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3 years ago

The moon has a mass of 7.34 . 102 kg and a radius of 1.74 . 106 meters. If you have a mass of 66 kg,

Physics

1 answer:

Ierofanga [76]3 years ago

3 0

Answer:

$F=1.06\times 10^{-18}\ N$

Explanation:

Given that,

Mass of the Moon, $M=7.34\times 10^2\ kg$

Mass of the person, m = 66 kg

The radius of Moon, $r=1.74\times 10^6\ m$

We need to find the force between the person and the Moon. The formula for the gravitational force is given by :

$F=G\dfrac{Mm}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{7.34\times 10^2\times 66}{(1.74\times 10^6)^2}\\\\=1.06\times 10^{-18}\ N$

So, the required force is $1.06\times 10^{-18}\ N$ .

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Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How fa

zloy xaker [14]

Answer:

150000000000 m

0.0000005 seconds

33.33 ns

Explanation:

Speed of electromagnetic waves through vacuum = $3\times 10^8\ m/s$

Echo time = 1000 seconds

Echo time is the time taken to reach the object and come back to the observer

Distance = Speed×Time

$Distance=3\times 10^8\times \frac{1000}{2}=150000000000\ m$

Venus is 150000000000 m away from Earth

Time = Distance / Speed

$Time=\frac{75}{3\times 10^8}=0.00000025\ seconds$

Echo time will be twice the time

$Echo\ time=2\times 0.00000025=0.0000005\ seconds$

The echo time will be 0.0000005 seconds

Difference in time = Difference in distance / Speed

$\Delta t=\frac{\Delta d}{v}\\\Rightarrow \Delta t=\frac{10}{3\times 10^8}\\\Rightarrow \Delta=33.33\ ns$

The accuracy by which I will be able to measure the echo time is 33.33 ns

5 0

3 years ago

James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a

Reptile [31]

Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

$y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

$v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

$y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\$

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

$y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]$

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

4 0

3 years ago

Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin

likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

Mg - T = Ma (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0 and T - mgcos57 - F = m(0) = 0

Mg - T = 0 (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0

3 years ago

Which of the following could possibly be predicted? Earthquakes Landslides Foreshocks Volcanic eruptions

NeTakaya

Answer:

Volcanic Eruptions

Explanation:

The volcano can start showing signs that it may be about to explode.

6 0

3 years ago

I have 2 bulbs one coverts 60% of energy to light 45% which is the most efficient and what happens to the rest of the energy

Leya [2.2K]

Answer:

The first one (60%)

Explanation:

The first one converts 15% more energy than the other one. Therefore, it is more efficient.

Hope this helps!

5 0

3 years ago

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