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3 years ago

A 10 kg ball is traveling at the same speed as a 1 kg ball. Compared to the 10 kg ball, the 1 kg ball has

1 answer:

Colt1911 [192]3 years ago

3 0

We can compare the two by their kinetic energies. The kinetic energy is the energy when an object is in motion. It is expressed as the product of the mass of the object and the square of the velocity divided by two. We assume a velocity of 1 m/s for this problem.<span>

KE = mv^2/2
KE1 = 10 (1)^2
KE1 = 10 J

KE2 = 1(1)^2
KE2 = 1 J

Therefore, c</span><span>ompared to the 10 kg ball, the 1 kg ball has lesser kinetic energy.</span>

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A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling hori

11111nata11111 [884]

Answer:

481 m

Explanation:

To fall 235 m, the time required is

t = √(2H/g)

t= √(2 $\times$ 235/9.8)

t=6.92 seconds.

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6.92 $\times$ 69.4 ≈ 481 m

Therefore, the goods must be dropped 481 m in advance of the recipients.

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3 years ago

What happens to the energy of gas particles when an elastic collision takes place?

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Answer:

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A roller coaster car of mass m= 300 kg is released from rest at the top of a 60 m high hill (position A), and rolls with a negli

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Answer: The principle of conservation of energy, angular speed and centripetal force

Explanation:

At point A, the car experienced maximum of potential energy

As it moves down the hill, the potential energy decreases while the kinetic energy increases.

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3 years ago

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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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3 years ago