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3 years ago

Which of the following is not true of a form?

Physics

1 answer:

Ludmilka [50]3 years ago

4 0

Answer:

It can only display one record at a time

Explanation:

Form ;

1. This is a document with spaces (also called placeholders or fields ) in which a series of documents with similar content can be written or selected.

2.This is the most popular method of data entry

3.It may contain images in the background.

4.This can be sorted data regardless of its source of information.

Only option C is wrong.

Therefore the answer C is correct.

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Starting from rest on a level road a girl can reach a speed of 5 m/s in 10s on her bicycle. Find average speed and distance trav

Mice21 [21]

Answer:

Average speed = 2.5 mph

Distance = 22.352 meters

Explanation:

Calculate speed, distance or time using the formula d = st, distance equals speed times time.

You can use this formula for any question like this

8 0

2 years ago

Steel train rails are laid in 12.0-m-long segments placed end to end. The rails are laid on a winter day when their temperature

Yanka [14]

Answer:

a) Space = $6.05 x 10^{-3} m = 0.605 cm$

b) Stress= $-100.8 x 10^{6} Pa$

Explanation:

1) Data Given

$L = 12 m , T_i = -9 C \degree, T_f = 33 C \degree$

2) Calculate the space using Linear thermal expansion formula

We need to use Linear thermal expansion formula since the space created would be a change on 1 dimension, the increase of the temperature will increase the length of the steel. The formula is given by:

$\Delta L = L_i \alpha_{steel} \Delta T$

We have everything except the $\alpha_{steel}$ , so we look for this on a book and we find that $\alpha_{steel} = 1.2 x 10^{-5} C^{-1}$ , so we can replace.

$\Delta L = 12 m (1.2 x 10^{-5} C^{-1}) (33 C \degree -(-9 C \degree)) = 12 m (1.2 x 10^{-5} C^{-1}) 42 C \degree =6.048 x 10^{-3} m = 0.6048cm$

3) Calculate the stress of the steel

The Stress is the ratio of applied force F to a cross section area - defined as

$\sigma = \frac{F_n}{A}$

Since we don't have the force and the Area, we need to look for another way to find the stress.

For this we can use the concept called Young's Modulus, defined as : "the mechanical property that measures the stiffness of a solid material", and the formula for this is given by:

$Y =\frac{F L}{A \Delta L}$ (1)

Solving $\frac{F}{A}$ from the previous formula we have this:

$\frac{F}{A}$ = (Y Δ L)/L (2)

From the Linear thermal expansion formula we can solve like this

$\frac{\Delta L}{L}$ = α ΔT (3)

And replacing equation (3) into equation (2) we have:

$\frac{F}{A}$ = Y α ΔT (4)

We have that the Young's Modulus for the steel is 20x10^{10} Pa, so replacing into equation (4)

$\frac{F}{A}$ = $20x10^{10}$ Pa (1.2x10^-5 C^-1) (42C) = $100.8 *10^{6}$ Pa

That represent the absolute value for the Stress, the sign on this case would be negative since there is a compression.

3 0

3 years ago

The energy of the electron in a hydrogen atom can be calculated from the Bohr formula: In this equation stands for the Rydberg e

labwork [276]

Answer:

Wavelength, $\lambda=657\ nm$

Explanation:

The energy of the electron in a hydrogen atom can be calculated from the Bohr formula as :

$E=\dfrac{-R}{n^2}$ .............(1)

Where

R is the Rydberg constant

n is the number of orbit

We need to find the wavelength of the line in the absorption line spectrum of hydrogen caused by the transition of the electron from an orbital with to an orbital with n₁ = 2 to an orbital with n₂ = 3.

Equation (1) can be re framed as :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$

$\dfrac{1}{\lambda}=1.096\times 10^7\times (\dfrac{1}{2^2}-\dfrac{1}{3^2})$

$\lambda=6.569\times 10^{-7}\ m$

$\lambda=657\ nm$

So, the the wavelength of the line in the absorption line spectrum is 657 nm. Hence, this is the required solution.

3 0

3 years ago

A wheel of radius 25cm has eight spokes. It is mounted on a fixed axle and is rotating at a constant angular speed w. You shoot

spayn [35]

Explanation:

We will assume that the rim of the wheel is also very thin, like the spokes. The distance s between the spokes along the rim is

$s = \frac{1}{8}C = \frac{1}{8}(2\pi)(0.25\:\text{m}) = 0.196\:\text{m}$

The 20-cm arrow, traveling at 6 m/s, will travel its length in

$t = \dfrac{0.2\:\text{m}}{6\:\text{m/s}} = \dfrac{1}{30}\:\text{s}$

The fastest speed that the wheel can spin without clipping the arrow is

$v = \dfrac{s}{t} = \dfrac{0.196\:\text{m}}{\left(\dfrac{1}{30\:\text{s}}\right)} = 5.9\:\text{m/s}$

The angular velocity $\omega$ of the wheel is given by

$\omega = \dfrac{v}{r} = \dfrac{5.9\:\text{m/s}}{0.25\:\text{m}} = 23.6\:\text{rad/s}$

In terms of rev/s, we can convert the answer above as follows:

$23.6\:\dfrac{\text{rad}}{\text{s}}×\dfrac{1\:\text{rev}}{2\pi\:\text{rad}} = 3.8\:\text{rev/s}$

As you probably noticed, I did the calculations based on the assumption that I'm aiming for the edge of the wheel because this is the part of the wheel where a point travels a longer linear distance compared to ones closer to the axle, thus giving the arrow a better chance to pass through the wheel without getting clipped by the spokes. If you aim closer to the axle, then the wheel needs to spin slower to allow the arrow to get through without hitting the spokes.

3 0

3 years ago

A baseball pitcher throws a ball with a speed of 41 m/s. Estimate the average acceleration of the ball during the throwing motio

dusya [7]

When t=0, v=0, d=0 When t=tf, v=41m/s, d=3.5m We have 2 formulas – the ones corresponding to uniformly accelerated linear movement: vf=a*t+vo d=(1/2)*a*t^2+vo*t Let’s put the data in the formulas: 41m/s=a*t+0=a*t 3.5m=(1/2)*a*t^2+0*t=1/2*a*t^2 You can use a variety of methods to find t and a. I will choose substitution. t=(41m/s)/a 3.5m=(1/2)*a*((41m/s)/a)^2=(1/2)*a*(41m/s)^2/a^2=(1/2)*(41m/s)^2/a a=(1/2)*(41m/s)^2/(3.5m)=(1/2)*41^2(m^2/s^2)/(3.5m) a=41^2(m/s^2)/( 2*3.5)=240m/s^2

8 0

3 years ago