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1 year ago

What are physical forms in which a substance can exist?

Physics

1 answer:

Alecsey [184]1 year ago

8 0

Physical forms are: gas,liquid,and solid

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A cart moves along a track at a velocity of 3.5 cm/s. When a force is applied to the cart, its velocity increases to 8.2 cm/s. I

zheka24 [161]

The first thing we are going to do for this case is write the equations of movement of the car.
We have then:
vf = a * t + vo
Substituting values:
8.2 = a * (1.5) + (3.5)
Clearing the acceleration we have:
a = (8.2-3.5) / (1.5)
a = 3.1 m / s ^ 2
Answer:
the acceleration of the cart is:
a = 3.1 m / s ^ 2

5 0

1 year ago

Read 2 more answers

Power is __________________. Power is __________________. the work done by a system the force required to push something the spe

DaniilM [7]

Answer:

The rate at which the energy of a system is transformed

Explanation:

Power is the rate at which energy of a system is transformed or the rate at which work is done. It is defined by Power = Workdone/time taken

Its unit is the Watt denoted by the letter W.

For example, assuming a work of 200 J is done in 10 s, then Power, P equals

P = workdone/time taken = 200 J/10 s = 20 J/s = 20 W

6 0

1 year ago

A raft is made of a plastic block with a density of 650 kg/m 3 , and its dimensions are 2.00 m Ã 3.00 m Ã 5.00 m. 1. what is the

cupoosta [38]

1) The volume of the raft is the product between the lenghts of its three dimensions:
$V = (2.00 m)(3.00m)(5.00m)=30 m^3$

2) The mass of the raft is the product between its density, d, and its volume, V:
$m=dV=(650 kg/m^3)(30 m^3)=19500 kg$

3) The weight of the raft is the product between its mass m and the gravitational acceleration, $g=9.81 m/s^2$ :
$W=mg=(19500 kg)(9.81 m/s^2)=1.91 \cdot 10^5 N$

4) The apparent weight is equal to the difference between the weight of the raft and the buoyancy (the weight of the displaced fluid):
$W_a = W- \rho_W V_{disp} g$
where $\rho _W = 1000 kg/m^3$ is the water density and $V_{disp}$ is the volume of displaced fluid.
The density of the raft ( $650 kg/m^3$ ) is smaller than the water density ( $1000 kg/m^3$ ), this means that initially the buoyancy (which has upward direction) is larger than the weight (downward direction) and so the raft is pushed upward, until it reaches a condition of equilibrium and it floats. At equilibrium, the weight and the buoyancy are equal and opposite in sign:
$W=B=\rho _W V_{disp} g$
and therefore, the apparent weight will be zero:
$W_a = W-B=W-W=0$

5) The buoyant force B is the weight of the displaced fluid, as said in step 4):
$B=\rho_W V_{disp} g$
When the raft is completely immersed in the water, the volume of fluid displaced $V_{disp}$ is equal to the volume of the raft, $V_{disp}=V$ . Therefore the buoyancy in this situation is
$B= \rho_W V g = (1000 kg/m^3)(30 m^3)(9.81 m/s^2)=2.94 \cdot 10^5 N$
However, as we said in point 4), the raft is pushed upward until it reaches equilibrium and it floats. At equilibrium, the buoyancy will be equal to the weight of the raft (because the raft is in equilibrium), so:
$B=W=1.91 \cdot 10^5 N$

6) At equilibrium, the mass of the displaced water is equal to the mass of the object. In fact, at equilibrium we have W=B, and this can be rewritten as
$mg = m_{disp} g$
where $m_{disp}= \rho_W V_{disp}$ is the mass of the displaced water. From the previous equation, we obtain that $m_{disp}=m=19500 kg$ .

7) Since we know that the mass of displaced water is equal to the mass of the raft, using the relationship $m=dV$ we can rewrite $m=m_{disp}$ as:
$d V =d_W V_{disp}$
and so
$V_{disp}= \frac{d V}{d_W}= \frac{(650 kg/m^3)(30m^3)}{1000kg/m^3}= 19.5 m^3$

8) The volume of water displaced is (point 7) $19.5 m^3$ . This volume is now "filled" with part of the volume of the raft, therefore $19.5 m^3$ is also the volume of the raft below the water level. We can calculate the fraction of raft's volume below water level, with respect to the total volume of the raft, $30 m^3$ :
$\frac{19.5 m^3}{30 m^3}\cdot 100= 65 \%$
Viceversa, the volume of raft above the water level is $30 m^3-19.5 m^3 = 10.5 m^3$ . Therefore, the fraction of volume of the raft above water level is
$\frac{10.5 m^3}{30 m^3}\cdot 100 = 35 \%$

9) Let's repeat steps 5-8 replacing $\rho _W$ , the water density, with $\rho_E=806 kg/m^3$ , the ethanol density.

9-5) The buoyant force is given by:
$B=\rho _E V_{disp} g = (806 kg/m^3)(30 m^3)(9.81 m/s^2)=2.37 \cdot 10^5 N$
when the raft is completely submerged. Then it goes upward until it reaches equilibrium and it floats: in this condition, B=W, so the buoyancy is equal to the weight of the raft.

9-6) Similarly as in point 6), the mass of the displaced ethanol is equal to the mass of the raft:
$m_E = m = 19500 kg$

9-7) Using the relationship $d= \frac{m}{V}$ , we can find the volume of displaced ethanol:
$V_E = \frac{m}{d_E} = \frac{19500 kg}{806 kg/m^3}=24.2 m^3$

9-8) The volume of raft below the ethanol level is equal to the volume of ethanol displaced: $24.2 m^3$ . Therefore, the fraction of raft's volume below the ethanol level is
$\frac{24.2 m^3}{30 m^3}\cdot 100 = 81 \%$
Consequently, the raft's volume above the ethanol level is
$30 m^3 - 24.2 m^3 = 5.8 m^3$
and the fraction of volume above the ethanol level is
$\frac{5.8 m^3}{30 m^3}\cdot 100 = 19 \%$

8 0

2 years ago

Which of the following quantities provide enough information to calculate the tension in a string of mass per unit length μ that

Bad White [126]

Answer:

A. the wave speed v and Wavelength

Explanation:

Given that

Mass density per unit length=μ

Frequency = f

The velocity V given as

$\mu=\dfrac{T}{V^2}\ kg/m$

$V=\sqrt{\dfrac{T}{\mu}}$

T=Tension

V=Velocity

V= f λ

λ=Wavelength

Therefore to find the tension ,only wavelength and speed is required.

The answer is A.

8 0

1 year ago

a student wants to push a box of books with the mass of 50 kg in 3 m horizontally towards the location of the shelves where the

irina1246 [14]

Answer:

The work done is 360 J.

Explanation:

Given that,

Mass = 50 kg

Distance =3 m

We need to calculate the work done

The work done is equal to the product of force and displacement.

Using formula of work done

$W = F\cdot d$

$W = Fd\cos\theta$

Where, F = force

D = distance

θ = Angle between force and displacement

Put the value into the formula

$W=120\times3\cos0^{\circ}$

$W=360\ J$

Hence, The work done is 360 J.

8 0

1 year ago