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schepotkina [342]
3 years ago
10

A heavy boy and a lightweight girl are balanced on a massless seesaw. The boy moves backward, increasing his distance from the p

ivot point. What happens to the seesaw? It is impossible to predict without knowing additional information. The side the boy is sitting on will tilt downward. The side the girl is sitting on will tilt downward. Nothing; the seesaw will remain balanced.
Physics
1 answer:
victus00 [196]3 years ago
6 0

Answer:

The side the boy is sitting on will tilt downward.

Explanation:

According to the law of moments when the same force is applied at a greater distance from the pivot then the effect of moment is greater about that point.

<u>Mathematically momentum is given as:</u>

M=F\times r

where:

F is the applied force at a distance 'r' acting in a direction perpendicular to the line joining the point of application and the hinge.

  • Moment is the rotational effect of the applied force on the body.

<em>When the boy of a heavier mass than the girl was sitting on a balanced see-saw then it is certain that he was closer to the hinge than the girl to balance the turning effect (in case of an unbiased see-saw). When the body moves farther his weight is same but the radial distance from the hinge increases which increases his moment of weight.</em>

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Answer:

39 m/s

Explanation:

Hi!

The equation of motion of the ball is:

y(t) = y_0 + v_{0y}t-(1/2)(9.8 m/s^2)t^2

Where y_0 is the initial horizontal position, which we will take as zero.

v_0y is the initial velocity and t is the time

To find the initial velocity we will set y(7.96s) = 0

0 = v_{0y} (7.96 s)-(4.9 m/s^2)(7.96 s)^2\\v_{0y} =(4.9 m/s^2) (7.96 s)\\v_{0y}=39.004 m/s

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4 years ago
A 12 kg box is pulled across the floor with a 48 N horizontal force. If the force of friction is 12 N, what is the acceleration
Oksi-84 [34.3K]

Answer:

The acceleration of the box is 3 m/s²

Explanation:

Given;

mass of the box, m = 12 kg

horizontal force pulling the box forward, Fx = 48 N

frictional force acting against the box in opposite direction, Fk = 12 N

The net horizontal force on the box, F = 48 N - 12 N

The net horizontal force on the box, F = 36 N

Apply Newton's second law of motion to determine the acceleration of the box;

F = ma

where;

F is the net horizontal force on the box

a is the acceleration of the box

a = F / m

a = 36 / 12

a = 3 m/s²

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3 years ago
(8 points) Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet
trapecia [35]

Answer:

a ) \dot m = 351.49 kg/s

b)  \dot m_{actual} = 1046.15 kg/s

Explanation:

given data:

pressure ration rp = 12

inlet temperature = 300 K

TURBINE inlet temperature  = 1000 K

AT the end of isentropic process (compression) temperature is

\frac{T_2'}{T_1} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{T_2'}{300} = 12^{\frac{1.4 -1}{1.4}}

T_2' = 610.181 K

AT the end of isentropic process (expansion) temperature is

\frac{T_3}{T_4'} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{1000'}{T_4'} = 12^{\frac{1.4 -1}{1.4}}

T_4' = 491.66 K

isentropic work is given as

w(compressor) = CP (T_2' -T_1)

w = 1.005(610.18 - 300)

w = 311.73 kJ/kg

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\dot m = \frac{70000}{510.88 - 311.73}

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b) actual mass flow rate uis given as

\dot m_{actual} = \frac{70000}{51.088\times 0.85 - \frac{311.73}{0.85}}

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