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Svetllana [295]

3 years ago

13

Which graph shows acceleration? A graph of position (meters) versus time (seconds) has a straight line running from 0 seconds 0

meters upward. A graph of position (meters) versus time (seconds) has a concave line running from 0 seconds 0 meters upward. A graph of position (meters) versus time (seconds) has a straight line running from 0 seconds positive number of meters downward to some later time 0 meters.

2 answers:

anzhelika [568]3 years ago

7 0

Answer:

its the middle one

Explanation:

i did it on edge 2021

charle [14.2K]3 years ago

6 0

It's the one that has a concave line running upward.

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An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

Anna35 [415]

Answer

Given,

Energy absorbed, $Q_H = 1.69\ kJ$

Energy expels, $Q_C = 1.25\ kJ$

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

$\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100$

$\eta = \dfrac{1.69 - 1.25}{1.69}\times 100$

$\eta =26.03 %$

b) Work done by the engine

$W = Q_H- Q_C$

$W =1.69 - 1.25$

$W = 0.44\ kJ$

c) Power output

t = 0.296 s

$P = \dfrac{W}{t}$

$P = \dfrac{0.44}{0.296}$

$P = 1.486\ kW$

8 0

3 years ago

An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su

s2008m [1.1K]

Answer: A. 23.59 minutes.

B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

mass of ice, m= 0.555 kg

temperature of ice, T= -16.6°C

rate of heat transfer, $q=820 J.min^{-1}$

specific heat of ice, $c_{i}= 2100 J.kg^{-1}.K^{-1}$

latent heat of fusion of ice, $L_{i}=334\times10^{3}J.kg^{-1}$

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

$Q=mc\Delta T$

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

$Q= 0.555\times2100\times16.6$

$Q= 19347.3 J$

Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

$t=\frac{Q}{q}$

$=\frac{19347.3}{820}$

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, $Q_{L}= mL_{i}$

$Q_{L}= 0.555\times334\times10^{3}= 185370 J$

<u>Now the time required for supply of 185370 J:</u>

$t=\frac{Q_{L}}{q}$

$t=\frac{185370}{820}$

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0

3 years ago

Why do different substances produce electromagnetic waves when they are heated up or energized?

Anit [1.1K]

Answer:

Electromagnetic radiation is made when an atom absorbs energy. The absorbed energy causes one or more electrons to change their locale within the atom. When the electron returns to its original position, an electromagnetic wave is produced. ... These electrons in these atoms are then in a high energy state.

Explanation:

4 0

4 years ago

Read 2 more answers

A skateboarder jumps horizontally off the top of a staircase at a speed of 14.5 – and lands at bottom of the

Marrrta [24]

Question:

A skateboarder jumps horizontally off the top of a staircase at a speed of 14.5 and lands at bottom of the stairs. The staircase has a horizontal length of 8.00 m. We can ignore air resistance. What is the skater's vertical displacement during the jump?

Answer:

y = 1.48 m

Explanation:

Projectile motion is a two dimensional motion experienced by an object or particle that is subjected near the Earth's surface and moves along a curved path under the influence of gravity only. The path followed by projectile motion is called projectile path.

As the skateboarder followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

Also Skateboard jumps horizontally, So initial velocity has only horizontal component.

Horizontal component of initial velocity = $v_{i_{x}}$ = 14.5 m/s

Horizontal displacement = x = 8.00 m

Vertical displacement = y = ?

Using the following formula

x = $v_{i_{x}}$ t

8 = (14.5)(t)

t = 0.55 s

As skateboarder jumps horizontally, So there is no vertical component of velocity.

According to 2nd equation of motion

y = $v_{i_{y}}t + \frac{1}{2}gt^{2}$

As $v_{i_{y}} = 0$

So

y = 0.5gt²

y = 0.5*9.8*(0.55)²

y = 1.48 m

8 0

3 years ago

Read 2 more answers

A positively charged particle 1 is at the origin of a Cartesian coordinate system, and there are no other charged objects nearby

8090 [49]

Answer:

P=(2 nm, 8mn)

Explanation:

Given :

Position of positively charged particle at origin, $O=(0\ nm,0\ nm)$

Position of desired magnetic field, $D\equiv(1\ nm,8\ nm)$

Magnitude of desired magnetic field, $E=0\ N.C^{-1}$

Let q be the positive charge magnitude placed at origin.

<u>We know the distance between the two Cartesian points is given as:</u>

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

<u>For the electric field effect to be zero at point D we need equal and opposite field at the point.</u>

$\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 } =\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 }$

$\therefore (1-0)^2+(8-0)^2=r^2$

$r^2=65\ nm$

$r=\sqrt{65}$

as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the given point.

assuming that the second charge is placed at (x,y) nano-meters.

Therefore,

$x=2\times 1=2\ nm$

and

$y=2\times 8=16\ nm$

3 0

4 years ago