please mark me as a brainliest

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For this exercise, use the position function s(t) = −4.9t2 + 250, which gives the height (in meters) of an object that has falle

nignag [31]

Answer:

t=7.14s

v=-69.972 m/s

Explanation:

Position function

$s(t)=-4.9*t^{2}+250$

Velocity is the derivative of position function

$V(t)=\frac{dx}{dt}\\V(t)=-2*4.9*t\\V(t)=-9.8*t$

The time the object hit the ground can be find by the given function know that the position is going to be 0m

$s(t)=-4.9*t^{2}+250$

$s(t)=0\\0=-4.9*t^{2} +250\\t=\sqrt{\frac{250}{4.9}}\\t=7.14s$

Check:

$s(7.14)=-4.9*(7.14s)^{2}+250\\ s(7.14)=-250+250\\s(7.14)=0m$

So the velocity can be find using the time discovery before and using the same function but with the derivate

$V(t)=-2*4.9*t\\V(7.14)=-2*4.9*(7.14)\\V(7.14)=-69.972 \frac{m}{s}$

The velocity is negative because the object is moving downward

6 0

3 years ago

Leto [7]

One eighth of a gram will remain

3 0

3 years ago

Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 4.0 m behind the apertur

devlian [24]

Answer:

d = 0.247 mm

Explanation:

given,

λ = 633 nm

distance from the hole to the screen = L = 4 m

width of the central maximum = 2.5 cm

2 y = 0.025 m

y = 0.0125 m

For circular aperture

$sin \theta = 1.22\dfrac{\lambda}{d}$

using small angle approximation

$\theta = \dfrac{y}{D}$

now,

$\dfrac{y}{D} = 1.22\dfrac{\lambda}{d}$

$y = 1.22\dfrac{\lambda\ D}{d}$

$d = 1.22\dfrac{\lambda\ D}{y}$

$d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}$

d =0.247 x 10⁻³ m

d = 0.247 mm

the diameter of the hole is equal to 0.247 mm

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3 years ago

A 1.0-kg block moving to the right at speed 3.0 m/s collides with an identical block also moving to the right at a speed 1.0 m/s

____ [38]

Answer:

Speed of both blocks after collision is 2 m/s

Explanation:

It is given that,

Mass of both blocks, m₁ = m₂ = 1 kg

Velocity of first block, u₁ = 3 m/s

Velocity of other block, u₂ = 1 m/s

Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{1\ kg\times 3\ m/s+1\ kg\times 1\ m/s}{2\ kg}$

v = 2 m/s

Hence, their speed after collision is 2 m/s.

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3 years ago

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zloy xaker [14]

Explanation:

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3 years ago