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3 years ago

If a force of 50 Newton’s was applied to an object with a mass of 500 grams, what will the objects acceleration be?

Physics

1 answer:

lapo4ka [179]3 years ago

5 0

Answer: 100 m/s^2

F=ma

Explanation:

50N = 50 kg*m/s^2

500g = 0.5 kg

F=ma

a = F/m

a = (50 kg*m/s^2)/(0.5 kg)

a = 100 m/s^2

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Because the block is not moving, the sum of the x components of the forces acting on the block must be zero. Find an expression

Viktor [21]

Answer:

$F_{xtotal} = F_{x} - \mu N$

Explanation:

Let the total force be F

The force that produces motion in the x direction is given as, $F_{x}$

Let the frictional force be given as:

$F_{fric} = \mu N$

Thus, the net force is given as:

$F_{xnet} = F_{x} - \mu N$

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3 years ago

How should variables be depicted on a graph?

nignag [31]

Answer:

D.) independent variable on the x-axis

Explanation:

Normally in an x-y plane, the value of x is taken as an independent variable, i.e. it is independent because it can assume any value (real number), while y depends on the value x. for example for the following equation.

y = x

Where:

x = independent value.

y = dependent value of x

This is the equation of a line, inclined with slope equal to 1.

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3 years ago

Everything in or on water pushes some water aside, even if it's just a little bit.

makvit [3.9K]

Answer:

This is called displacement.

Everything in or on water pushes some water aside, even if it’s just a little bit. This is called displacement.

8 0

4 years ago

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Science is hard sometimes help please!

dolphi86 [110]

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3 years ago

A pendulum of length L = 1 m is released from an initial angle of 15° . After 1000 s, its amplitude is reduced to 2.5º. What is

jonny [76]

Answer:

The value of the time constant is 558.11 sec.

Explanation:

Given that,

Pendulum length = 1 m

Initial angle = 15°

Time = 1000 s

Reduced amplitude = 2.5°

We need to calculate the value of the time constant

Using formula of damping oscillation

$\theta=\theta_{0}e^{\dfrac{t}{\tau}}$

Where, $\theta$ =amplitude

$\theta_{0}$ =amplitude at t = 0

Put the value into the formula

$2.5=15e^{\dfrac{-1000}{\tau}}$

$\dfrac{1}{6}=e^{\dfrac{-1000}{\tau}}$

$ln\dfrac{1}{6}=\dfrac{-1000}{\tau}$

$\tau=\dfrac{1000}{-ln\dfrac{1}{6}}$

$\tau=558.11\ sec$

Hence, The value of the time constant is 558.11 sec.

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3 years ago