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Alenkinab [10]
4 years ago
8

Students should only talk to friends when struggling to make a decision rather than a parent, teacher, or school counselor. Plea

se select the best answer from the choices provided. T F
Physics
2 answers:
velikii [3]4 years ago
7 0

Answer:

false

Explanation: took quiz

mojhsa [17]4 years ago
3 0
False, some adults will know more than your friends as they are already grown up and may have experienced the problem already
You might be interested in
A block of 250-mm length and 48 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b
Marrrta [24]

Answer:

153.6 kN

Explanation:

The elastic constant k of the block is

k = E * A/l

k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m

0.12% of the original length is:

0.0012 * 0.25 m = 0.0003  m

Hooke's law:

F = x * k

Where x is the change in length

F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)

The compressive load will generate a stress of

σ = F / A

F = σ * A

F = 80*10^6 * 0.048 * 0.04 = 153.6 kN

The smallest admisible load is 153.6 kN

8 0
3 years ago
On a transverse wave, the crest isa. the highest part of the wave.
Greeley [361]
I think it is C) the resting position of the wave.
3 0
3 years ago
Can any1 tell if my answer is right
jok3333 [9.3K]

Answer:

The correct answer would be C. 5.0kg

Explanation:

The mass of an object never changes unless parts of the object are taken away. In other words, although the gravitational force is different on the moon then on the earth the mass of the object would remain the same.

6 0
2 years ago
Which of the following contributes to the phases of the Moon as seen from Earth?
balandron [24]

Answer: Varying amounts of the Moon's lit surface being visible from Earth.

Explanation:

Phases of the moon can be defined as the different shapes of the moon visible from the Earth. This happens because sun lits up the face of moon and due to different position of moon in the orbit around earth, varying portion of the lit surface of the moon is visible from Earth. Refer to the diagram below:            

4 0
3 years ago
Read 2 more answers
A tire is filled with air at 10 ∘C to a gauge pressure of 250 kPa. Part A If the tire reaches a temperature of 45 ∘C, what fract
Alex_Xolod [135]

Answer:

The air fraction to be removed is 0.11

Given:

Initial temperature, T = 10^{\circ} = 283 K

Pressure, P = 250 kPa

Finally its temperature increases, T' = 45^{\circ} = 318 K

Solution:

Using the ideal gas equation:

PV = mRT

where

P = Pressure

V = Volume

m = no. of moles of gas

R = Rydberg's Constant

T = Temperature

Now,

Considering the eqn at constant volume and pressure, we get:

mT = m'T'

Thus

\frac{m}{m'} = \frac{T'}{T}                      (1)

Now, the fraction of the air to be removed for the maintenance of pressure at 250 kPa:

y = \frac{m - m'}{m} = 1 - \frac{m'}{m}

From eqn (1):

y = 1 - \frac{T}{T'}

y = 1 - \frac{283}{318} = 0.11

6 0
4 years ago
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