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IRISSAK [1]
1 year ago
9

A thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center. A force F is exerted tangentially to th

e hoop. If the hoop’s radius is 2.0 m and it is rotating with an angular acceleration of 2.5 rad/s2, calculate the magnitude of F.
a. 35 N
a. 20 N
c. 25 N
d. 8 N
Physics
1 answer:
Stella [2.4K]1 year ago
5 0

Given :

Thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center.

A force F is exerted tangentially to the hoop. If the hoop’s radius is 2.0 m and it is rotating with an angular acceleration of 2.5 rad/s².

To Find :

The magnitude of F.

Solution :

Torque on hoop is given by :

\tau =F\times R\\\\I\alpha = FR\\\\MR^2\alpha = FR\\\\F = MR\alpha( Moment of Inertia of hoop is MR² )

Putting  value of M, R and α in above equation, we get :

F=5\times 2\times 2.5\ N\\\\F = 25  \ N

Therefore, the magnitude of force F is 25 N.

Hence, this is the required solution.

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A charge Q is placed on the x axis at x = +4.0 m. A second charge q is located at the origin. If Q = +75 nC and q = −8.0 nC, wha
Stells [14]

Answer:

23.1 N/C

Explanation:

OP = 3 m , OQ = 4 m

PQ = \sqrt{4^{2}+3^{2}}=5 m

q = - 8 nC, Q = 75 nC

Electric field at P due to the charge Q is

E_{1}=\frac{KQ}{PQ^{2}}=\frac{9\times 10^{9}\times 75\times 10^{-9}}{25}=27 N/C

Electric field at P due to the charge q is

E_{2}=\frac{Kq}{PO^{2}}=\frac{9\times 10^{9}\times 8\times 10^{-9}}{9}=8 N/C

According to the diagram, tanθ = 3/4

Resolve the components of E1 along x axis and along y axis.

So, Electric field along X axis, Ex = - E1 Cos θ

Ex = - 27 x 4 / 5 = - 21.6 N/C

Electric field along y axis, Ey = E1 Sinθ - E2

Ey = 27 x 3 /5 - 8 = 8.2 N/C

The resultant electric field at P is given by

E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-21.6)^{2}+(8.2)^{2}}=23.1 N/C

3 0
2 years ago
A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at 11.0
Serjik [45]

Answer:

The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.

Explanation:

By looking at the images below wee see that the airspeed on one side of the rectangular plate decreases the statical pressure over this side. Since over the downside, the pressure still bein the atmospheric pressure. This difference in pressure produces a lift force in the plate. The list force is the net force obtained between the difference of the forces that produce the pressure over the upside and the downside:

F_{lift}=F_{up} - F_{dw}=0.5*p*V^2

Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.

When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:

F_{lift}=0.5*p*V^2=W

By replacing the known values it is possible to find the plate's weight:

F_{lift}=0.5*1.2 \frac{kg}{m^{3}}*(11 m/s)^2=W

W=72.6 N

When the plate kept to 30° from the vertical the moment equation balance is written as:

F_{lift}=0.5*p*V^2=W*sen(30\°)

The sine of 30° is due to the weight is 30° oriented, therefore the new value for the airspeed is:

V=\sqrt(W*sen(30\°)/0.5p)

V=\sqrt(\frac{72.6 N * 0.5}{0.5*1.2 kg/m^3})

V=\sqrt(60.5 \frac{N}{kg/m^3})

V=\sqrt(60.5 \frac{kg.m/s^2}{kg/m^3})

V=\sqrt(60.5 \frac{m^2}{s^2})

V= 7.78 m/s

7 0
2 years ago
Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th
Oksana_A [137]
<h2>The K.E of the charge is 1.02 x 10⁻¹⁷ J</h2>

Explanation:

When the charge of 2e is placed in between the plates .

The force applied on this charge by plates is = q E

here q is the magnitude of charge = 2 e = 2 x 1.6 x 10⁻¹⁹ C

and E is the magnitude of electric field intensity

The work done = Force x displacement

Thus W = q E x S

here S is displacement

Therefore W = 2 x 1.6 x 10⁻¹⁹ x 4 x 8

= 1.02 x 10⁻¹⁷ J

This work will be converted into the kinetic energy of charge .

Thus K.E = 1.02 x 10⁻¹⁷ J

3 0
2 years ago
Neha and Suhani are playing with two identical pendulums. They leave the bob from a certain position and wait for it to return t
Anika [276]

Answer:

The correct option is C

Explanation:

The pendulum bob would return at the same time because the initial angle a pendulum bob is dropped does not affect it's period (the time it takes for the pendulum to move back and forth), however the one with a larger angle move faster but would eventually arrive at the same "starting point" due to varying displacements made.

6 0
2 years ago
Suppose a certain scale is not calibrated correctly, and as a result, the mass of any object is displayed as 0.75 kilogram less
Papessa [141]

Answer

A certain scale is not calibrated correctly,

Mass displayed  as 0.75 kilogram less than its actual mass.

The scale of display will be equal to '1'

The correlation between actual mass and displayed mass is 1 because correlation is independent of change of origin and scale and correlation between any value and value minus 0.75 will be one.

So, the correct answer will be "1"

7 0
2 years ago
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