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3 years ago

Describe the changes of motion when marble is moved from one viewpoint to another.

Physics

1 answer:

slavikrds [6]3 years ago

5 0

Answer:

Explanation:

according to Newton First Law of Motion (Law of Inertia); An object at rest will stay at rest, forever, as long as nothing pushes or pulls on it. An object in motion will stay in motion, traveling in a straight line, forever, until something pushes or pulls on it.

the marble will move in a straight line

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A car traveling east increases its velocity from 4 m/s to 19 m/s in 5 s. What is the acceleration of the car?

Virty [35]

It’s going 3m/s. If we have 5 seconds to work with then we can find the acceleration by adding speed and how fast it going every second. So like the 7-10-13-16-19 so we go 3m/s faster ever second

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3 years ago

A. A PH202 student lives next to a construction site and sees a crane with a wrecking ball demolish the building next door. The

netineya [11]

Answer:

The length of cable is 12.5 m

Explanation:

Since, the wrecking ball completes 1 oscillation, in the same time, as it takes for the rock to complete 5 oscillations.

Therefore,

Time Period of Wrecking Ball = 5 (Time Period of Rock)

Since,

Time Period of Pendulum = 2π√(L/g)

Therefore,

2π√(L₁/g) = 5[2π√(L₂/g)]

√L₁ = 5√L₂

Squaring on both sides:

L₁ = 25 L₂

where,

L₁ = Length of Cable = ?

L₂ = Length of string = 0.5 m

Therefore,

L₁ = 25 (0.5 m)

8 0

3 years ago

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yulyashka [42]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

a freight train travels at v=60(1-e^-t) ft/s, where t is the elapsed time in seconds. Determine the distance traeled in tree sec

lesantik [10]

Answer

given,

$v = 60(1-e^{-t})\ ft/s$

t = 3 s

we know,

$v = \dfrac{dx}{dt}$

$a = \dfrac{dv}{dt}$

position of the particle

$dx = v dt$

integrating both side

$\int dx =\int 60(1-e^{-t}) dt$

$x = 60 t + 60 e^{-t}$

Position of the particle at t= 3 s

$x = 60\times 3+ 60 e^{-3}$

x = 182.98 ft

Distance traveled by the particle in 3 s is equal to 182.98 ft

now, particle’s acceleration

$a = \dfrac{dv}{dt}$

$a = \dfrac{d}{dt}60(1-e^{-t})$

$a = 60 e^{-t}$

at t= 3 s

$a = 60 e^{-3}$

a = 2.98 ft/s²

acceleration of the particle is equal to 2.98 ft/s²

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3 years ago

First step in using a balance beam scale?

prohojiy [21]

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