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2 years ago

Accumulation of excess electric charge on an object

1 answer:

5 0

Ch7 and 8 stguideQuestionAnswerThe rate at which electrical energy is converted into another form of energy is calledElectric powerA material in which electrons are able to move easily is a conductorThe accumulation of excess electric charge on an objectStatic electricity

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An object is traveling at a constant velocity of 15 m/s when it experiences a constant acceleration of 3.5 m/s2 for a time of 11

Luda [366]

Vf=Vi+at=15m/s+(3.5m/s^2)(11s)=53.5m/s

7 0

2 years ago

What is the change in internal energy (in J) of a system that does 4.50 ✕ 105 J of work while 3.20 ✕ 106 J of heat transfer occu

Dmitrij [34]

Answer:

$-3.25\times 10^6 J$

Explanation:

We are given that

Work done by the system= $4.5\times 10^5$ J

Heat transfer into the system= $U_1=3.2\times 10^6$ J

Heat transfer to the environment= $U_2=6\times 10^6$ J

We have to find the change in internal energy

By first law of thermodynamics

$\Delta Q=\Delta U+w$

$\Delta Q=U_1-U_2=3.2\times 10^6-6\times 10^6=-2.8\times 10^6J$

Substitute the values then we get

$-2.8\times 10^6=\Delta U+4.5\times 10^5$

$\Delta U=-2.8\times 10^6-4.5\times 10^5=-28\times 10^5-4.5\times 10^5=-32.5\times 10^5=-3.25\times 10^6 J$

Hence, the change in internal energy = $-3.25\times 10^6 J$

7 0

2 years ago

From the circuit above, predict which bulb (or bulbs) will be the brightest. Why do you think that?

inysia [295]

Answer:

the middle

Explanation:

the left one bulb gets power from the outher bulb

the one on right has more bulbs

7 0

2 years ago

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A block with mass m = 0.250 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth

ehidna [41]

Answer:

13.54 N/m

0.6 m

4.37 m/s

32.496 m/s²

Explanation:

m = Mass of block = 0.25 kg

k = Spring constant

A = Amplitude

x = Compression of spring = 0.24 m

a = Acceleration = -13 m/s²

v = Velocity = 4 m/s

The weight of the block and force on spring is equal

$ma=-kx\\\Rightarrow k=-\frac{ma}{x}\\\Rightarrow k=-\frac{0.25\times -13}{0.24}\\\Rightarrow k=13.54\ N/m$

The spring's force constant is 13.54 N/m

Total energy of the system is given by

$E=\frac{1}{2}(mv^2+kx^2)\\\Rightarrow E=\frac{1}{2}(0.25\times 4^2+13.54\times 0.24^2)\\\Rightarrow E=2.39\ J$

At maximum displacement v = 0

$E=\frac{1}{2}(mv^2+kA^2)\\\Rightarrow E=\frac{1}{2}(0+kA^2)$

$\\\Rightarrow E=\frac{1}{2}kA^2\\\Rightarrow A=\sqrt{\frac{E2}{k}}\\\Rightarrow A=\sqrt{\frac{2\times 2.39}{13.54}}\\\Rightarrow A=0.6\ m$

The amplitude of the motion is 0.6 m

Speed of the block

$E=\frac{1}{2}mv_m^2+0\\\Rightarrow v_m=\sqrt{\frac{2E}{m}}\\\Rightarrow v_m=\sqrt{\frac{2\times 2.39}{0.25}}\\\Rightarrow v_m=4.37\ m/s$

The maximum speed of the block during its motion is 4.37 m/s

Forces in the spring

$ma_m=kA\\\Rightarrow a_m=\frac{kA}{m}\\\Rightarrow a_m=\frac{13.54\times 0.6}{0.25}\\\Rightarrow a_m=32.496\ m/s^2$

Maximum magnitude of the block's acceleration during its motion is 32.496 m/s²

5 0

2 years ago

A free falling object has the velocity time graph shown. What is the objects displacement between 0.0 and 6.0s

zloy xaker [14]

For free fall motion the displacement can be found by graphically as well as by kinematics equation

Here acceleration of object is constant as it fall due to gravity so we can use

$d = v_i * t + \frac{1}{2}at^2$

here if body starts with zero initial speed then we can say

$d = 0 + \frac{1}{2}*9.8*t^2$

here we need to find the displacement from t = 0 to t = 6s

so we can say

$d = \frac{1}{2}*9.8*6^2$

$d = 176.4 m$

so the displacement will be 176.4 m

in order to find the displacement from the graph of velocity and time we need to find the area under the graph for given time interval that will also give us same displacement for given period of time.

6 0

2 years ago

Read 2 more answers