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andrew-mc [135]
3 years ago
11

A light, inextensible cord passes over a light, frictionless pulley with a radius of 15 cm. It has a(n) 12 kg mass on the left a

nd a(n) 4.1 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 3.3 m apart. The acceleration of gravity is 9.8 m/s 2 . 3.3 m 15 cm ω 12 kg 4.1 kg At what rate are the two masses accelerating when they pass each other? Answer in units of m/s 2 .
Physics
1 answer:
Lerok [7]3 years ago
7 0

Answer:

5.04m/s^2

Explanation:

We define the variables like this,

m_1 = 15Kg

m_2 = 4.8Kg

g= 9.8m/s^2

a= Acceleration

We know that the rate  is gived for the equation,

a = \frac{F_{net}}{m_{Total}}

a= \frac{ ( m1 - m2 ) g}{ (m1 + m2 )}

a= \frac {10.2 ( 9.8 )}{19.8} = 5.04m/s^2

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Answer: chihuahua

Explanation:

7 0
4 years ago
Read 2 more answers
URGENT PLEASE HELP ME ASAP
saul85 [17]

Answer:

The correct answer is - Denying the pleasure of an immediate reward so that you can receive a greater reward later.

Explanation:

Delayed gratification is a gratification method or the process that helps in improving self-control and helps in goals that take time to achieve. It is referred to as the ability to stop or prevent oneself from something more rewarding or pleasurable immediately, in order to get something more pleasurable or rewarding later.

For instance, an individual watch tv before a big exam or studying for the exam so can watch tv after getting good grades.

The correct answer is - option D.

7 0
3 years ago
If 61.4 cm of copper wire (diameter = 1.08 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendic
harkovskaia [24]

Answer:

the thermal energy generated in the loop = 6.64*10^{-4}  \ W

Explanation:

Given that;

The length of the copper wire L = 0.614 m

Radius of the loop  r = \frac{L}{2 \pi}

r = \frac{0.614}{2 \pi}

r = 0.0977 m

However , the area of the loop is :

A_L = \pi r^2

A_L = \pi (0.0977)^2

A_L = 0.02999 \ m^2

Change in the magnetic field is \frac{dB}{dt}= 0.0914 \ T/s

Then the induced emf e = A_L \frac{dB}{dt}

e = 0.02999 * 0.0914

e = 2.74 × 10⁻³ V

resistivity of the copper wire \rho = 1.69* 10^{-8} Ω m

diameter of the wire = 1.08 mm

radius of the wire = 0.54 mm = 0.54 × 10⁻³  m

Thus, the resistance of the wire  R = \frac {\rho L}{\pi r^2}

R =  \frac{(1.69*10^{-8})(0.614)}{   \pi (0.54*10^{-3})^2}

R = 1.13× 10⁻² Ω

Finally,  the thermal energy generated in the loop (i.e the power) = \frac{e^2}{R}

= \frac{(2.74*10^{-3})^2}{1.13*10^{-2}}

= 6.64*10^{-4}  \ W

8 0
3 years ago
a 55.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 130 kg football players at a height of 0.800 m . the diamet
suter [353]

The diameter of the football player's piston is 0.55 m

Given that the mass of the cheerleader(m) is 55 kg, mass of football player to be hold (M) is 130 kg, height of the players (h) is 1.30 m, radius of the piston corresponding to the diameter (r) is 0.09 m, Diameter of football player's piston (R), P1 is Pressure on the cheerleader's side, P2 is Pressure on the football player's side

Using Pascal's law,

This law states that if there is a change at a point of a body immersed in a fluid then that change will spread thoroughly to each and every point of the body.

The formula of hydraulic system is,

P1= P2

F1/A1 = F2/A2

mg/πr^2 = 4Mg/πR^2

m/r^2=4M/R^2

R^2=4M×r^2/m

By plugging the values, we get.,

R^2=(4×130×0.09^2)/55

R^2=4.21/55=0.076

R=√0.076 = 0.275 m

Hence, diameter of football player's piston is 0.55 m

Learn more about Hydraulic lift here

brainly.com/question/19052774

#SPJ4

8 0
1 year ago
During a breath, the average human lung expands by about 0.50 l. part a if this expansion occurs against an external pressure of
saveliy_v [14]
Work, in thermodynamics, is the amount of energy that is transferred from one system to another system without transfer of entropy. It is equal to the external pressure multiplied by the change in volume of the system. It is expressed as follows:<span>

W = PdV

Integrating and assuming that P is not affected by changes in V or it is constant, then we will have:

W = P (V2 - V1)

Substituting the given values:
P = 1.0 atm = 101325 Pa
(V2 - V1) = 0.50 L = 
W = 101325 N/m^3 ( 0.50) (1/1000) m^3
W = 50.66 N-m or 50.66 J
<span>
So, in the expansion process about 50.66 J of work is being done.
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8 0
3 years ago
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