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rosijanka [135]
3 years ago
8

What is the resistance of a voltage of 65 V and a current of 2.2 A? Include units.

Physics
1 answer:
kozerog [31]3 years ago
4 0

Answer:

29.5 ohms

Explanation:

R= V/I

= 65 / 2.2

= 29.5 ohms

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Which of the following is an example of an immiscible liquid?
Nonamiya [84]

Answer:

the answer is 1, oil and water

4 0
3 years ago
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Calculate the time period of simple pendulum whose length is 98.2cm
alekssr [168]

The time period resulting in oscillations will be 1.986 seconds.

<h3>What is the period of oscillation?</h3>

The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.

The time period of the oscillation is;

\rm T = 2 \pi\sqrt{\frac{L}{g}} \\\\ \rm T = 2 \times 3.14 \times \sqrt{ \frac{98.2 \  \times 10^{-2} \ m}{9.81 \ m/s^2}} \\\\ T= 1.986 \ sec

Hence the time period resulting oscillations will be 1.986 seconds.

To learn more about the time period of oscillation refer to the link;

brainly.com/question/20070798

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5 0
2 years ago
Identify the physical property of a material that is NOT a good conductor of heat.
EleoNora [17]

Answer:D.no of the above

Explanation:get right with Christ

4 0
3 years ago
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
DiKsa [7]

Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

7 0
3 years ago
At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown strai
zhannawk [14.2K]

Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

<u>Motion of 0.50 kg ball:</u>

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

                 s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

<u>Motion of 0.25 kg ball:</u>

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

                 s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

          \bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

          m₁ = 0.50 kg

          x₁ = 5.38 m        

          m₂ = 0.25 kg

          x₂ = 10.38 m    

Substituting

         \bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m

The center of mass of the two-ball system is 7.05 m above ground.  

8 0
3 years ago
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