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Mashutka [201]
2 years ago
12

Can a physical change, change what a substance is

Physics
1 answer:
Elden [556K]2 years ago
7 0
No the substance will remain the same substance as before.
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Which simple machine is a doorknob?
Anestetic [448]

The answer is wheel and axle

3 0
3 years ago
A proton moves perpendicular to a uniform magnetic field B at a speed of 2.30 107 m/s and experiences an acceleration of 1.70 10
anzhelika [568]

Answer:

Explanation:

Given the following :

Speed (V) = speed of 2.30×10^7 m/s

Acceleration (a) = 1.70×10^13 m/s^2

Using the right hand rule provided by  Lorentz law:

B = F / qvSinΘ

Where B = magnitude of the magnetic field

v = speed of the particle

Θ = 90° (perpendicular to the field)

q = charge of the particle

SinΘ = sin90°  = 1

Note F = ma

Therefore,

B = ma / qvSinΘ

Mass of proton = 1.67 × 10^-27

Charge = 1.6 × 10^-19 C

B = [(1.67 × 10^-27) × (1.70 × 10^13)] / (1.6 × 10^-19) × (2.30 × 10^7) × 1

B = 2.839 × 10^-14 / 3.68 × 10^-12

B = 0.7715 × 10^-2

B = 7.72 × 10^-3 T

2) Magnetic field will be in the negative y direction according to the right hand thumb rule.

Since Velocity is in the positive z- direction, acceleration in the positive x - direction, then magnetic field must be in the negative y-direction.

5 0
3 years ago
What is the mechanical advantage of a 8 m ramp that rises 2 m to a stage?
Neko [114]

Answer:

Mechanical advantage = 4

Explanation:

Given the following data;

Distance of effort, de = 8m

Distance of ramp, dr = 2m

To find the mechanical advantage;

Mechanical advantage = de/dr

Substituting into the equation, we have;

Mechanical advantage = 8/2

Mechanical advantage = 4

5 0
2 years ago
When the sun and moon work together ____________ are formed on Earth.
Paha777 [63]

Answer:

C. Spring Tides is the answer.

7 0
2 years ago
A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me
loris [4]

To solve this exercise we will use the concept related to heat loss which is mathematically given as

Q = mC_p \Delta T

Where,

m = mass

C_p= Specific Heat

\Delta T = Change in temperature

Replacing with our values we have that

m = 0.1g

C_p = 139J/Kg\cdot K \rightarrow Specific heat of mercury

\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K

Replacing

Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J

Therefore the heat lost by mercury is 0.09J

5 0
3 years ago
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