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1 year ago

7

What method could I use to test this hypothesis? If the mass and the volume of and object are known, then its density can be cal

culated dividing the object's mass by its volume.

1 answer:

Nitella [24]1 year ago

7 0

Answer:

The scientific method

Explanation:

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Each value in nature has a number part, called its____<br> and a dimension, or unit

saul85 [17]

Answer:

<u>Magnitude</u>

Explanation:

Each value in nature has a number part, called its magnitude and a dimension called its unit.

For example,

The length of an object is 10 cm. It means that 10 shows the magnitude of length and cm shows its unit.

7 0

2 years ago

6. A Cadillac Escalade has a mass of 2 569.6 kg, if it accelerates at 4.65m/S<br> force on the car?

SVETLANKA909090 [29]

Answer:

F= 2569.6 X 4.65 = 11,948.64

*Multiply the mass and the acceleration to find the force

Explanation:

5 0

3 years ago

Read 2 more answers

High school and collage students answers only. <br> Are all compounds molecules? Why or why not?

Shkiper50 [21]

All compounds are molecules but not all molecules are compounds. A molecule is two atoms joined together. A compound is two different atoms joined together.

5 0

2 years ago

Consider a block on frictionless ice. Starting from rest, the block travels a distance din

sweet [91]

Answer:

<em>The distance is now 4d</em>

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The acceleration can be calculated by solving for a:

$\displaystyle a=\frac{F}{m}$

Once we know the acceleration, we can calculate the distance traveled by the block as follows:

$\displaystyle d = vo.t+\frac{at^2}{2}$

If the block starts from rest, vo=0:

$\displaystyle d = \frac{at^2}{2}$

Substituting the value of the acceleration:

$\displaystyle d = \frac{\frac{F}{m}t^2}{2}$

Simplifying:

$\displaystyle d = \frac{Ft^2}{2m}$

When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:

$\displaystyle a'=\frac{4F}{m}$

And the distance is now:

$\displaystyle d' = \frac{4Ft^2}{2m}$

Dividing d'/d:

$\displaystyle \frac{d' }{d}=\frac{\frac{4Ft^2}{2m}}{\frac{Ft^2}{2m}}$

Simplifying:

$\displaystyle \frac{d' }{d}=4$

Thus:

d' = 4d

The distance is now 4d

3 0

2 years ago

A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance,

Lisa [10]

Answer:

Explanation:

The two charges are q and Q - q. Let the distance between them is r

Use the formula for coulomb's law for the force between the two charges

$F = \frac{Kq_{1}q_{2}}{r^{2}}$

So, the force between the charges q and Q - q is given by

$F = \frac{K\left ( Q-q \right )q}}{r^{2}}$

For maxima and minima, differentiate the force with respect to q.

$\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )$

For maxima and minima, the value of dF/dq = 0

So, we get

q = Q /2

Now $\frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}$

the double derivate is negative, so the force is maxima when q = Q / 2 .

6 0

3 years ago