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Alex777 [14]
3 years ago
8

A marble rolls across a table for 6 meters in 2.3 seconds. What is the velocity of the marble?

Physics
2 answers:
inn [45]3 years ago
7 0

Answer:

\boxed {\tt v\approx2.61 \ m/s}

Explanation:

Velocity can be found by dividing the displacement by the time.

v=\frac{d}{t}

The displacement is 6 meters and the time is 2.3 seconds.

d= 6 \ m\\t= 2.3 \ s

Substitute the values into the formula.

v=\frac{6 \ m}{2.3 \ s}

Divide.

v=2.60869565 \ m/s

Let's round to the nearest hundredth. The 8 in the thousandth place tells us to round the 0 to a 1.

v\approx2.61 \ m/s

The velocity of the marble is about 2.61 meters per second.

monitta3 years ago
5 0

Answer:

2.61 \frac{m}{s}

Explanation:

v= d/t

v = 6m/2.3 s

v=2.61 \frac{m}{s}

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A small compass is placed near a current-carrying wire. The wire is vertical, and the compass is on a horizontal surface. The re
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further, The red end of the compass needle points in the direction of external magnetic field. As the red end of the needle is pointing away from us, the external magnetic field at its location should also be directed away from us. Using the right hand thumb rule, we can see that this is only possible if the current in the wire is flowing upwards.

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Earthquakes are usually caused by tectonic plates slipping past each other. An earthquake generates seismic waves of energy that
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4 years ago
What is the force on an object with a mass of 25 kg and an acceleration of 5 m/s/s? 1 point
Oduvanchick [21]

Answer:125N

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7 0
3 years ago
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
4 years ago
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