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2 years ago

Which planet is smaller than Earth’s core?

Physics

2 answers:

makvit [3.9K]2 years ago

7 0

Answer:

The answer <em><u>is C. Mars</u></em>. Mars and Mercury are both smaller than Earth's core. Hope this helps you :)

worty [1.4K]2 years ago

4 0

Answer:

C. Mars

Explanation:

Size of a planetary object depends on when it forms and where it forms.

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The asteroid 2007 VK184 has one chance in 2940 of hitting the Earth between the years 2048 and 2057. The asteroid is 130 meters

N76 [4]

Answer:

The average death rate for this type of event is closest to 350 people per event.

Explanation:

Between the years 2048 and 2057

Is 10 years. Since there is only one chance of event between these years, there is no point to consider it.

The total number of events = 2940 chances.

average death rate per event = total number of dead people divided by total number of events of occurrence.

1000000/2940 = 340.134

The average death rate for this type of event is therefore closest to 350 people per event

4 0

2 years ago

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

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

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0

3 years ago

A 67 kg soccer player uses 5100 kJ of energy during a 2.0 h match. What is the

Oksanka [162]

The average power produced by the soccer player is 710 Watts.

Given the data in the question;

Mass of the soccer player; $m = 67kg$

Energy used by the soccer player; $E = 5100KJ = 5100000J$

Time; $t = 2.0h = 7200s$

Power; $P =\ ?$

Power is simply the amount of energy converted or transferred per unit time. It is expressed as:

$Power = \frac{Energy\ converted }{time}$

We substitute our given values into the equation

$Power = \frac{5100000J}{7200s}\\\\Power = 708.33J/s \\\\Power = 710J/s \ \ \ \ \ [ 2\ Significant\ Figures]\\\\Power = 710W$

Therefore, the average power produced by the soccer player is 710 Watts.

Learn more: brainly.com/question/20953664

8 0

2 years ago

Which change indicates that the universe is expanding?

Aleonysh [2.5K]

Red shift of distant galaxies

5 0

2 years ago

Read 2 more answers

You lay a mirror flat on the floor with one edge against a wall and aim a laser at the mirror. The ray reflects from the mirror

joja [24]

Answer:

the angle of incidence θ is 45.56 º

Explanation:

Given data

strikes the mirror before wall x = 30.7 cm

reflected ray strikes the wall y = 30.1 cm

to find out

the angle of incidence θ

solution

let us consider ray is strike at angle θ so after strike on surface ray strike to wall at angle 90 - θ

we will apply here right angle triangle rule that is

tan( 90 - θ) = y /x

tan( 90 - θ) = 30.1 / 30.7

90 - θ = tan^-1 (30.1/30.7)

90 - θ = 44.4345

θ = 45.56 º

the angle of incidence θ is 45.56 º

4 0

3 years ago