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Oksi-84 [34.3K]
3 years ago
6

A)Watt (W) is a derived unit.

Physics
1 answer:
Aleks04 [339]3 years ago
4 0

Answer:

a) Watt is a derived unit because it's not a “base unit.”

b) The atomic clock is suitable for measuring time because natural atomic oscillations happen at a higher frequency and as a result they are more stable.

c) There is fluctuation in time measured by a pendulum clock because the pendulum rod changes in length slightly with changes in temperature, causing changes in the rate of the clock. An increase in temperature causes the rod to expand, making the pendulum longer, so its period increases and the clock loses time.

D) The weight of an object is more at pole and less at equator of the earth because the distance of the pole is less than the distance of the equator from the centre of the earth.

e) The weight of an object on the moon is about one-sixth the weight it would have on the earth because the moon's gravitation force is determined by the mass and the size of the moon. Hence, the weight of an object on the moon is 1/6th its weight on the earth.

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Rounded to the nearest whole number what is the atomic mass of platinum?
NISA [10]

195 is the rounded atomic atomic mass of platinum

6 0
3 years ago
A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform ra
sergey [27]

Answer:

\frac{di}{dt}  = 7.31 \  A/s

Explanation:

From the question we are told that  

     The  number of turns is  N =  450 \  turns

      The  radius is  r =  1.17 \ cm =  0.0117 \ m

       The  position from the center consider is  x =  3.45 cm  =  0.0345 m

       The  induced emf is  e  =  8.20 *10^{-6} \  V/m

Generally according to Gauss law

        \int\limits { e } \, dl  =  \mu_o *  N  *  \frac{di}{dt }  *  A

=>    e *  2\pi x  =  \mu_o  *  N  *  \frac{d i }{dt }  *  A

Where A is the  cross-sectional area of the solenoid which is mathematically represented as

                A =  \pi r ^2

=>      e *  2\pi x  =  \mu_o  *  N  *  \frac{d i }{dt }  *  \pi r^2

=>       \frac{di}{dt}  =  \frac{2e * x  }{\mu_o * N  * r^2}ggl;

Here  \mu_o is the permeability of free space with value

          \mu_o  =  4\pi * 10^{-7} \  N/A^2

=>     \frac{di}{dt}  =  \frac{2 *  8.20*10^{-6} *  0.0345  }{ 4\pi * 10^{-7} * 450  * (0.0117)^2}

=>      \frac{di}{dt}  = 7.31 \  A/s

6 0
4 years ago
If you travel 5 miles north then turn and travel 5 miles south, you are now _____ miles from where you started.
horsena [70]
0 miles from where you started
3 0
3 years ago
Read 2 more answers
A massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface. The mass hang
Romashka-Z-Leto [24]

Both masses will have the same acceleration. The cart accelerates to the right with a magnitude of 4.9 m/s^{2}. The correct answer is 4.90 m/s^{2}

Given that a massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface.

Let M = 1kg and m = 3 kg

Since the horizontal surface is frictionless, the tension in the string will be the same. when the mass is hanged over a frictionless pulley, the tension will also be the same.

When the mass is released, the cart accelerates to the right can be calculated  from Newton' second law of motion. That is,

M( g + a) = m(g - a)

1(9.8 + a) = 3( 9.8 - a)

9.8 +a = 29.4 - 3a

collect the like terms

4a = 19.6

a = 19.6/4

a = 4.9 m/s^{2}

Therefore, the cart accelerates to the right with a magnitude of 4.9 m/s^{2}. The correct answer is 4.90 m/s^{2}

Learn more about dynamics here: brainly.com/question/24994188

5 0
3 years ago
A child is playing on a swing. As long as he does not swing too high the time it takes him to complete one full oscillation will
Aleks [24]

Answer:

We know that for a pendulum of length L, the period  (time for a complete swing) is defined as:

T = 2*pi*√(L/g)

where:

pi = 3.14

L = length of the pendulum

g = gravitational acceleration = 9.8 m/s^2

Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.

Then the period is independent of:

The mass of the child

The initial angle

Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.

7 0
3 years ago
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