1) In which of these examples is the greatest movement occurring? A) A B) B C) C D) D

ch4aika [34]

Explanation:

Boyle's law can be stated as the "volume of a fixed mass of a gas varies inversely as the pressure changes if the temperature is constant". It is mathematically expressed as;

P1 V1 = P2 V2

P1 is the initial pressure

V1 is the initial volume

P2 is the final pressure

V2 is the final volume

Charles's law states that the volume of a fixed mass of a gas varies directly as its absolute temperature if the pressure is constant.

It is mathematically expressed as;

$\frac{V1}{T1}$ = $\frac{V2}{T2}$

V and T are volume and temperature respectively

1 and 2 are the initial and final states

8 0

3 years ago

10%) Problem 7: Water flows through a water hose at a rate of Q1 = 620 cm3/s, the diameter of the hose is d1 = 1.99 cm. A nozzle

Semenov [28]

Answer:

a) $A_1 = \frac{\pi d_1^2}{4}$

Explanation:

a) the cross-sectional area of the hose would be the square of radius times pi. And since the sectional radius is half of its diameter d. We can express the cross-sectional area A1 in term of diameter d1

$A_1 = \pi r_1^2 = \pi (d_1/2)^2 = \frac{\pi d_1^2}{4}$

6 0

3 years ago

A particle with kinetic energy equal to 282 J has a momentum of magnitude 26.4 kg · m/s. Calculate the speed (in m/s) and the ma

masha68 [24]

Answer:

$v=21.36\,\,\frac{m}{s}\\$

$m=1.2357\,\,kg$

Explanation:

Recall the formula for linear momentum (p):

$p = m\,v$ which in our case equals 26.4 kg m/s

and notice that the kinetic energy can be written in terms of the linear momentum (p) as shown below:

$K=\frac{1}{2} m\,v^2=\frac{1}{2} \frac{m^2\,v^2}{m} =\frac{1}{2}\frac{(m\,v)^2}{m} =\frac{p^2}{2\,m}$

Then, we can solve for the mass (m) given the information we have on the kinetic energy and momentum of the particle:

$K=\frac{p^2}{2\,m}\\282=\frac{26.4^2}{2\,m}\\m=\frac{26.4^2}{2\,(282)}\,kg\\m=1.2357\,\,kg$

Now by knowing the particle's mass, we use the momentum formula to find its speed:

$p=m\,v\\26.4=1.2357\,v\\v=\frac{26.4}{1.2357} \,\frac{m}{s} \\v=21.36\,\,\frac{m}{s}$

4 0

3 years ago

Calcular el módulo del vector resultante de dos vectores fuerza de 9 [N] y 12 [N] concurrentes en un punto o, cuyas direcciones

sdas [7]

Answer:

a) 20.29N

b) 19.43N

c) 15N

Explanation:

To find the magnitude of the resultant vectors you first calculate the components of the vector for the angle in between them, next, you sum the x and y component, and finally, you calculate the magnitude.

In all these calculations you can asume that one of the vectors coincides with the x-axis.

a)

$F_R=(9cos(30\°)+12)\hat{i}+(9sin(30\°))\hat{j}\\\\F_R=(19.79N)\hat{i}+(4.5N)\hat{j}\\\\|F_R|=\sqrt{(19.79N)^2+(4.5N)^2}=20.29N$

b)

$F_R=(9cos(45\°)+12)\hat{i}+(9sin(45\°))\hat{j}\\\\F_R=(18.36N)\hat{i}+(6.36N)\hat{j}\\\\|F_R|=\sqrt{(18.36N)^2+(6.36N)^2}=19.43N$

c)

$F_R=(9cos(90\°)+12)\hat{i}+(9sin(90\°))\hat{j}\\\\F_R=(12N)\hat{i}+(9N)\hat{j}\\\\|F_R|=\sqrt{(12N)^2+(9N)^2}=15N$

7 0

3 years ago

A 22-turn circular coil of radius 3.00 cm and resistance 1.00 Ω is placed in a magnetic field directed perpendicular to the plan

zloy xaker [14]

Answer:

23.5 mV

Explanation:

number of turn coil 'N' =22

radius 'r' =3.00 cm=> 0.03m

resistance = 1.00 Ω

B= 0.0100t + 0.0400t²

Time 't'= 4.60s

Note that Area'A' = πr²

The magnitude of induced EMF is given by,

lƩl =ΔφB/Δt = N (dB/dt)A

=N[d/dt (0.0100t + 0.0400 t²)A

=22(0.0100 + 0.0800(4.60))[π(0.03)²]

=0.0235

=23.5 mV

Thus, the induced emf in the coil at t = 4.60 s is 23.5 mV

8 0

3 years ago