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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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What would be the speed of an object just before hitting the ground if dropped 91.5 meters

Aleks04 [339]

Answer:

about 42.35 m/s

Explanation:

Use the equation for accelerated motion (g), and with zero initial velocity that doesn't include time:

$v_f^2=v_i^2+2\,a\,\Delta x$

which for our case would reduce to:

$v_f^2=v_i^2+2\,a\,\Delta x\\v_f^2=0+2\,9.8\,(91.5)\\v_f^2= 1793.4\\v_f=\sqrt{1793.4} \\v_f \approx 42.35$

then the velocity just before hitting would be about 42.35 m/s

5 0

3 years ago

Josh has a toy car of mass 3 kg tied to a string of length 2 m. He ties the string to a pole and has the toy car drive in a circ

gulaghasi [49]

Part a)

Centripetal acceleration is defined as

$a_c = \frac{v^2}{R}$

now here we know that

$v = 3m/s$

$R = 2 m$

m = 3 kg

now from above formula we have

$a_c = \frac{3^2}{2}$

$a_c = 4.5 m/s^2$

Part b)

Maximum possible tension in the string is given as

$T = 50 N$

now by force equation we have

$F = ma$

$50 = 3 a$

$a = \frac{50}{3} m/s^2$

now again by above formula

$\frac{v^2}{R} = \frac{50}{3}$

$v = \sqrt{\frac{50 \times 2}{3}}$

$v = 5.77 m/s$

5 0

3 years ago

Read 2 more answers

Why is conserving energy resources not a simple problem with a simple solution?

anyanavicka [17]

Answer:

Because the uncontrolled and unreasonable use of energy resources in the past few decades has led to the depletion of conventional type of resources. Although, measures are taken at the government levels to control the energy consumption, these attempts only meet with little success. Public awareness is the only way to successfully deal with this problem.

Explanation:

6 0

3 years ago

Read 2 more answers

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.