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3 years ago

The mass of the Earth is about 6 x 1024 kg

1 answer:

3 0

Well it seems that you did not give answer choices, but that its fine since we can use newtons law of universal gravitational, Fg = GM1M2/r^2. So G is the gravitational constant, which is 6.67*10^-11, we can plug in 6*1024 for M1, and 7*1022 for M2, and 3.8*108 for r. Which then we get 1.74 * 10^8 N as the force of attraction between the Earth and the moon.

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You push a box with a force of 80 n. if the net force on the box is 50 n, what is the force on the box due to sliding friction?

Elodia [21]

The force of the sliding friction is 30 N.

3 0

3 years ago

6) Set the battery to a value between 0.0 V and 1.5 V. Now drag the voltage meter toward the capacitor and move the red and blac

KIM [24]

Answer:

A) 1.5 v

B) Top plate is at higher voltage than the bottom plate

Explanation:

Battery value set between 0.0 V and 1.5 V

a) The potential difference between the plates

Δ V = V1( potential at top plate) - V2( potential at lower plate )

potential at top plate = 1.5 V

potential at lower plate = 0.0 V

hence potential difference = 1.5 V

b ) The top plate is always connected to the positive terminal of the DC source ( which is at a higher potential )while the bottom plate is connected to the negative terminal of the DC source ( which is at a lower potential )

hence the Top plate is at higher voltage than the bottom plate

7 0

3 years ago

For both resonance curves and Fourier spectra, amplitude is plotted vs frequency, but these two types of plots are not the same.

andrey2020 [161]

Answer:

he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

Explanation:

In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values where some natural frequency of the system coincides with the excitation frequency.

In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.

From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

7 0

3 years ago

The table below shows the distance from the coast of four locations in Florida.

weeeeeb [17]

<span>LOCATION Z, because it is only 2 away from the coast.

The rest are farther inland

hope this helps</span>

6 0

3 years ago

An air-track glider with a mass of 239 g is moving at 0.81 m/s on a 2.4 m long air track. It collides elastically with a 513 g g

HACTEHA [7]

Answer:

Glider it stops just when it reaches the end of the runway

Explanation:

This is a shock between two bodies, so we must use the equations of conservation of the amount of movement, in the instant before the crash and the subsequent instant, with this we calculate the second glider speed, as the shock that elastic is also keep it kinetic energy

Po = pf

Ko = Kf

Before crash

Po = m1 Vo1 + 0

Ko = ½ m1 Vo1²

After the crash

Pf = m1 Vif + Vvf

Kf = ½ m1 V1f² + ½ m2 V2f²

m1 V1o = m1 V1f + m2 V2f (1)

m1 V1o² = m1 V1f² + m2 V2f² (2)

We see that we have two equations with two unknowns, so the system is solvable, we substitute in 1 and 2

m1 (V1o -V1f) = m2 V2f (3)

m1 (V1o² - V1f²) = m2 V2f²

Let's use the relationship (a + b) (a-b) = a² -b²

m1 (V1o + V1f) (V1o -V1f) = m2 V2f²

We divide with 3 and simplify

(V1o + V1f) = V2f (4)

Substitute in 3, group and clear

m1 (V1o - V1f) = m2 (V1o + V1f)

m1 V1o - m2 V1o = m2 V1f + m1 V1f

V1f (m1 -m2) = V1o (m1 + m2)

V1f = V1o (m1-m2 / m1+m2)

We substitute in (4) and group

V2f = V1o + (m1-m2 / m1 + m2) V1o

V2f = V1o [1+ + (m1-m2 / m1 + m2)]

V2f = V1o (2m1 / (m1+m2)

We calculate with the given values

V1f = 0.81 (239-513 / 239 + 513)

V1f = 0.81 (-274/752)

V1f = - 0.295 m/s

The negative sign indicates that the planned one moves in the opposite direction to the initial one

V2f = 0.81 [2 239 / (239 + 513)]

V2f = 0.81 [0.636]

V2f = 0.515 m / s

Now we analyze in the second glider movement only, we calculate the energy and since there is no friction,

Eo = Ef

Where Eo is the mechanical energy at the lowest point and Ef is the mechanical energy at the highest point

Eo = K = ½ m2 vf2²

Ef = U = m2 g Y

½ m2 v2f² = m2 g Y

Y = V2f² / 2g

Y = 0.515²/2 9.8

Y = 0.0147 m

At this height the planned stops, let's use trigonometry to find the height at the end of the track of the track

tan θ = Y / x

Y = x tan θ

The crash occurs in the middle of the track whereby x = 1.2 m

Y = 1.2 tan 0.7

Y = 0.147 m

As the two quantities are equal in glider it stops just when it reaches the end of the runway

7 0

3 years ago