All categories

3 years ago

The volume of a piece of metal of a mass 6 gram is 15cm3. What is the density of the metal piece?

Physics

2 answers:

Marizza181 [45]3 years ago

6 0

Density = (mass) / (volume)

Density = (6 g) / (15 cm³)

Density = (6/15) (g/cm³)

<em>Density = 0.4 g/cm³</em>

This 'metal' has less than half the density of water, and it floats on water ! It cannot be of Earthly origin, and must have been dropped here by aliens.

Either that, or else the metal has been hammered and stretched and beaten out and shaped so that it has an artificially large volume ... something like the way sailing ships are made out of thick plates of steel.

torisob [31]3 years ago

4 0

Answer:

0.4 g/cm^3

Explanation:

The density of an object can be found using the following formula.

d= m/v

where m is the mass and v is the volume.

The mass of the metal is 6 grams and the volume is 15 centimeters^3

m=6 g

v= 15 cm^3

Substitute these into the formula.

d= 6 g/ 15 cm^3

Divide 6 g by 15 cm^3 (6/15=0.4)

d= 0.4 g/ cm^3

The density of the metal is 0.4 grams per cubic centimeter.

You might be interested in

How does the balanced chemical equation show the conservation of matter in the chemical reaction?

Dmitry [639]

C. The number of F atoms in the reactants equals the number of F atoms in the products.

4 0

3 years ago

Using Excel, or some other graphing software, plot the values of y as a function of x. (You will not submit this spreadsheet. Ho

Evgesh-ka [11]

Answer:

a) > x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept) x

1.10 0.98

b) $y = 0.98 x +1.10$

And if we compare this with the general model $y = mx +b$

We see that the slope is m= 0.98 and the intercept b = 1.10

Explanation:

Part a

For this case we have the following data:

x: 1,2,3,4,5

y: 1.9,3.5,3.7,5.1, 6

For this case we can use the following R code:

> x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept) x

1.10 0.98

Part b

For this case we have the following trend equation given:

$y = 0.98 x +1.10$

And if we compare this with the general model $y = mx +b$

We see that the slope is m= 0.98 and the intercept b = 1.10

7 0

3 years ago

A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f

Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, $F=70\ N$

Mass of the dog is, $m=25\ kg$

Angle of inclination is, $\theta =30$ °

Acceleration due to gravity is, $g=9.8\ m/s^2$

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

$F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N$

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

$F_g=mg=25\times 9.8=245\ N$

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

$N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N$