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Triss [41]
2 years ago
12

Two point charges of magnitudes +5.00 μC, and +7.00 μC are placed along the x-axis at x = 0 cm and x = 100 cm, respectively. Whe

re must a third charge be placed along the x-axis so that it does not experience any net force because of the other two charges? Two point charges of magnitudes +5.00 μC, and +7.00 μC are placed along the x-axis at x = 0 cm and x = 100 cm, respectively. Where must a third charge be placed along the x-axis so that it does not experience any net force because of the other two charges? 50 cm 45.8 cm 9.12 cm 91.2 cm 4.58 cm
Physics
1 answer:
joja [24]2 years ago
8 0

Answer:

45.8 cm

Explanation:

To solve this, we will use the formula

5 / x² = 7/(1 - x)²

5 / x² = 7 / (1 - 2x + x²)

5 / 7 = x² / (1 - 2x + x²)

x = 0.5 * (√(35) - 5) meters

x = 0.5 * (5.916 - 5)

x = 0.5 * (0.916)

x = 0.458 or x = 45.8

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A 92-kg skier is sliding down a ski slope that makes an angle of 30 degrees above the horizontal direction. The coefficient of k
9966 [12]

Answer:

a = 4.05 m/s²

Explanation:

Known data

m= 92 kg  : mass of the  skier

θ =30°  :angle θ of the ski slope  with respect to the horizontal direction

μk= 0.10 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the skier

W: Weight of the skier : In vertical direction

N : Normal force : perpendicular to the ski slope

f : Friction force: parallel to the ski slope

Calculated of the W

W= m*g

W=  92kg* 9.8 m/s² = 901,6 N

x-y weight components

Wx= Wsin θ= 901,6 N *sin 30° = 450.8 N

Wy= Wcos θ = 901,6 N *cos 30° =780.8 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - Wy = 0

N = Wy

N = 780.8 N

Calculated of the f

f = μk* N=  0.10*780.8 N  

f = 78.08 N

We apply the formula (1) to calculated acceleration of the skier:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx - f = m*a

450.8- 78.08 = ( 92)*a

372.72 =  (92)*a

a = (372.72)/ (92)

a = 4.05 m/s²

6 0
2 years ago
You are pulling your little sister on her sled across an icy (frictionless) surface. When you exert a constant horizontal force
Tpy6a [65]

Answer:

Mass of Little Sister = 44.17 kg

Explanation:

From Newton's second law of motion, the magnitude of force applied on the sled is given by the following formula:

F = ma

where,

F = Force Applied = 120 N

a = Acceleration = 2.3 m/s²

m = Mass of Sled + Mass of Little Sister = 8 kg + Mass of Little Sister

Therefore,

120 N = (2.3 m/s²)(8 kg + Mass of Little Sister)

(120 N)/(2.3 m/s²) = 8 kg + Mass of Little Sister

Mass of Little Sister = 52.17 kg - 8 kg

<u>Mass of Little Sister = 44.17 kg</u>

4 0
2 years ago
You are participating in a NASA traineeship, working with a group planning a new landing on Mars. Your supervisor has come up wi
aivan3 [116]

Answer:

h=17005.8 km

Explanation:

Newton's law of universal gravitation states that the force experimented by a satellite of mass m orbiting Mars, which has mass M=6.39\times10^{23} kg at a distance r will be:

F=\frac{GMm}{r^2}

where G=6.67\times10^{-11}Nm^2/kg^2 is the gravitational constant.

This force is the centripetal force the satellite experiments, so we can write:

F=ma_{cp}=mr\omega^2=mr(\frac{2\pi}{T})^2=\frac{4\pi^2mr}{T^2}

Putting all together:

\frac{GMm}{r^2}=\frac{4\pi^2mr}{T^2}

which means:

r=\sqrt[3]{\frac{GM}{4\pi^2}T^2}

Which for our values is:

r=\sqrt[3]{\frac{(6.67\times10^{-11}Nm^2/kg^2)(6.39\times10^{23} kg)}{4\pi^2}(1.026\times24\times60\times60s)^2}=20395282m=20395.3km

Since this distance is measured from the center of Mars, to have the height above the Martian surface we need to substract the radius of Mars R=3389.5 km , which leaves us with:

h=r-R=20395.3km-3389.5 km=17005.8 km

6 0
2 years ago
the grid in a triode is kept negatively charged to prevent… a. the variations in voltage from getting too large. b. electrons be
Nezavi [6.7K]
D:the electrons from being attracted to the grid instead of the anode
5 0
2 years ago
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The famous cliff divers of Acapulco leap from a perch 35 m above the ocean. How fast are they moving when they reach the surface
Rus_ich [418]

1) 26.2 m/s

The mechanical energy of the divers at any point of their vertical motion is sum of the kinetic energy and the gravitational potential energy:

E=K+U = \frac{1}{2}mv^2 + mgh

where

m is the mass of the diver

v is the speed

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height above the water

When the diver is on the cliff, v = 0 (he is at rest), so K=0 and the initial mechanical energy is just potential energy:

E_i = mgh

where h=35 m is the height of the cliff.

When the diver hits the water above, h = 0, so U=0 and the final mechanical energy is just kinetic energy:

E_f = \frac{1}{2}mv^2

since the total mechanical energy is conserved, we have

E_i = E_f\\mgh = \frac{1}{2}mv^2

And solving the equation for v, we find the speed when they reach the surface of the water:

v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(35 m)}=26.2 m/s

2) It is converted into thermal energy of the water

When the diver enters the water, he suddenly feels another force acting against the motion of the diver: the resistance of the water. The resistance of the water acts upward, slowing down the diver until he stops.

In this process, the speed of the diver (v) decreases, and therefore the kinetic energy of the diver decreases as well, until it becomes zero.

However, this does not mean that the conservation of energy has been violated. In fact, the kinetic energy of the diver has been converted into thermal energy of the molecules of water surrounding the diver.

8 0
2 years ago
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