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Shkiper50 [21]
3 years ago
12

Four point charges have the same magnitude of 2.4×10^−12C and are fixed to the corners of a square that is 4.0 cm on a side. Thr

ee of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.

Physics
1 answer:
Gwar [14]3 years ago
6 0

Answer:

7.2N/C

Explanation:

Pls see attached file

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A 50-kg box is being pushed a distance of 8.0 m across the floor by a force P with arrow whose magnitude is 159 N. The force P w
Georgia [21]

Answer:

Wp = 1,272 J

Wf = -940.8 J

Wn = 0

Wg = 0

Explanation:

  • Applying the definition of work, as the product of the component of the  force applied, along the direction of the displacement, times the displacement, we find that due to the normal force is always perpendicular to the surface, it does no work, as it has not a component in the direction of the displacement, so Wn = 0.
  • As the weight goes directly downward, it has no component in the direction of the displacement either, so Wg = 0.
  • We can get the work done by the force applied P, simply as follows:

        W_{p} = F_{p} * d * cos \theta = 159 N * 8.0 m * cos 0 = 1,272 J (1)

  • Finally, we have the work done by the force of friction that always opposes to the displacement, so it has negative sign.
  • The frictional force , can be written as follows:

        F_{fr} = \mu k * F_{n} (2)

        where μk = coefficient of kinetic friction = 0.24

         Fn = Normal Force

  • In this case, since the surface is level and horizontal, and there is no acceleration of the box in the vertical direction, this means that the normal force (in magnitude) must be equal to the weight:
  • Fn = m*g = 50 kg * 9.8 m/s2 = 490 N
  • Replacing these values in (2), we get:

       F_{fr} = 0.24 * 50 kg* 9.8 m/s2 = 117.6 N

  • Applying the definition of work, we can get the work done by the frictional force, as follows:

        W_{ffr} = F_{fr} * d * cos \theta = F_{fr} * d * cos (180) = - F_{fr} * d = -117.6 N * 8.0 m \\ =  W_{ffr} = -940.8 N

4 0
3 years ago
neptune is an average distance of 4.5×10^12m from the sun. Estimate the length of the Neptunian year.
Vikentia [17]

As per Kepler's third law we know that

\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}

now here we know that

T_1 = year of Neptune

T_2 = year of Earth

R_1 = distance of Neptune from Sun

R_2 = Distance of Earth from Sun

so now we will have

\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}

T_1^2 = 27000

T_1 = 164.3 years

so length of year of Neptune is 164.3 years

6 0
3 years ago
How much heat do you need to raise the temperature of 150 g of oxygen from -30c to -15c?
Naddika [18.5K]
The amount of heat needed to raise the temperature of a substance by \Delta T is given by
Q=m C_s \Delta T
where
m is the mass of the substance
Cs is its specific heat capacity
\Delta T is the increase in temperature

For oxygen, the specific heat capacity is approximately 
C_s = 0.92 J/(g K)
The variation of temperature for the sample in our problem is 
\Delta T= -15^{\circ}C-(-30^{\circ} C)=+15^{\circ}C=15 K
while the mass is m=150 g, so the amount of heat needed is
Q=m C_s \Delta T=(150 g)(0.92 J/g K)(15 K)=2070 J
4 0
3 years ago
Read 2 more answers
Parachutes utilize maximum air ____ in order to limit the acceleration of the fall.
eduard

Answer: Resistance

Explanation:

8 0
3 years ago
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A force of 50 newtons causes a sled to accelerate at a rate of 5 meters per second. What is the mass of the sled.
notka56 [123]
F=ma
50=m(5)
m=10kg
hence,ans is B
5 0
3 years ago
Read 2 more answers
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