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AveGali [126]
3 years ago
14

Rank the average gravitational forces from greatest to least between (a) the Sun and Mars, (b) the Sun and the Moon, and (c) the

Sun and Earth. Rank from greatest to least. To rank items as equivalent, overlap them.
Physics
1 answer:
Vera_Pavlovna [14]3 years ago
7 0

Answer:

b,c,a.

Explanation:

We know that mass of sun > mass of Earth> mass of mars>mass of mars.  

And Gravitation force F =\frac{GmM}{r^2}

G= gravitational constant

m= mass of lighter body

M= mass of heavier body.

r= distance between them.

Although the mass of earth is greater than that of moon , its distance from the sun is larger hence, gravitational force on moon by sun is greater than that on Earth.

Hence the correct order will be

b,c,a.

You might be interested in
An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below
grin007 [14]
<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let P be the pressure at a point.

Let p be the density fluid at a point.

Let v be the velocity of fluid at a point.

Bernoulli's equation states that P+\frac{1}{2}pv^{2}+pgh=constant for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let p_{up} be the pressure of a point just above the wing.

Let p_{do} be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}

So,p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa

Force is given by the product of pressure difference and area.

Given that area is 27ms^{2}.

So,lifting force is 16398.48\times 27=442758.96N

6 0
3 years ago
An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N
algol13

Answer:

K.E = 15.57 x 10⁻¹⁷ J

Explanation:

First, we find the acceleration of the electron by using the formula of electric field:

E = F/q

F = Eq

but, from Newton's 2nd Law:

F = ma

Comparing both equations, we get:

ma = Eq

a = Eq/m

where,

E = electric field intensity = 120 N/C

q = charge of electron = 1.6 x 10⁻¹⁹ C

m = Mass of electron = 9.1 x 10⁻³¹ kg

Therefore,

a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)

a = 2.11 x 10¹³ m/s²

Now, we need to find the final velocity of the electron. Using 3rd equation of motion:

2as = Vf² - Vi²

where,

Vf = Final Velocity = ?

Vi = Initial Velocity = 1.4 x 10⁷ m/s

s = distance = 3.5 m

Therefore,

(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²

Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)

Vf = 1.85 x 10⁷ m/s

Now, we find the kinetic energy of electron at the end of the motion:

K.E = (0.5)(m)(Vf)²

K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

<u>K.E = 15.57 x 10⁻¹⁷ J</u>

4 0
2 years ago
Suppose there is a uniform magnetic field, B, pointing into the page (so your index finger will point into the page). If the vel
Naily [24]

Answer:

Explanation:

We shall show all given data in vector form and calculate the direction of force with the help of following formula

force F = q ( v x B )

q is charge , v is velocity and B is magnetic field.

Given B = - Bk ( i is  right  , j is  upwards  and k is straight up the page  )

v = v j

F = q ( vj x - Bk )

= -Bqvi

The direction is towards left .

a ) If velocity is down

v = - v j

F = q ( - vj x - bk )

= qvB i

Direction is right .

b ) v = v i

F = q ( vi x - Bk )

= qvB j

force is upwards

c ) v = - vi

F = q ( -vi x - Bk )

= -qvBj

force is downwards

d ) v = - v k

F = q( - vk x -Bk  )

= 0

No force will be created

e ) v =  v k

F = q(  vk x -Bk  )

= 0

No force will be created  

3 0
3 years ago
A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho
fiasKO [112]

Answer:

v_f=8.17\frac{m}{s}

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}

The box is initially at rest, so v_i=0. Solving for v_f:

v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}

5 0
2 years ago
how much chemical energy must be supplied to a car engine if we want it to produce 5000J of kinetic energy and it is 45% efficie
vagabundo [1.1K]

Here i state the conservation of energy rule and use that to justify my answer. I showed how to manipulate percentages to get the final answer of 11000J (2sf). Hope I'm right xx

3 0
3 years ago
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