All categories

3 years ago

The tidal bulge on the side of the earth opposite the moon is due to _______ .

Physics

2 answers:

butalik [34]3 years ago

8 0

Due to the moon's gravitational force and inertias counterbalance.

Zepler [3.9K]3 years ago

5 0

Answer:

Due to the moon's gravitational force and inertias counterbalance.

You might be interested in

A plane traveled west for 4.0 hours and covered a distance of 4,400 kilometers. What was its velocity? 18,000 km/hr

olga55 [171]

Velocity is the same as the formula for speed, the only difference is velocity has direction. Velocity is distance over time. Given is 4,400 kilometers travelled west in 4 hours. Applying the given equation, we wil have 4,400/4 = 1,100 km/hr west

5 0

3 years ago

Read 2 more answers

A force F→=(cx-3.00x2)iˆ acts on a particle as the particle moves along an x axis, with F→ in newtons, x in meters, and c a cons

DerKrebs [107]

Answer:

Explanation:

Work done = ∫Fdx

= ∫(cx-3.00x²) dx

[ c x² / 2 - 3 x³ / 3 ]₀²

= change in kinetic energy

= 11-20

= - 9 J

[ c x² / 2 - x³ ]₀² = - 9

c x 2² / 2 - 2³ = -9

2c - 8 = -9

2c = -1

c = - 1/2

6 0

3 years ago

Read 2 more answers

What are the scientific mesurments ordered from greatest to least

Sveta_85 [38]

Meters Micrometers centimeters millimeters

5 0

3 years ago

A box full of charged plastic balls sits on a table. The electric force exerted on a ball near one upper corner of the box has c

tatuchka [14]

We have that the values for F north, F east, F up are

$F_N=1.09090909*10^{-5}$
$F_E=5.18181818*10^{-6}$
$F_E=2*10^{-6}$

From the Question we are told that

electric force $F_1 = 1.2 x 10^{-3} N(N)$

electric force , $F_2=5.7 x 10^{-4} N(E)$

electric force , $F_3=2.2 x 10^{-4} N (U)$

charge on this ball one $q_1= 110 nC.$

charge on this ball two $q_2= -50 nC.$

Generally the equation for the F north is mathematically given as

$F_N=\frac{F_1}{q_1}\\\\F_N=\frac{ 1.2 * 10^{-3} )}{110}$

$F_N=1.09090909*10^{-5}$

For F East

$F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}$

$F_E=5.18181818*10^{-6}$

For F UP

$F_U=\frac{F_3}{q_1}\\\\F_U=\frac{2.2 x 10^-4 }{110}$

$F_E=2*10^{-6}$

For more information on this visit

brainly.com/question/21811998

5 0

2 years ago

A charge feels a 7.89 x 10-7 N force when it moves 2090 m/s at a 29.4° angle to a 4.23 x 10-3 T magnetic field. What is the amou

tamaranim1 [39]

We have that the amount of the charge q is

$q=1.8*10^{-7}$

From the Question we are told that

Force $F=7.89 x*10^{-7}$

Velocity $V=2090m/s$

Angle $\theta=29.4$

Magnetic field $B=4.23 * 10^{-3} T$

Generally, the equation for Force F is mathematically given by

$F=qVBsin\theta\\\\q=\frac{F}{VBsin\theta}$

$q=\frac{7.89 x*10^{-7}}{4.23 * 10^{-3} T*sin29.4*2090m/s}$

$q=1.8*10^{-7}$

In conclusion

The amount of the charge q is

$q=1.8*10^{-7}$

For more information on this visit

brainly.com/question/1470439

6 0

2 years ago

Read 2 more answers