1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
prisoha [69]
3 years ago
12

(1) Arectangularmetal block with mass 18 kg has

Physics
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

Volume=0.003m^3 density=6000kg/m^3

Explanation:

mass=18kg

volume=length x width x height

Volume of block=0.1x0.15x0.2

volume of block=0.003m^3

Density=mass ➗ volume

Density=18 ➗ 0.003

Density=6000kg/m^3

You might be interested in
I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......
pav-90 [236]

Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100cm^{2} x 10cm = 1000cm^{3}

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

\frac{1g}{cm^{3} } x 1000cm^{3} = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

b)

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ \frac{8g}{cm^{3} } = 2.5 cm^{3}

c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5cm^{3} overflowed. So now we the same process as in number a) just with a few adjustments.

\frac{1g}{cm^{3} } x (1000cm^{3} - 2.5cm^{3}) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0
3 years ago
A cylinder of compressed gas has a pressure of 488.2 kPa. The next day the cylinder of gas has a
Luda [366]

Answer:

20 °C

Explanation:

Ideal gas law:

PV = nRT

Rearranging:

P / T = nR / V

Since n, R, and V are constant:

P₁ / T₁ = P₂ / T₂

488.2 kPa / T = 468 kPa / 281.15 K

T = 293.29 K

T = 20.1 °C

Rounded, the temperature was 20 °C.

6 0
3 years ago
Consider two waves defined by the wave functions y1(x,t)=0.50msin(2π3.00mx+2π4.00st) and y2(x,t)=0.50msin(2π6.00mx−2π4.00st). Wh
guapka [62]

Answer:

They two waves has the same amplitude and frequency but different wavelengths.

Explanation: comparing the wave equation above with the general wave equation

y(x,t) = Asin(2Πft + 2Πx/¶)

Let ¶ be the wavelength

A is the amplitude

f is the frequency

t is the time

They two waves has the same amplitude and frequency but different wavelengths.

4 0
3 years ago
A boy and a girl are on a spinning merry-go-round. The boy is at a radial distance of 1.2 m from the central axis; the girl is a
Licemer1 [7]

Answer:

E) True. The girl has a larger tangential acceleration than the boy.

Explanation:

In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.

Angular and linear quantities are related

          v = w r

          a = α r

the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m

as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different

          v₁ = w 1,2 (boy)

          v₂ = w 1.8 (girl)

whereby

          v₂> v₁

reviewing the claims we have

          a₁ = α 1,2

          a₂ = α 1.8

          a₂> a₁

A) False. Tangential velocity is different from zero

B) False angular acceleration is the same for both

C) False. It is the opposite, according to the previous analysis

D) False. Angular acceleration is equal

E) True. You agree with the analysis above,

8 0
2 years ago
WHOEVER GET IT RIGHT GETS 50 POINTS
AVprozaik [17]

Answer:

ANSWER BELOW I

                             I

                            V

Remember that w=mg where w is weight in Newtons, m is mass in kilograms, and g is gravity in

m/s2

. For example, for Earth, 445 N = 45.4 × 9.8

m/s2

:Notice that the x-axis values will be gravity in

m/s2

, which is already given in the table, and the y-axis values will be the weight in Newtons. Remember to round your weights to a whole number, and to enter the points starting with the lowest gravity (moon, then Mars, then Venus, then Earth).

3 0
3 years ago
Other questions:
  • Dark clothing feels warmer than white clothing on a sunny day. Which interaction causes the dark clothing to feel warm?
    15·2 answers
  • Which of these is the best explaination for why two negatively charged balloons move apart without ever touching?
    13·2 answers
  • Which statements accurately describe mechanical waves? Check all that apply.
    6·2 answers
  • A charged particle of mass m = 5.00g and charge q = -70.0μC moves horizontally to the right at a constant velocity of v = 30.0 k
    10·2 answers
  • A net force of 15 N is applied to a cart with a mass of 2.1 kg. a. What is the acceleration of the cart? b. How long will it tak
    5·1 answer
  • What is the mass of one sheet of printer paper?
    8·1 answer
  • What is the mass in kg of a leopard in a tree if the tree branch is 36 m up and the leopard's gravitational potential energy is
    13·1 answer
  • calculate the pressure exerted on the floor when a elephant who weighs 6000N stands on 1 foot which has a area of 20m
    15·1 answer
  • A racing car on a straight accelerates from 100mph to 316 mph in three seconds what is the acceleration?
    10·1 answer
  • The acceleration of a block attached to a spring is given by a=−(0.324m/s2)cos([2.50rad/s]t) a = − ( 0.324 m / s 2 ) c o s ( [ 2
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!