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3 years ago

Displacement vs distance

1 answer:

6 0

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position

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Light from a laser strikes a diffraction grating that has 5500 lines per centimeter. The central and first-order maxima are sepa

WINSTONCH [101]

.Answer:

491.4 nm

Explanation:

The distance between central and first maxima is,

$y=0.455m$

And the distance between screen abnd grating is,

$L=1.62 m$

Now the angle can be find as,

$tan\theta=\frac{y}{L} \\\theta=tan^{-1}(\frac{0.455}{1.62}) \\\theta=15.68^{\circ}$

Now the grating distance is,

$d=\frac{1}{5500} cm\\d=1.82\times 10^{-6}m$

Now with m=1 condition will become,

$\lambda=dsin\theta$

So,

$\lambda=1.82\times 10^{-6}m\times sin(15.68^{\circ})\\\lambda=1.82\times 10^{-6}m\times 0.270\\\lambda=491.4\times 10^{-9}m\\\lambda=491.4 nm$

Therefore the wavelength of laser light is 491.4 nm.

3 0

3 years ago

You are using a bucket of water in a science experiment. You conclude that water is a liquid because

Harrizon [31]

Well you could see it's physical properties have the same descriptions corresponding to a liquid. (Sorry if this doesn't help you..)

8 0

4 years ago

When a battery is attached to an electrical circuit, the electrical energy produced comes from stored ___ energy.

Llana [10]

Answer:

Chemical energy

Explanation:

6 0

3 years ago

Read 2 more answers

The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the ca

Xelga [282]

The question is incomplete. The complete question is :

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .

Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:

1. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.

Solution :

Given :

$A = 10 \ cm^2$

$=0.0010 \ m^2$

d = 10 mm

= 0.010 m

Then, Capacitance,

$C=\frac{k \epsilon_0 A}{d}$

$C=\frac{8.85 \times 10^{12} \times 3 \times 0.0010}{0.010}$

$C=2.655 \times 10^{12} \ F$

$U_1 = \frac{1}{2}CV^2$

$U_1 = \frac{1}{2} \times 2.655 \times 10^{-12} \times (15V)^2$

$U_1=2.987 \times 10^{-10}\ J$

Now,

$C_k=\frac{1}{2} \frac{k \epsilon_0}{d} \times \frac{A}{2}$

And

$C_{air}=\frac{1}{2} \frac{\epsilon_0}{d} \times \frac{A}{2}$

In parallel combination,

$C_{eq}= C_k + C_{air}$

$C_{eq} = \frac{1}{2} \frac{\epsilon_0 A}{d}(1+k)$

$C_{eq} = \frac{1}{2} \times \frac{8.85 \times 10^{-12} \times 0.0010}{0.01} \times (1+3)$

$C_{eq} = 1.77 \times 10^{-12}\ F$

Then energy,

$U_2 =\frac{1}{2} C_{eq} V^2$

$U_2=\frac{1}{2} \times 1.77 \times 10^{-12} \times (15V)^2$

$U_2=1.99 \times 10^{-10} \ J$

b). Now the charge on the $\text{capacitor}$ is :

$Q=C_{eq} V$

$Q = 1.77 \times 10^{-12} \times 15 V$

$Q = 26.55 \times 10^{-12} \ C$

Now when the capacitor gets disconnected from battery and the $\text{dielectric}$ is slowly $\text{removed the rest}$ of the way out of the $\text{capacitor}$ is :

$C_3=\frac{A \epsilon_0}{d}$

$C_3 = \frac{0.0010 \times 8.85 \times 10^{-12}}{0.01}$

$C_3=0.885 \times 10^{-12} \ F$

$C_3 = 0.885 \times 10^{-12} \ F$

Without the dielectric,

$U_3=\frac{1}{2} \frac{Q^2}{C}$

$U_3=\frac{1}{2} \times \frac{(25.55 \times 10^{-12})^2}{0.885 \times 10^{-12}}$

$U_3=3.98 \times 10^{-10} \ J$

4 0

3 years ago

To make a position-time graph, you plot position relative to the starting point on the y-axis against the corresponding_______on

galina1969 [7]

Answer:

time.

Explanation:

Motion can be defined as a change in the location (position) of a physical object or body with respect to a reference point.

This ultimately implies that, motion would occur as a result of a change in location (position) of an object with respect to a reference point or frame of reference i.e where it was standing before the effect of an external force.

A reference point refers to a location or physical object from which the motion (movement) of another physical object or body can be determined.

Mathematically, the motion of an object is described in terms of acceleration, time, distance, speed, velocity, position, displacement, etc.

A graph can be defined as the graphical representation of data (informations) on horizontal and vertical lines i.e x-axis and y-axis respectively.

A position-time (x-t) graph is a graph of the position of an object against (versus) time.

To make a position-time graph, you should plot the values of the position of an object or body with respect to the starting point on the y-axis against the corresponding time on the x-axis.

Generally, the slope of the line of a position-time (x-t) graph is typically used to determine or calculate the velocity of an object.

An instantaneous velocity can be defined as the rate of change in position of an object in motion for a short-specified interval of time. Thus, an instantaneous velocity is a quantity that can be found by measuring the slope of a line that is tangent to a point on the graph.

In conclusion, the x - t graph also referred to as the position-time graph is used for determining the instantaneous velocity from the slope.

4 0

3 years ago