Answer:
whirl a rock at the end of a string and it follows a circular path. if the string breaks, the tendency of the rock is to stop
Answer:
The outside layer is the wettest.
Explanation:
Potential energy does.
A book on the floor has negative potential energy relative to the seat
of your chair, zero potential energy relative to the floor, and a whole
bunch of positive potential energy relative to the ground if the floor
is the floor of the passenger jet in which you are cruising and dropped
your book on the floor.
Answer:
x-component of velocity: 7.5 m/s
y-component of velocity: 13 m/s
Explanation:
This problem is pure trigonometry. Assuming you know trig, there are only a couple of steps to solving this problem. First, split the velocity into components; recall that any vector not directed along an axis has x and y components. Then, remember that sinΘ = opposite/hypotenuse. Applying this to your scenario, you get sin60° = vy/15. Multiplying this out gives you vy=15sin60. Put this into a calculator (make sure it's set to degree mode because the angle in this problem is in degrees) and you should get 12.99, which you can round up to 13 m/s. This is the velocity in the y-direction.
The procedure to find the x-velocity is very similar, but instead of using sine, we will use the cosine of theta. Recall that cosΘ=adjacent/hypotenuse. Once again plugging this scenario's numbers into that, you end up with cos60 = vₓ/15. Multiplying this out gives you vₓ = 15cos60. Once again, plug this into your calculator. 7.5 m/s should be your answer. This is the velocity in the x-direction.
By the way, a quick way to find the components of a vector, whether it's velocity, force, or whatever else, is to use these functions. Generally, if the vector points somewhere that's not along an axis, you can use this rule. The x-component of the vector is equal to hypotenuse*cosΘ and the y-component of the vector is equal to hypotenuse*sinΘ.
Answer:
+1.46×10¯⁶ C
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C
Force (F) = 0.615 N
Distance apart (r) = 0.750 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Charge 2 (q₂) =?
The value of the second charge can be obtained as follow:
F = Kq₁q₂ / r²
0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²
0.615 = 236700 × q₂ / 0.5625
Cross multiply
236700 × q₂ = 0.615 × 0.5625
Divide both side by 236700
q₂ = (0.615 × 0.5625) / 236700
q₂ = +1.46×10¯⁶ C
NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C
