Please help me with my question

Which is the largest and most dense of the terrestrial planets? A.Mercury B.Venus C.Earth D.Mars

Effectus [21]

C.Earth is the largest and most dense of the terrestrial planets

7 0

3 years ago

Please Help!!!!

Len [333]

Answer:

v = 0.92 c

Explanation:

Here, we will use the time dilation formula from Einstein's theory of relativity to find the speed of traveling of the friend:

$t =\frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}} \\\\\\\sqrt{1-\frac{v^2}{c^2}}=\frac{t_o}{t}\\\\$

where,

v = speed of traveling = ?

c = speed of light

t = time of return = 10 years

t₀ = time passed on earth = 4 years

Therefore,

$\sqrt{1-\frac{v^2}{c^2}} = \frac{4\ years}{10\ years}\\\\ 1-\frac{v^2}{c^2}=(\frac{2}{5})^2\\\\\frac{v^2}{c^2} = 1-\frac{4}{25}\\\\\frac{v^2}{c^2} = \frac{21}{25}\\\\v^2 = 0.84c^2\\\\$

8 0

2 years ago

our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three

Licemer1 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The torque produced by the pile of rocks is $\tau = 35.63\ N \cdot m$

b

The distance of the single for equilibrium to occur is $r_s =1.62 \ m$

Explanation:

From the question we are told that

The mass of the left rock is $m_s = 2.25 \ kg$

The mass of the rock on the right $m_p = 10.1 kg$

The distance from fulcrum to the center of the pile of rocks is $r_p = 0.360 \ m$

Generally the torque produced by the pile of rock is mathematically represented as

$\tau = m_p * g * r_p$

Substituting values

$\tau = 10.1 * 9.8 * 0.360$

$\tau = 35.63\ N \cdot m$

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

The torque due to the single rock is

$\tau = m_s * g * r_s$

At equilibrium the both torque are equal

$35.63 = m_s * r_s * g$

Making $r_s$ the subject of the formula

$r_s = \frac{35.63 }{m_s * g}$

Substituting values

$r_s = \frac{35.63 }{2.25 * 9.8}$

$r_s =1.62 \ m$

6 0

3 years ago

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor

const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is $\tau = 34.3 \ N\cdot m$

Explanation:

From the question we are told that

The mass of the steel ball is $m = 3.0 \ kg$

The length of arm is $l = 70 \ cm = 0.7 \ m$

The mass of the arm is $m_a = 4.0 \ kg$

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

$r = \frac{l}{2}$

=> $r = \frac{ 0.7}{2}$

=> $r = 0.35 \ m$

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

$\tau = m_a * g * r + m * g * L$

=> $\tau = 4 * 9.8 * 0.35 + 3 * 9.8 * 0.70$

=> $\tau = 34.3 \ N\cdot m$

5 0

2 years ago

if you drop a stone from height of 2.5m. what is the speed of the stone right before it hits the ground?

KonstantinChe [14]

Since the stone was dropped from height, its initial velocity = 0 m/s

Using v² = u² + 2gs.

Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m

v² = u² + 2gs

v² = 0² + 2*10*2.5

v² = 0 + 50

v² = 50

v = √50

v ≈ 7.07 m/s

Hence velocity just before hitting the ground is ≈ 7.07 m/s

6 0

3 years ago