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kramer
3 years ago
5

A particle has a charge of 5.2..*.10ȹ© coulombs and experiences a force of 9.5..*.10ȹ¢ newtons when it travels through a magne

tic field with strength 2.2..*.10ȹ tesla. What is the speed of the particle?
Physics
1 answer:
Ira Lisetskai [31]3 years ago
6 0
The answer is 22.4 is what they going to get althoug is diffrent

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If you blow across the open end of a soda bottle and produce a tone of 250 Hz, what will be the frequency of the next harmonic h
Naddika [18.5K]

Answer:

Generally, the lowest overtone for a pipe open at one end and closed would be at  y / 4  where y represents lambda, the wavelength.

Since F (frequency) = c / y       Speed/wavelength

F2 / F1 = y1 / y2      because c is the same in both cases

F2 = y1/y2 * F1

F2 = 3 F1 = 750 /sec

Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe

and y1 = 3 y2

8 0
3 years ago
Which is the best example of the use of imagery in a sentence?
Talja [164]

Answer:

Common Examples of Imagery

Taste: The familiar tang of his grandmother's cranberry sauce reminded him of his youth. Sound: The concert was so loud that her ears rang for days afterward. Sight: The sunset was the most gorgeous they'd ever seen; the clouds were edged with pink and gold.

I hope it's helpful!

6 0
3 years ago
A worker pushes a box across the floor to the right at a constant speed with a force of 25N. What can you conclude about the box
kotykmax [81]

Answer:

C. Friction between the box and the floor is 25N to the left

Explanation:

7 0
3 years ago
Read 2 more answers
The atomic mass of gold is 0.197 kg/mole. how many moles are in 0.566 kg of gold.
iogann1982 [59]

Explanation:

0.566kg *(1mol/0.197 kg)= 2.87 mol gold

note how the units cancel out, if the units do not cancel out (kg/kg=1) then u did something wrong

7 0
3 years ago
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A heat-conducting rod consists of an aluminum section, 0.30 m long, and a copper section, .70m long. Both sections have a cross-
Igoryamba
Thermal conductions
K= QL/ART
Aluminium T₁ = 10 + 273.15
                    T₂ = 283.15k
205 = 2.0  × 0.30/4× 10⁻⁴ × (T₂ - 283.15)
Copper
385 = Q × 0.70/4×10⁻⁴ ×(433.15 - T₂)
Where T₃ = 160 + 273.15
T₃ = 433.15K
From 2 to 3
205/385 = 0.30/0.70 × 433.15 - T₂/T₂ - 283.15
= 0.53T₂ -150.06 = 181.92 - 0.42 T₂
→ 0.95T₂ = 331.98 ⇒ T₂ = ₂349.45k
T₂ = 76.3°c
=77°c.

6 0
3 years ago
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