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3 years ago

A 23.0 kg iron weightlifting plate has a volume of 2920 cm3 . what is the density of the iron plate in g/cm3?

1 answer:

8 0

The first thing you should know for this case is that density is defined as the quotient between mass and volume:
D = M / V
In addition, you should keep in mind the following conversion:
1Kg = 1000g
Substituting the values we have:
D = (23.0 * 1000) / (2920) = 7.88 g / cm ^ 3
answer
the density of the iron plate is 7.88 g / cm ^ 3

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For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is s=29.0cm long)

Lesechka [4]

The question is missing a diagram of the ray reflection. I attached a diagram which comes from a similar question in the answer section. The full question should be as follows:

Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point d = 10.0cmfrom their point of intersection, as shown in the figure. For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is s=29.0cm long) after reflecting from the first mirror?

Answer:

34.6°

Explanation:

To strike the midpoint of the second mirror, the ray light will have to travel half of the distance vertically

i.e. 29/2 = 14.5

We can solve this through trigonometry.

Let the angle between the ray and the vertical plane mirror is known as α

tan α = 10/14.5

α = $tan^{-1} (10/14.5)$ = 34.6°

The angle of incidence is the angle between the ray and the normal line of the mirror.

Let angle of incidence of first mirror be β

β = α = 34.6

6 0

3 years ago

Type the correct answer in the box. use numerals instead of words. anne has a sample of a substance. its volume is 20 cm3, and i

CaHeK987 [17]

The density of sample is 5 g/cm3

Given:

volume of sample = 20 cm3

mass of sample = 100 grams

To Find:

density of sample

Solution: Density is the measure of how much “stuff” is in a given amount of space. For example, a block of the heavier element lead (Pb) will be denser than the softer, lighter element gold (Au). A block of Styrofoam is less dense than a brick. It is defined as mass per unit volume

density = mass/volume

d = 100/20

d = 5 g/cm3

So, density of sample is 5 g/cm3

Learn more about Density here:

brainly.com/question/1354972

#SPJ4

6 0

1 year ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago

Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an

ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

$Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega$

Power factor is given by

$F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802$

The power factor is 0.28802

The average power to the circuit is given by

$P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W$

The average power to the circuit is 2.57162 W

Power to resistor

$P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W$

Power to resistor is 14.28 W

Power to inductor

$P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W$

Power to the inductor is 53.55 W

Power to the capacitor

$P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W$

The power to the capacitor is 6.07142 W

8 0

3 years ago

Which statement correctly explains scientific theories?

Elan Coil [88]

<span>Scientific theories are tested and proven over time; they are then considered scientific laws.

Sometimes however, they are proven wrong, and so they do not become laws

hope this helps</span>

4 0

3 years ago

Read 2 more answers