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JulsSmile [24]
3 years ago
12

The acceleration (a) of a particle moving with uniform speed (v) in a circle of radius (r) is found to be propotional to r^n and

v^m, where n and m are constants. Determine the values of n and m and write the simplest form of an equation of acceleration
Physics
2 answers:
KATRIN_1 [288]3 years ago
6 0

Answer:

n=-1 and m=2

Explanation:

A particle is moving uniformly with acceleration "a"

Uniform speed is v

And radius of circle is r

The acceleration is proportional to

r^n and v^m

i.e

a∝ rⁿ

a ∝v^m

Combing the two

Then, a∝rⁿv^m

Let k be constant of proportionality

Then, a=krⁿv^m. Equation 1

So, we know that the centripetal acceleration keeping an object in circular path is given as

a=v²/r

Rearranging

a=v²r^-1. Equation 2

So comparing this to the proportional

Equating equation 1 and 2

krⁿv^m = v²r^-1

This shows that,

k=1

rⁿ = r^-1

Then, n =-1

Also, v^m =v²

Then, m=2

Therefore,

n=-1 and m=2

Zanzabum3 years ago
5 0

Answer:

m=2

n= -1

a=Kv²/r

Explanation:

using dimensional analysis

V = meter/second = m/s

r = meter = m

a = meter/second²=m/s²

If a is proportional to r^n v^m

we have a=r^n v^m =(m)^n (m/s)^m = m/s²

from law of indices;

m^n+m/s^m =m/s²

 using system of equations

n+m=1

m=2

so n= -1

then a=kr^n v^m

a=Kv²/r

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Pls ans 10 no. From laws of motion
Vlad1618 [11]

Answer:

40N

Explanation:

Since both weights are connected to one string, you can say that the tensions above each are equal to each other.

If you do the sum of forces for the 4kg mass, then the tension comes out to 40N (if we take gravity to be 10m/s²). But that seemed too good to be true, so I decided to do the work for the 7kg mass as well [which included finding the normal force (N) and plugging it into the sum of forces for the 7kg mass] to find that it also gives 40N as the answer.

If I were to put my process into steps:

  1. Write out the sum of Forces for both masses
  2. Set them equal to each other to find normal force (because this is the only unknown)
  3. Calculate and compare the two tensions to see if they are equal

*This all seems to line up perfectly, but do let me know if my answer doesn't match up with what you might find to he the answer later on.

4 0
3 years ago
MATHPHYS CAN U HELP ME PLEASE
ludmilkaskok [199]

Explanation:

(1) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.041 kg) (2090 J/kg/°C) (0°C − (-11°C)) = 942.59 J

The heat added to melt the ice is:

q = mL = (0.041 kg) (3.33×10⁵ J/kg) = 13,653 J

The heat added to warm the water to 100°C is:

q = mCΔT = (0.041 kg) (4186 J/kg/°C) (100°C − 0°C) = 17,162.6 J

The heat added to evaporate the water is:

q = mL = (0.041 kg) (2.26×10⁶ J/kg) = 92,660 J

The heat added to warm the steam to 115°C is:

q = mCΔT = (0.041 kg) (2010 J/kg/°C) (115°C − 100°C) = 1236.15 J

The total heat needed is:

q = 942.59 J + 13,653 J + 17,162.6 J + 92,660 J + 1236.15 J

q = 125,654.34 J

(2) When the first two are mixed:

m C₁ (T₁ − T) + m C₂ (T₂ − T) = 0

C₁ (T₁ − T) + C₂ (T₂ − T) = 0

C₁ (6 − 11) + C₂ (25 − 11) = 0

-5 C₁ + 14 C₂ = 0

C₁ = 2.8 C₂

When the second and third are mixed:

m C₂ (T₂ − T) + m C₃ (T₃ − T) = 0

C₂ (T₂ − T) + C₃ (T₃ − T) = 0

C₂ (25 − 33) + C₃ (37 − 33) = 0

-8 C₂ + 4 C₃ = 0

C₂ = 0.5 C₃

Substituting:

C₁ = 2.8 (0.5 C₃)

C₁ = 1.4 C₃

When the first and third are mixed:

m C₁ (T₁ − T) + m C₃ (T₃ − T) = 0

C₁ (T₁ − T) + C₃ (T₃ − T) = 0

(1.4 C₃) (6 − T) + C₃ (37 − T) = 0

(1.4) (6 − T) + 37 − T = 0

8.4 − 1.4T + 37 − T = 0

2.4T = 45.4

T = 18.9°C

(3) Heat gained by the ice = heat lost by the tea

mL + mCΔT = -mCΔT

m (3.33×10⁵ J/kg) + m (2090 J/kg/°C) (30.8°C − 0°C) = -(0.176 kg) (4186 J/kg/°C) (30.8°C − 32.8°C)

m (397372 J/kg) = 1473.472 J

m = 0.004 kg

m = 4 g

4 grams of ice is melted and warmed to the final temperature, which leaves 128 grams unmelted.

(4) The heat added to warm the ice to 0°C is:

q = mCΔT = (0.028 kg) (2090 J/kg/°C) (0°C − (-67°C)) = 3920.84 J

The heat added to melt the ice is:

q = mL = (0.028 kg) (3.33×10⁵ J/kg) = 9324 J

The heat added to warm the melted ice to T is:

q = mCΔT = (0.028 kg) (4186 J/kg/°C) (T − 0°C) = (117.208 J/°C) T

The heat removed to cool the water to T is:

q = -mCΔT = -(0.505 kg) (4186 J/kg/°C) (T − 27°C)

q = (2113.93 J/°C) (27°C − T) = 57076.11 J − (2113.93 J/°C) T

The heat removed to cool the copper to T is:

q = -mCΔT = -(0.092 kg) (387 J/kg/°C) (T − 27°C)

q = (35.604 J/°C) (27°C − T) = 961.308 J − (35.604 J/°C) T

Therefore:

3920.84 J + 9324 J + (117.208 J/°C) T = 57076.11 J − (2113.93 J/°C) T + 961.308 J − (35.604 J/°C) T

13244.84 J + (117.208 J/°C) T = 58037.418 J − (2149.534 J/°C) T

(2266.742 J/°C) T = 44792.58 J

T = 19.8°C

(5) Kinetic energy of the hammer = heat absorbed by ice

KE = q

½ mv² = mL

½ (0.8 kg) (0.9 m/s)² = m (80 cal/g × 4.186 J/cal × 1000 g/kg)

m = 9.68×10⁻⁷ kg

m = 9.68×10⁻⁴ g

(6) Heat rate = thermal conductivity × area × temperature difference / thickness

q' = kAΔT / t

q' = (1.09 W/m/°C) (4.5 m × 9 m) (10°C − 4°C) / (0.09 m)

q' = 2943 W

After 10.7 hours, the amount of heat transferred is:

q = (2943 J/s) (10.7 h × 3600 s/h)

q = 1.13×10⁸ J

q = 113 MJ

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