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mrs_skeptik [129]
3 years ago
7

A small but dense 2.0-kg stone is attached to one end of a very light rod that is 1.2 m long. The other end of the rod is attach

ed to a frictionless pivot. The rod is raised until it is vertical, with the stone above the pivot. The rod is released and the stone moves in a vertical circle with no air resistance. What is the tension in the rod as the stone moves through the bottom of the circle? (a) 20 N (b) 40 N (c) 60 N (d) 80 N (e) 100 N

Physics
1 answer:
tigry1 [53]3 years ago
8 0

Answer:

Option B is the correct answer.

Explanation:

Refer the figure we have centripetal force at bottom of circle

       F_c=F_t-F_w\\\\\frac{mv^2}{r}=F_t-mg\\\\F_t=m\left ( \frac{v^2}{r}+g\right )

We have mass, m = 2 kg

Radius, r = 1.2 m

For circular motion to occur we have tension at top = 0

That is

        \frac{mv^2}{r}=mg\\\\v=\sqrt{rg}

Now let us find tension at bottom point

       F_t=2\times \left ( \frac{rg}{r}+g\right )=4g=40N

Option B is the correct answer.  

         

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A uniformly charged sphere has a total charge of 300uc and a radius of 8cm. Find the electric field density at A point 16cm from
s2008m [1.1K]

E = <u>kQ</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

(r + h)²

where,

k = 9 × 10^9Nm²C^-2

Q = total charge, 300uC = 300 × 10^ -6C

r = 8 × 10^ -2m

h = 16 × 10^ -2m

then,

E = <u>9</u><u>e</u><u>9</u><u> </u><u>*</u><u> </u><u>3</u><u>0</u><u>0</u><u>e</u><u>^</u><u>-</u><u>6</u><u> </u><u> </u><u> </u><u> </u>

(8e^-2 + 16e^-2)²

E = 4687500N/C

6 0
2 years ago
A projectile is fired with initial speed vo at an angle of 45o above the horizontal. Assume no air resistance.
katrin2010 [14]

Answer:

The correct answer is a

Explanation:

At projectile launch speeds are

X axis     vₓ = v₀ = cte

Y axis     v_{y} = v_{oy} –gt

The moment is defined as

         p = mv

For the x axis

         pₓ = mvₓ = m v₀ₓ

As the speed is constant the moment is constant

For the y axis

        p_{y} = m v_{y} = m (v_{oy} –gt) = m v_{oy} - m (gt)

Speed ​​changes over time, so the moment also changes over time

Let's examine the answer

i   True

ii False.  The moment changes with time

The correct answer is a

7 0
3 years ago
The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve
Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

\varepsilon _t=\dfrac{D-d}{d}

\varepsilon _t=\dfrac{32-30}{30}

\varepsilon _t=0.0667

We know that

n=-\dfrac{\epsilon _t}{\varepsilon _{long}}

\varepsilon _{long}=-\dfrac{\epsilon _t}{n}

\varepsilon _{long}=-\dfrac{0.0667}{0.45}

\varepsilon _{long}=-0.148

So the axial pressure

P=E\times \varepsilon _{long}

P=5\times 0.148

P=740 KPa

The movement in the sleeve

\Delta =\varepsilon _{long}\times L

\Delta =0.148\times 50

Δ=7.4 mm

6 0
3 years ago
1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t
tiny-mole [99]

Explanation:

Charges,q_1=8\ \mu C=8\times 10^{-6}\ C

q_2=-5\ \mu C=-5\times 10^{-6}\ C

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.

6 0
3 years ago
Two horses are pulling a box in two different directions as shown in the below image. The image shows one 30.0 N force due north
Novay_Z [31]

Explanation:

'What is the magnitude of the force needed to stop the horses and bring the box into equilibrium?' ≈42N; according to the vectors rules.

'Where would you locate the rope to apply the force?' - in point D.

PS. zoom out the attached picture.

4 0
3 years ago
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