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mestny [16]
3 years ago
12

A boy throws a stone vertically in the air with an initial speed of 40m/sec.At the instant the stone is thrown,a monkey at the t

op of a tree h m high falls freely.How long will it take for the stone to hit the monkey if they meet at a point mid way between the ground and the top of tree?
Physics
1 answer:
mote1985 [20]3 years ago
6 0

∞infinaty stones∞

because why not

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What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?
Daniel [21]

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

F=\frac{kq_{1}q_{2}  }{r^{2} }

Here, q_{1} andq_{2} are the charges on the pith balls, r is the separation between the charges and k is constant and its value is  8.99\times 10^9 N m^2/C^2.

Given  q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C and  r=8 cm=8\times10^{-2} m.

Substituting these values in above formula we get,

F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9}  N\\\\F=8.8\times10^{-4}N

Thus, the repulsive force between two pith balls is 8.8\times10^{-4}N.

7 0
3 years ago
Read 3 more answers
What is the minimum coeffecient of static friction μmin required between the ladder and the ground so that the ladder does not s
mars1129 [50]

Answer:

\mu = \frac{1}{2tan\theta}

Explanation:

let the ladder is of mass "m" and standing at an angle with the ground

So here by horizontal force balance we will have

\mu N_1 = N_2

by vertical force balance we have

N_1 = mg

now by torque balance about contact point on ground we will have

mg(\frac{L}{2}cos\theta) = N_2(L sin\theta)

so we will have

N_2 = \frac{mg}{2tan\theta}

now from first equation we have

\mu (mg) = \frac{mg}{2tan\theta}

\mu = \frac{1}{2tan\theta}

3 0
3 years ago
Estimate the Joule-Thomson coefficient of refrigerant-134a at 0.70 MPa and 50°C. Assume the second state will be selected for a
leva [86]

Answer:

\mu = 0.018\,\frac{^{\textdegree}C}{kPa}

Explanation:

The Joule-Thomson coefficient is the ratio of the change of temperature to the change of pressure under isoenthalpic conditions:

\mu = \left(\frac{\Delta T}{\Delta P}\right)_{h}

Initial and final properties are:

T_{1} = 50^{\textdegree}C, P_{1}=700\,kPa, h_{1}=288.54\,\frac{kJ}{kg}. Superheated Vapor.

T_{2} = 48.186^{\textdegree}C, P_{2}=600\,kPa, h_{2}=288.54\,\frac{kJ}{kg}. Superheated Vapor.

The Joule-Thomson coefficient is approximately:

\mu = \frac{50^{\textdegree}C-48.186^{\textdegree}C}{700\,kPa-600\,kPa}

\mu = 0.018\,\frac{^{\textdegree}C}{kPa}

5 0
4 years ago
A flywheel of mass M is rotating about a vertical axis with angular velocity ω0. A second flywheel of mass M/5 is not rotating a
Contact [7]

Answer:

0.83 ω

Explanation:

mass of flywheel, m = M

initial angular velocity of the flywheel, ω = ωo

mass of another flywheel, m' = M/5

radius of both the flywheels = R

let the final angular velocity of the system is ω'

Moment of inertia of the first flywheel , I = 0.5 MR²

Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²

use the conservation of angular momentum as no external torque is applied on the system.

I x ω = ( I + I') x ω'

0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'

0.5 x MR² x ωo = 0.6 MR² x ω'

ω' = 0.83 ω

Thus, the final angular velocity of the system of flywheels is 0.83 ω.

5 0
4 years ago
State any five branches of physics​
prisoha [69]

Answer:

  1. Thermodynamics
  2. Quantum mechanics
  3. Nuclear physics
  4. Mechanics
  5. Astrophysics

7 0
3 years ago
Read 2 more answers
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