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3 years ago

15

If the cart has potential energy of 5,000 J and kinetic energy of 2,750 J. How much mechanical energy does the cart have?

1 answer:

Dahasolnce [82]3 years ago

6 0

Answer: 7750 J

Explanation:

Mechanical energy is potential energy added to kinetic energy.

5000 + 2750 J = 7750 J

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Consider a solid metal sphere (S) a few centimeters in diameter and a feather (F). For each quantity in the list that follows, i

Roman55 [17]

Answer:

A) Gravitational Force is greater in S.

B) Time taken to fall a given distance in air will be greater for F.

C) Both will take same time to fall in a vacuum.

D) Total force is greater in S.

Explanation:

(a) In this case, the gravitational force of S will be greater, because Newton's Second Law states that - F = ma, or weight =mg. g is constant. And mass of the solid metal is heavier.

(b) In this case, the time it will take for F to fall from a given distance in air will be greater than that of S, since the air resistance is not negligible (as in the case of S).

(c) In this, It will take same time for S and F because in a vacuum, there are no air particles, so there is no air resistance and gravity is the only force acting and so objects fall at the same rate in a vacuum.

(d) The total force will be greater in S than F because Force=ma and S is of heavier mass than F.

5 0

3 years ago

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pshichka [43]

Answer:a

Explanation: the plates shifts cracking making earthquakes

8 0

3 years ago

In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6

Rasek [7]

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

$\lambda = \dfrac{h}{mv}$

$1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}$

$v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}$

v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

$E = \dfrac{1}{2}mv^2$

$E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2$

E = 2.5 x 10⁻¹⁴ J

8 0

3 years ago

A particle released during the fission of uranium-235 is a(n) ________

Kaylis [27]

<span>A particle released during the fission of uranium-235 is a "Neutron"</span>

6 0

3 years ago

A. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kine

Korolek [52]

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = $\frac{4}{3}\pi(\frac{d}{2})^3$

or

Volume of the balloon = $\frac{4}{3}\pi(\frac{0.296}{2})^3$

or

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

R is the ideal gas constant = 8.314 kg⋅m²/s²⋅K⋅mol

on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

or

n = 0.563

1 mol = 6.022 × 10²³ atoms

Thus,

0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms

b) Average kinetic energy = $\frac{3}{2}\times K_BT$

where,

Boltzmann constant, $K_B=1.3807\times10^{-23}J/K$

Average kinetic energy = $\frac{3}{2}\times1.3807\times10^{-23}\times292$

or

Average kinetic energy = 6.04 × 10⁻²¹ J

c) rms speed = $\frac{3RT}{m}$

where, m is the molar mass of the Helium = 0.004 Kg

or

rms speed = $\frac{3\times8.314\times292}{0.004}$

or

rms speed = 1349.35 m/s

5 0

3 years ago