All categories

3 years ago

Help please thank you whoever helps

Physics

1 answer:

Delicious77 [7]3 years ago

8 0

Answer:

24a. WX, XW

24b. XY

24c. WV

24d. UV, VU

24e. XY, YX

24f. YX

24g. VW

24h. YZ, ZY

24i. VW, WV

24j. WX

25a. ≈ 20°C

25b. ≈ 290°C

25c. ≈ 20°C

25d. ≈ 290°C

Explanation:

Let's look at the graph and see what phase changes happen as we go from U to Z.

UV: the substance is a solid. As heat is added, temperature increases.

VW: the solid begins to melt. The temperature stays constant as the combination of solid and liquid becomes more and more liquid.

WX: the solid is done melting, and there's only liquid now. As heat is added, temperature increases.

XY: the liquid begins to boil. The temperature stays constant as the combination of liquid and gas becomes more and more gas.

YZ: the liquid is done boiling, and there's only gas now. As heat is added, temperature increases.

If we reverse the direction, YX represents condensation instead of boiling, and WV represents freezing instead of melting.

Kinetic energy of molecules increases with temperature. The greatest increase in temperature is also the greatest increase in kinetic energy.

You might be interested in

A lawyer most likely has __________. a.spatial intelligence b.practical intelligence c.analytical intelligence d.creative intell

ArbitrLikvidat [17]

The type of intelligence required by a profession depends on its job description. Lawyers most likes have analytical intelligence.

A renowned psychologist named Howard Gardner devised a theory of multiple intelligence. This means that nobody is really not intelligent because a person must posses one or more of these multiple intelligence.

A lawyer has a lot to do with analytic reasoning and comparison of evidence to determine their validity. This task requires a strong analytical mind hence a lawyer requires analytical intelligence.

Learn more: brainly.com/question/4158015

3 0

2 years ago

A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s.

Oksana_A [137]

Answer:9.75 m/s

Explanation:

Given

Length of ladder $(L)=5 m$

Foot the ladder is moving away with speed of $\frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s$

From diagram

$x^2+y^2=L^2$ ------1

at $x=3$

$y^2=25-9=16$

$y=4 m$

Now differentiating equation 1 w.r.t time

$2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0$

$x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}$

$3\times 13=-4\times \frac{\mathrm{d} y}{\mathrm{d} t}$

$\frac{\mathrm{d} y}{\mathrm{d} t}=-\frac{3\times 13}{4}=-9.75 m/s$

negative indicates distance is decreasing with time

5 0

3 years ago

A 0.250 kg block on a vertical spring with spring constant of 4.45 ✕ 103 N/m is pushed downward, compressing the spring 0.080 m.

Elza [17]

Answer:h=5.81 m

Explanation:

Given

Mass of block(m)=0.250 kg

Spring Constant $k=4.45\times 10^3 N/m$

Initial elongation =0.080 m=8 cm

Thus Initial Potential Energy stored =Final Potential Energy stored in Block

$P_i=\frac{kx^2}{2}$

$P_i=\frac{4.45\times 10^3\times 64\times 10^{-4}}{2}=14.24 J$

$P_f=mgh=0.25\times 9.8\times h$

$P_i=P_f$

$14.24 =0.25\times 9.8\times h$

$h=\frac{14.24}{0.25\times 9.8}=5.81 m$

6 0

3 years ago

A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0

3 years ago

An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu

Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

$\alpha = \frac{\omega_f - \omega_i}{t}$

where we have

$\alpha = -80.0 rad/s^2$ is the angular acceleration (negative since the rotor is slowing down)

$\omega_f$ is the final angular speed

$\omega_i = 2000 rad/s$ is the initial angular speed

t = 10.0 s is the time interval

Solving for $\omega_f$ , we find the final angular speed after 10.0 s:

$\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s$

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

$\alpha = \frac{\omega_f - \omega_i}{t}$

If we re-arrange it for t, we get:

$t = \frac{\omega_f - \omega_i}{\alpha}$

where here we have

$\omega_i = 2000 rad/s$ is the initial angular speed

$\omega_f=0$ is the final angular speed

$\alpha = -80.0 rad/s^2$ is the angular acceleration

Solving the equation,

$t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s$

6 0

3 years ago