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3 years ago

Graphs are representations of equations. A. True B. False

Physics

2 answers:

zhuklara [117]3 years ago

6 0

Answer:

Hi there!

The answer is: A. True

Talja [164]3 years ago

5 0

Most graphs are representations of equations but not all are.

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Drop a rock from a 5-m height and it accelerates at 9.81 m/s^2 and strikes the ground 1s later. Drop the sane rock from a height

KiRa [710]

Answer:

A. The same amount

Explanation:

The acceleration at which objects free falls to the ground on Earth is constant and its value is always $9.81 m/s^2$ , regardless of their mass (in this problem we neglect air resistance).

So, it doesn't matter if the two rocks are different or they are launched from different heights: their acceleration will be exactly the same.

This can be proved this way: first of all, the force of gravity exerted on every object is equal to the weight of the object,

$F=mg$

where m is the mass of the object and g the acceleration of gravity.

However, we also know for Newton's second law that

$F=ma$

where a is the acceleration of the object.

Combining the two equations,

$ma=mg\\a=g$

So, the acceleration of an object in free fall is exactly the acceleration of gravity.

8 0

3 years ago

Suppose that the electric potential outside a living cell is higher than that inside the cell by 0.063 V. How much work is done

Varvara68 [4.7K]

Answer:

$W=1.008\times 10^{-20}\ J$

Explanation:

Given that

Potential difference ,ΔV= 0.063 V

Charge q

$q=1.6\times 10^{-19}\ C$

The work done by electric force given as

W = ΔV . q

$W=0.063\times 1.6\times 10^{-19}\ J$

$W=1.008\times 10^{-20}\ J$

5 0

3 years ago

An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 49.1 N,

balu736 [363]

When the spring is stretched by 15.2 cm = 0.152 m, the spring exerts a restorative force with magnitude (due to Hooke's law)

$F = kx$

where $k$ is the spring constant. Solve for $k$ .

$49.1\,\mathrm N = k (0.152\,\mathrm m) \implies k \approx 323 \dfrac{\rm N}{\rm m}$

The amount of work required to stretch or compress a spring by $x\,\mathrm m$ from equilibrium length is

$W = \dfrac12 kx^2$

Then the work needed to stretch the spring by 15.2 cm is

$W_1 = \dfrac12 \left(343\dfrac{\rm N}{\rm m}\right) (0.152\,\mathrm m)^2 \approx 3.73\,\mathrm J$

and by 15.2 + 13.7 = 28.9 cm is

$W_2 = \dfrac12 \left(343\dfrac{\rm N}{\rm m}\right) (0.289\,\mathrm m)^2 \approx 13.5\,\mathrm J$

so the work needed to stretch from 15.2 cm to 28.9 cm from equilibrium is

$\Delta W = W_2 - W_1 \approx \boxed{9.76\,\mathrm J}$

4 0

2 years ago

What is also known as watered carbons

faltersainse [42]

The name carbohydrate means "watered carbon" or carbon with attached water molecules. Many carbohydrates have empirical formuli which would imply about equal numbers of carbon and water molecules. For example, the glucose formula C6H12O6 suggests six carbon atoms and six water molecules.

5 0

3 years ago

Read 2 more answers

You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a

allochka39001 [22]

Answer:

Explanation:

If we consider the ball,the tension is not balance by weight then the force acting will be non zero.It means that train car in not inertial frame of reference.And the same time the train car must accelerated with respected to the earth.It is not moving straight path.When train is in rest position but the string will be at the constant angle.

So the following option are correct:

6 0

3 years ago