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Bess [88]
2 years ago
6

A plane flew at a speed of 600 km/hr for a distance of 1500 km,

Physics
1 answer:
Zarrin [17]2 years ago
4 0

Explanation:

600 ÷ 60 = 10

1500 ÷ 10 = 150 minutes

I don't know how you should express the answer in:

So it's either 150 minutes or 2 hours and 30 minutes.

(This is kind of like mathematics, not physics)

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You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. a) What fraction of its initial energy is lost
tangare [24]
The fraction of energy that is lost is 25%, it depends how fast the ball was going until it lost 25% of its energy, the gravitational energy was transferred into the kinetic energy that helped the ball bounce back
4 0
3 years ago
A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.
marishachu [46]

Answer:

a) the magnitude of the force is

F= Q(\frac{kqs}{r^3}) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b)  the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = \frac{kq}{r^{2} }

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = \frac{kp}{r^{3} } , where p = q × s

E(r) = \frac{kqs}{r^{3} } where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q(\frac{kqs}{r^3})

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall  F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

8 0
3 years ago
What is the equation used to calculate the total amount of energy used by an appliance?
charle [14.2K]

Answer:

Power = Current × Voltage

Explanation:

Units:

Power = Watts

Current = Àmperes

Voltage = Volts

8 0
3 years ago
A Carnot air conditioner operates between an indoor temperature of 20°C and an outdoor temperature of 39°C. How much energy does
Gelneren [198K]

Answer:

D. 130 J

Explanation:

The coefficient of performance for a machine that is being used to cool, is given by:

COP=\frac{Q_C}{W}=\frac{T_C}{T_H-T_C}

Here Q_C  is the heat removed from the cold reservoir, W is the work required, that is, the energy required to remove the heat from the interior of the house, T_C is the cold temperature and T_H is the hot temperature. Recall use absolutes temperatures(273.15+^\circ C). Replacing and solving for W:

W=Q_c\frac{T_H-T_C}{T_C}\\W=2000J\frac{312.15K-293.15K}{293.15K}\\W=129.63J

8 0
3 years ago
A 5kg object is moving at a height of 2 m. The potential energy of the object is closest to ___ j
elena55 [62]

Answer:

In this case, a body of mass 5 kg kept at a height of 10 m. So the potential energy is given as 5 * 10 *10 = 500 J.

6 0
3 years ago
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