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3 years ago

Calculate ine gravitational potential energy of the ball using pe=m×g×h.(use g=9.8 n/kg)

Physics

1 answer:

Neko [114]3 years ago

5 0

Answer:

58.8J

Explanation:

Given parameters;

Mass of ball = 4kg

Height above the floor = 1.5m

g = 9.8n/kg

Unknown:

Potential energy = ?

Solution:

The potential energy of a body is the energy due to the position of the body.

It is mathematically expressed as:

Potential energy = mass x acceleration due to gravity x height

Potential energy = 4 x 9.8 x 1.5 = 58.8J

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in a cricket match there are 5000 spectators counted 10 by 10 the number of significant figure in the measurement will be

BabaBlast [244]

Answer:

$\boxed{3 \ Significant \ figures}$

Explanation:

Total spectators = 5000

Counted by the groups of ten, So at last the result will be:

=> 5000/10 = 500

Significant figures in 500 are 3

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2 years ago

Two friends are playing a version of proton golf where the hole is marked by a single proton. The first friend reads his meter,

yarga [219]

Answer:

the electric field strength on the second one is 2.67 N/C.

Explanation:

the electric fiel on the first one is:

E1 = k×q/(r^2)

r^2 = k×q/(E1)

= (9×10^9)×(q)/(24.0)

= 375000000q

then the electric field on the second one is:

E2 = k×q/(R^2)

we know that R = 3r

R^2 = 9×r^2

E2 = k×q/(9×r^2)

= k×q/(9×375000000q)

= k/(9×375000000)

= (9×10^9)/(9×375000000)

= 2.67 N/C

Therefore, the electric field strength on the second one is 2.67 N/C.

5 0

3 years ago

The mussels are not native to the lakes, and they have begun to kill native mussels by attaching to them and covering them up. T

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The zebra mussels are an invasive species

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3 years ago

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac

alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = $47\times 10^{-9}\ C/m^2$

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

$E=\dfrac{\sigma}{2\epsilon_0}$

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

$E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Now, plug in $47\times 10^{-9}\ C/m^2$ for 'σ' and $8.85\times 10^{-12}\ F/m$ for $\epsilon_0$ and solve for the electric field. This gives,

$E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C$

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

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2 years ago

a train has an initial velocity of 30 m/s. If the train accelerates uniformly at a rate of 6.3 m/s ^ for 2.8 seconds what is the

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Answer:

the velocity is a second final to initial velocity of 39

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2 years ago