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Alborosie
3 years ago
7

Click on the reset button, and stack one 50kg. crate on top of the other, so that the total mass is 100kg. The Friction should b

e set to halfway between None and Lots. Adjust the Applied Force (force exerted) slowly, and determine when the crates begin accelerating. What is the minimum force that must be exerted on the crates so that they accelerate?
Physics
1 answer:
Helen [10]3 years ago
6 0

Answer:

490.5 N

Explanation:

Coefficient of friction is 0.5 since friction force is set to halfway between none and lots. Minimum force is given by multiplying the weight and coefficient of friction

F= kN where k is coefficient of friction while N is weight. Also, N=mg where m is mass and g is acceleration due to gravity.

F=kmg=0.5*100*9.81=490.5 N

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An X-ray tube emits X-rays with a wavelength of 1.0 x 10-11 m. Calculate the potential that must be applied across the X-ray tub
sergey [27]

Answer:

1.24 x 10 to the 5 ev = 124,000 ev          its B

Explanation:

E = hc/lambda = 1.24 ev-micrometer/1.0x10 to the -5 micrometers = 1.24 x 10 to the 5 ev = 124,000 ev

h = Planck's constant = 6.626 × 10 to the -34 joule·s

c = speed of light = 2.998 × 10 to the 8 m/s

lambda is the given wavelength

E is the desired photon energy

3 0
3 years ago
Read 2 more answers
Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g
ollegr [7]

Answer:

The  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

<u>velocity of wave on the string with greater tension;</u>

v_1 = \sqrt{\frac{T_1}{\mu }

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} =  \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

v_2^2 = \frac{T_2v_1^2}{T_1}  \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s

<u>Fundamental frequency of wave on the string with greater tension;</u>

<u />f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1}  =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz<u />

Beat frequency = F₂ - F₁

                          = 270.6 - 258

                          = 12.6 Hz

Therefore, the  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0
3 years ago
Which tool would you use to measure the amount of rainfall
mariarad [96]
A rain gauge! Hope this helps!
3 0
3 years ago
Read 2 more answers
20 POINTS PLUS BRAINLIEST What work is done by a system on its environment when an ideal gas is maintained at a constant pressur
Oliga [24]

Say you have 60 cubic meters of an ideal gas at a pressure of 200 pascals. You heat the gas until it expands to a volume of 120 cubic meters. How much work does the gas do? All you have to do is plug in the numbers:

The gas does 12,000 joules of work as it expands under constant pressure.


4 0
3 years ago
Read 2 more answers
A man paddles a canoe in a long, straight section of a river. The canoe moves downstream with constant speed 5 m/s relative to t
nikklg [1K]

Answer:

Explanation:

We shall consider all movement with respect to water assuming that river is at rest or motionless .

speed of canoe = 5 m /s

in five minutes , distance between hat and canoe = 5 x 60 x 5

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This distance will be covered by man in return journey . In this case his speed is 5 m /s again considering river constant .

So this will be covered at 5 m /s

time taken = 1500 / 5 = 300 s

= 5 minutes .

so it will take 5 minutes to row back to reclaim his hat .

7 0
3 years ago
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