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3 years ago

The magnitude of the centripetal force acting on

1 answer:

5 0

1 - Radius of the path is increased
The formula for centripetal force is Fc=mv^2/r, therefore if r increases, a the divisor is larger and hence the centripetal force will be smaller.

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Six friends are at a pizzeria. They want to order enough pizza so that each person can eat at least 2/5 . start fraction, 2, d

Lyrx [107]

Friend #1 gets at least 2/5 of a pizza.

Friend #2 gets at least 2/5 .

Friend #3 gets at least 2/5 .

Friend #4 gets at least 2/5 .

Friend #5 gets at least 2/5 .

Friend #6 gets at least 2/5 .

Sum . . . . . . . . . at least 12/5 of a pizza.

Simplify . . . . . . at least 2.4 pizzas.

-- If pizzas can be bought by the half, they should order at least 2-1/2 pizzas.

-- If only whole pizzas have to be ordered, then they should order at least 3 pizzas.

4 0

4 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago

What distance is required for a train to stop if its initial velocity is 23 m/s and its

Varvara68 [4.7K]

-- The train starts at 23 m/s and slows down by 0.25 m/s every second.

So it'll take (23/0.25) = 92 seconds to stop.

-- Its average speed during that time will be (1/2)(23+0) = 11.5 m/s

-- Moving at an average speed of 11.5 m/s for 92 sec, the train will cover

(11.5 m/s) x (92 sec) = 1,058 meters .

7 0

3 years ago

Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. one force is 9.0 n, acting due east, an

Zolol [24]

The first thing you should do for this case is to find the horizontal and vertical components of the forces acting on the body.
We have then:
Horizontal = 9-9.2cos (58) = 4.124742769 N.
Vertical = 9.2sin (58) = 7.802042485 N
Then, the resulting net force is:
F = √ ((4.124742769) ^ 2 + (7.802042485) ^ 2) = 8.825268826 N
Then by definition:
F = m * a
Clearing the acceleration:
a = F / m
a = (8.825268826) / (3.0) = 2.941756275 m / s ^ 2
answer:
The magnitude of the body's acceleration is
2.941756275 m / s ^ 2

6 0

3 years ago

How much longer will a 300 m steel bridge be on a 30C day than a –10C night?

RideAnS [48]

Answer:

Explanation: A bar of an unknown metal has a length of 0.975 m at 45°C and a length of 0.972 m at. 23°C.

6 0

3 years ago