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3 years ago

The data table depicts an object moving with changing speed. True False

Physics

2 answers:

Nimfa-mama [501]3 years ago

6 0

The answer is true. The table does show an object moving with changing speed.

hoa [83]3 years ago

5 0

Answer:

TRUE

Explanation:

Uniform speed means the object will travel in such a way that it must cover equal distance in equal interval of time.

So after each equal interval of time the distance moved by the object is same then it is uniform speed but if the object travel unequal distance in equal interval of time then it is representing the changing speed.

So here we can see that object covers 1 m distance in first interval of 1 s while in next 1 s interval from t= 1 to t = 2 it moved total distance of d = 5 - 1 = 4 m

In next interval of 1 s from t = 2 to t =3 it moved total distance of d = 25 - 5 = 20 m

So here we can say that speed of the object is changing

So above statement is TRUE

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A spy satellite uses a telescope with a 1.7-m-diameter mirror. It orbits the earth at a height of 180 km.

WINSTONCH [101]

Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

Explanation:

Given that;

diameter of the mirror d = 1.7 m

height h = 180 km = 180 × 10³ m

wavelength λ = 500 nm = 5 × 10⁻⁹ m

Now Angular separation from the peak of the central maximum is expressed as;

sin∅= 1.22 λ / d

sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7

sin∅ = 3.588 × 10⁻⁷

we know that;

sin∅ = object separation / distance from telescope

object separation = sin∅ × distance from telescope

object separation = 3.588 × 10⁻⁷ × 180 × 10³

object separation =6.45 × 10⁻² m

then we convert to centimeter

object separation = 6.45 cm

Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

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3 years ago

A real heat engine operates between temperatures TcTcT_c and ThThT_h. During a certain time, an amount QcQcQ_c of heat is releas

Nookie1986 [14]

The maximum amount of work performed is

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

Explanation:

The efficiency of a real heat engine is given by the equation:

$\eta = 1-\frac{T_C}{T_H}$ (1)

where

$T_C$ is the temperature of the cold reservoir

$T_H$ is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

$\eta = \frac{W_{max}}{Q_H}$

where

$W_{max}$ is the maximum work done

$Q_H$ is the heat absorbed from the hot reservoir

$Q_H$ can be written as

$Q_H=W_{max}+Q_C$

where

$Q_C$ is the heat released to the cold reservoir

So the previous equation can be also written as

$\eta=\frac{W_{max}}{W_{max}+Q_C}$ (2)

By combining eq.(1) and (2) we get

$1-\frac{T_C}{T_H}=\frac{W_{max}}{W_{max}+Q}$

And re-arranging the equation and solving for $W_{max}$ , we find

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

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Answer:

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with this info, we can estimate the current going through branch A using Ohm's Law, and the information that the power source is 6 V:

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Than means that the lowest current will be registered through branch D where the 50 $\Omega$ resistor is.

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