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Tatiana [17]
1 year ago
7

Ejbsjca bdsj dbcsc j ecsabkfskbj jcbaskjbb

Physics
1 answer:
alexandr1967 [171]1 year ago
3 0

Sorry but I can't help you cuz the words you use makes no sens if you are going to ask a question pls do and dont joke around about it thank you very much and have a great day! ^^

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5p - 14 = 8 p + 4 find p
Digiron [165]

Answer:

Explanation:

5p - 14 = 8p + 4

5p = 8p + 18 <-- Moving constants to one side; add the same number of +14 to both sides.

-3p = 18. <-- The same thing with the variable itself.

p = -6 <-- Divide both sides by negative 3.

6 0
3 years ago
A ball is rolled uphill a distance of 12 meters before it slows, stops, and begins to roll back. The ball rolls downhill 20 mete
Sedaia [141]
The ball rolled a distance of
d = 12m + 20m.
But the change of position is
x = + 12m - 20m
5 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
A firecracker in a coconut blows the coconut into three pieces. two pieces of equal mass fly off south and west, perpendicular t
netineya [11]
Here we have mass that moves at ceratin speed. This means that we have momentum. The law that must be observed is law of conservation of momentum. It states that momentum before certain event is equal to a momentum after that event. Here we have three masses so we can write this as:
m_{1}  v_{1i} + m_{2}  v_{2i} + m_{3}  v_{3i} = m_{1}  v_{1f} + m_{2}  v_{2f} + m_{3}  v_{3f}
Before the firecracker blows a coconut does not move, so left side is equal to 0:
0 = m_{1} v_{1f} + m_{2} v_{2f} + m_{3} v_{3f}

We know that m1=m2=m and m3=2m. Also we are asked to find v3f so we can rewrite formula:
v_{3f} = -  \frac{m_{1}  v_{1f}  + m_{2} v_{2f} }{ m_{3} }

We must take in consideration that two parts with same mass do not move in same direction. The center of mass of these two parts moves between them at angle of 45° with respect to both south and west. The speed of a center of mass is:
v_{f} = \sqrt{ v_{1f}^{2}+ v_{2f}^{2} } \\ \\ v_{f} = 33.9m/s
This speed we can insert into formula for v3f:
v_{3f} = - \frac{m*33.9+m*33.9 }{ 2m } \\  \\ v_{3f} = - \frac{2m*33.9 }{ 2m }  \\  \\ v_{3f} = - 33.9m/s

We can see that part of a coconut with biggest mass has same speed as center of mass of two other parts. Negative sign shows that direction is opposite to direction of two pats. Biggest part moves towards north-east.
8 0
3 years ago
Read 2 more answers
The Stronger the force that is exerted on an object,the ____ the change in velocity will be.
leonid [27]

osisisksj

#sorryneedpoints

5 0
3 years ago
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