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erma4kov [3.2K]
3 years ago
14

A spaceprobe has an 29.0 m length when measured at rest. What length

Physics
1 answer:
elixir [45]3 years ago
3 0

Answer:

The observer sees the space-probe 9.055m long.

Explanation:

Let L_0 be the length of the space-probe when measured at rest, and L be its length as observed by an observer moving at velocity v, then

(1).\: \: L = L_0\sqrt{1-\dfrac{v^2}{c^2} }

Now, we know that L_0 = 29.0m and v = 0.95c, and putting these into (1) we get:

L = 29\sqrt{1-\dfrac{(0.95c)^2}{c^2} }

L = 29\sqrt{1-0.95^2 }

\boxed{L = 9.055m}

Thus, an observer moving at 0.95c observes the space-probe to be 9.055m long.

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DETERMINE THE BCD EQUIVALENT OF 1100111110101001
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Answer: It states that the BCD equivalent would be 0001000100000000000100010001000100010000000100000001000000000001.

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Which of the following is an example of changing momentum?!
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B!!!

Explanation:

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A cylindrical storage tank has a radius of 1.35 m. When filled to a height of 3.45 m, it holds 14,014 kg of a liquid industrial
Kruka [31]

Answer:

709.93 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³

From the question above,

D = m/v.............................. Equation 1

Where D = density, m = mass, v = volume.

Note: The volume of the liquid is equal to the volume of the height occupied by the liquid in the container

Since the tank is cylindrical,

v = πr²h........................ Equation 2

Where r = radius of the the tank, h = height of the liquid in the tank

Substitute equation 2 into equation 1

D = m/(πr²h)............... Equation 3

Given: m = 14014 kg, r = 1.35 m, h = 3.45 m, π = 3.14

Substitute into equation 3

D = 14014/(3.14×1.35²×3.45)

D = 14014/19.74

D = 709.93 kg/m³

6 0
3 years ago
A roadway for stunt drivers is designed for racecars moving at a speed of 40 m/s. A curved section of the roadway is a circular
Fynjy0 [20]

Answer:

Bank angle = 35.34o

Explanation:

Since the road is frictionless,

Tan (bank angle) = V^2/r*g

Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2

Tan ( bank angle) = 40^2/(230*9.81)

Tan (bank angle) = 0.7091

Bank angle = tan inverse (0.7091)

Bank angle = 35.34o

3 0
3 years ago
A 6.50-g sample of copper metal at 25.0°C is heated by the addition of 84.0 J of energy. The final temperature of the copper is
Mekhanik [1.2K]

Answer:

Final temperature of the copper is 59 degrees Celsius

Explanation:

It is given that,

Mass of the sample of copper metal, m = 6.5 g

Initial temperature of the metal, T_i=25^{\circ}\ C=298\ K

Heat generated, Q = 84 J

The specific heat capacity of liquid water is 0.38 J/g-K

Let T_f is the final temperature of the copper. It can be calculated using the definition of specific heat of any substance. It is given by :

Q=mc\Delta T

Q=mc(T_f-T_i)

T_f=\dfrac{Q}{mc}+T_i

T_f=\dfrac{84}{6.5\times 0.38}+298

T_f=332\ K

or

T_f=59^{\circ}C

So, the final temperature of the copper is 59 degrees Celsius. Hence, this is the required solution.

4 0
3 years ago
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