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kotegsom [21]
3 years ago
13

A force of 1200 N is applied to a drum of radius 80 cm which has a 12 kg mass, m1, attached by a cord

Physics
1 answer:
maksim [4K]3 years ago
4 0

Answer:

1.022 x 103 N.m

Explanation:

Solution

Given:

The weight of the block of mass m₂ is :

w₂ = m₂*g

Where

w₂ = 39 x 9.8 = 382.2 N

Then,

The weight of the block of mass m₁

w₁= m₁*g;

so,

w₁ = 12 x 9.8 = 117.5 N

Thus,

The tension wrapped in cord on drum (80 cm) T₁ = F - w₁

Now,

T₁ = 1200 - 117.5

T₁ = 1082.5 N

The tension  wrapped in the cord on drum (41 cm) T₂ = w₂;

T₂ = 382.2 N

Hence,

We calculate net torque on the center of the drum:

The net torque = T₁ x 0.8 + T₂ x 0.41;

= 1082.5 x 0.8 + 382.2 x 0.41;

= 1.022 x 103 N.m

Therefore, the resulting torque applied to the system is 1.022 x 103 N.m

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Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

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- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

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What does it mean when it says a scientific question must be supportable
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An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 8.00 m/s from a height of 1.70 m above
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Answer:

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Explanation:

given data

base = 3.60 m

speed u = 8 m/s

height = 1.70 m

to find out

check change in speed

solution

we know here formula for v  that is

v² = u² - 2gh      ............1        for upward speed

v² = u² + 2gh     ............2        for projected speed

so here put all value and find v with h = 3.60 - 1.70 = 1.9 m

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v² = 8² + 2(9.8) 1.9   = 101.24

v = 5.173  m/s    ..............3

v = 10.061 m/s   ...................4

so change in speed form 3 and 4 equation

change in speed = v - u = 8 - 5.173  = 2.827 m/s     .................5

change in speed = v - u = 10.061 - 8 = 2.061 m/s     ..................6

so now we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

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Answer:

9.8

Explanation:

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