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makkiz [27]
4 years ago
14

An object was dropped from the plane with 1000 km above the ground with a mass of 300 kg. Find the acceleration of the object ea

ch second until it reaches the ground.
Physics
1 answer:
Oxana [17]4 years ago
8 0

Answer:

9.8m/s^2

Explanation:

The acceleration due to gravity on earth is 9.8m/s^2 regardless of the mass of the object. This means, in the absence of air resistance, both a 3000 kg object and a 300 kg will accelerate at the same rate of 9.8m/s^2 towards  earth.

The reason the acceleration due to gravity is the same for all masses is that by grace of nature, the inertial mass m of the object from F= ma  equals the gravitational mass m from F =G \dfrac{mM}{R^2}, resulting

ma=G\dfrac{mM}{R^2}

\boxed{a=G\dfrac{M}{R^2}}

which is the acceleration that only depends on the mass of earth and its radius, and not on the mass of the object.

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A battery with an emf of 24.0 V is connected to a resistive load. If the terminal voltage of the battery is 16.1 V and the curre
lions [1.4K]

Answer:

2.03 Ω

Explanation:

EMF: This can be defined as the potential difference of a cell when it is not delivering any current. The S.I unit of Emf is Volt.

The formula of emf is given as,

E = I(R+r)............................ Equation 1

Where E = Emf, I = current, R = External resistance, r = internal resistance.

Make r the subject of the equation

r = (E-IR)/I........................ Equation 2

Note: From ohm's law, V = IR.

r = (E-V)/I........................ Equation 3

Where V = Terminal voltage

Given: E = 24 V, I = 3.9 A, V = 16.1 V.

Substitute into equation 3

r = (24-16.1)/3.9

r = 7.9/3.9

r = 2.03 Ω

6 0
3 years ago
A 68 kg student runs up a flight of stairs that is 7 m high in 9 seconds. Calculate
NeTakaya
Power = Work/time
Work= energy
Potential energy =mxgxh
Power=mxgxh/t
Power=68x10x7/9
Power=529 watts
8 0
4 years ago
Read 2 more answers
Must all solutions be liquids? Explain. Please help.
hichkok12 [17]
I'm not sure, but I'm pretty sure it depends.
7 0
3 years ago
When you connect an unknown resistor across the terminals of a 1.50 V D-cell battery having negligible internal resistance, you
g100num [7]

Answer:

Explanation:

Using ohm's law

a) V = IR where V is voltage in Volt, I is current in Ampere and R is resistance in ohms

R = V / I = 1.50 V/ ( 2.05 /1000) A = 731.71 ohms

b) Power = IV = \frac{V}{R} × v = \frac{V^{2} }{R} = \frac{9^{2} }{731.71} = 0.1107 W

c) E = IR + Ir = ( 731.71 × 0.0036) + ( 35 × 0.0036) = 2.76 V

d) Power use by the resistor = I²R = 0.0036² × 731.71 = 0.00948 W = 0.00948 W = 0.000009483 kw × ( 18 / 60 ) H = 2.84 × 10⁻⁶ KW-H

5 0
3 years ago
Use this technique to find a formula for the intensity I of a sound, in terms of the sound level β and the reference intensity I
mezya [45]

The problem is basically asking us to find a way to find the sound intensity I, in terms dependent on the sound level and the reference intensity I_0.For this purpose we can start from the unit used in the scale logarithmic decibel, that is

\beta = 10log_{10}\frac{I}{I_0}

Where

I = Acoustic intensity on the linear scale

I_0 = Hearing threshold

Using the logarithmic properties of the exponents the above expression can be described as:

(\frac{I}{I_0})^{10} = 10^{\beta}

I = I_0 10^{\frac{\beta}{10}} \righarrow that is the expression or technique to find the intensity of sound.

8 0
3 years ago
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