How can you say gravitational force is a field force

Answer:

Explanation:

Givens

t = 10 seconds

vi = 0

vf = 3600 mph

a = ?

d = 10000 feet

Formula

a = (vf - vi)/t

Solution

vf = 3600 mph * 1 hr / 3600 seconds * 5280 feet / 1 mile = 5280 ft / sec

a = (5280 - 0)/10

a = 528 ft/sec^2

Answer:

Remember the relation:

Speed*Time = Distance.

We can estimate that the speed at which an earthquake "moves", in the surface, is:

S = 6km/s  (this is a low estimation actually)

Then if the wave is detected 12.5 minutes after the earthquake, we know that it traveled for 12.5 minutes before reaching the detector.

So we know the speed of the wave and the time it took to reach the detector, then we can use the equation:

Speed*Time = Distance.

to find the distance.

First, we should write the time in seconds

we know that:

1 min = 60 s

then:

12.5 min = 12.5*(60 s) = 750 s

Then, the wave traveled with a speed of 6 km/s for 750 seconds until it reached the detector, then the distance that it traveled is:

(6km/s)*750s = 4500 km

The distance between the detector and the site of the quake is around 4500 km.

Answer:

D

Explanation:

i just did it trust me

[tex]\boxed{\sf E=QV^2}[/tex]

[tex]\\ \sf\longmapsto Q=\dfrac{E}{V^2}[/tex]

[tex]\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{24^2}[/tex]

[tex]\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{576}[/tex]

[tex]\\ \sf\longmapsto Q=0.006\times 10^{-4}C[/tex]

[tex]\\ \sf\longmapsto Q=6\times 10^{-1}C[/tex]

[tex]\\ \sf\longmapsto Q=0.6C[/tex]

Now

[tex]\boxed{\sf Q=CV}[/tex]

[tex]\\ \sf\longmapsto C=\dfrac{Q}{V}[/tex]

[tex]\\ \sf\longmapsto C=\dfrac{0.6}{24}[/tex]

[tex]\\ \sf\longmapsto C=0.025F[/tex]

Note:-

Answer:

Explanation:

We are basically needing to solve for the time in the equation d = rt, where d is the distance around Mars (aka the circumference), r is the velocity, and t is time. We need to find the circumference and the velocity. We will begin with the velocity.

Because the gravitational attraction between Phobos and Mars provides the centripetal acceleration necessary to keep Phobos in its (sort of) circular path, the equation we use for this is:

[tex]F_g=F_c[/tex] which says that Force supplied by gravity is equal to the centripetal force. Expanding that:

[tex]\frac{Gm_{Phobos}m_{Mars}}{r^2}=\frac{m_{Phobos}v^2}{r}[/tex]

When we move that around mathematically to solve for the velocity value, what we end up with is:

[tex]v=\sqrt{\frac{Gm_{Mars}}{r}[/tex] and filling in:

[tex]v=\sqrt{\frac{(6.67*10^{-11})(6.42*10^{23})}{9.38*10^6} }[/tex] and we get that

v = 2100 m/s

Now for the circumference:

C = 2πr and

C = 2(3.1415)(9.38 × 10⁶) so

C = 5.9 × 10⁷

Putting that all together in the C = vT equation:

5.9 × 10⁷ = 2100T so

T = 2.8 × 10⁴ sec or 7.8 hours

[tex]\\ \sf\longmapsto 300-70.5=229.5kJ[/tex]

We know

[tex]\boxed{\sf Efficiency=\dfrac{Used\:Energy}{Supplied\:Energy}\times 100}[/tex]

[tex]\\ \sf\longmapsto Efficiency=\dfrac{229.5}{300}\times 100[/tex]

[tex]\\ \sf\longmapsto Efficiency=\dfrac{229.5}{3}[/tex]

[tex]\\ \sf\longmapsto Efficiency=76.5\%[/tex]

Answer:

Yes, this is true.

-- While the apple is falling, the same gravitational force acts on both the apple and the Earth.

-- The mass of the apple is somewhere in the neighborhood of 1/4 kg.

-- The mass of the Earth is about 5.972 x 10²⁴ kg.

-- Since the Earth has roughly 2.389 x 10²⁶ times as much mass as the apple has, the apple has roughly 2.389 x 10²⁶ greater acceleration than the Earth has, and moves roughly 2.389 x 10²⁶ times as far down as the Earth moves up, before they smack together.

-- That's why you don't notice the Earth's motion.

-- Also, you're standing on the Earth, moving up with it, toward the apple.  Maybe it would be different if you were sitting on the apple, riding it down to the ground, and you were able to notice the motion of the ground coming up to meet you at a speed that's 0.00000000000000000000000000419 of YOUR speed.  

Answer From Gauth Math

Momentum is a quantity that we can say is typically conserved in collisions.

Momentum is the quantity that that we see typically conserve in the collision,therefore the correct answer is the option A

It is the type of collision in which the total momentum  as well as the kinetic energy on the system is conserved .The momentum before the collision is equal to the momentum after the collision in other words one can say in the elastic collision momentum of the quantity is conserved.

In the inelastic collision the momentum is conserved but the kinetic energy of the quantity is not conserved.

Momentum of a quantity is conserved irrespective of the type of collision  whether it will be elastic collision or inelastic collision the momentum of the quantity will always be conserved.

Hence,we can say that momentum is the quantity that is conserved in collision,The correct answer for the given problem is option A.

Learn more about the collision and from here

https://brainly.com/question/2356330

#SPJ2

Answer:

could and largejsjisj and we look like they can get to

When,a man stands still on a moving walkway that is going at a speed of 0.2 m/s to the west,then the velocity of the man according to a stationary observer would be 0.2 m/s west,Therefore, the correct answer would be option B.

The total displacement covered by any object per unit of time is known as velocity. The velocity of an object is depend on the magnitude as well as the direction of the object.

the mathematical expression for velocity is given by

velocity = total displacement /time

The speed and velocity of the object are very different from each other because speed does not depend upon the direction of the moving  object but velocity depends on the magnitude of speed as well as the direction of the moving object.

As per given in the problem a man stands still on a moving walkway that is going at a speed of 0.2 m/s to the west therefore from the perspective  of a stationary observer the velocity of man would be 0.2 m/s towards West.

Hence the correct options for the problem would be option B.

Learn more about Velocity from here

brainly.com/question/18084516

#SPJ2

Answer:

Explanation:

        16093.4

Answer:

250 Pa or N/m^2

Explanation:

Pressure(P)=F/A=500/2=250 Pa or N/m^2

Answer:

38500

Explanation:

I looked it up so it may be wrong

Explanation:

Volume is the quantity of three-dimensional space enclosed by a closed surface

The si unit of mass is kilogram (kg) and the si unit of length is metre(m)

Our sun is 27 million times larger than our moon. Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth. If tidal forces were based solely on comparative masses, the sun should have a tide-generating force that is 27 million times greater than that of the moon. However, the sun is 390 times further from the Earth than is the moon. Thus, its tide-generating force is reduced by 3903, or about 59 million times less than the moon. Because of these conditions, the sun’s tide-generating force is about half that of the moon.

Answer:

E_O = 96J

Explanation:

E_O = E_I*(%e/100%)

E_O = 120J*(80%/100%)

E_O = 96J

The approximate temperature of the star as determined is D) 5500 K.

The Wien's displacement law relates the maximum wavelength of a body to its absolute temperature. Wien's displacement law states that:

λ = [tex]\frac{b}{T}[/tex]

where λ is the maximum wavelength of the body, b is the constant of proportionality and T is the absolute temperature.

Thus from the given question, λ = 525 nm (525 x [tex]10^{-9}[/tex]), and b = 2.9 x [tex]10^{-3}[/tex] mK.

So that,

525 x [tex]10^{-9}[/tex] = [tex]\frac{2.9*10^{-3} }{T}[/tex]

Make T the subject of the formula to have;

T = [tex]\frac{2.9*10^{-3} }{525*10^{-9} }[/tex]

  = 5523.81

T = 5523.81 K

T ≅ 5500.00 K

The approximate temperature of the star in Kelvin is 5500 K.

For more clarifications, kindly visit: https://brainly.com/question/20038918

Answer:

The acceleration due to gravity at Pluto is 0.0597 m/s^2.

Explanation:

Length, L = 1 m

10 oscillations in 257 seconds

Time period, T = 257/10 = 25.7 s

Let the acceleration due to gravity is g.

Use the formula of time period of simple pendulum

[tex]T = 2\pi\sqrt{\frac{L}{g}}\\\\25.7 = 2 \times 31.4\sqrt{\frac{1}{g}}\\\\g = 0.0597 m/s^2[/tex]

The height of the window above the ground is 11.025m

The first step is to write out the parameters given in the question

U(initial velocity)= 0

Time= 1.5 seconds

Acceleration due to gravity= 9.8

Therefore the height of the window can be calculated as follows

S= ut + 1/2gt²

= 0(1.5) + 1/2(9.8)(1.5²)

= 0 + 1/2(9.8)(2.25)

=  1/2(22.05)

= 0.5×22.05

= 11.025

Hence the height of the window above the ground is 11.025m

Click on the link below to learn more

https://brainly.com/question/11853995?referrer=searchResults

Answer:

I THINK IT'S D....

HOPE SO

Answer:

42g

Explanation:

the mass of a body is given by

m=weight/gravity

=420/10

=42

I hope this helps

Answer:

Current- the flow of free charges, such as electrons and ions

Drift velocity- the average speed at which these charges move

Answer:

An electric current is the stream of changed particals, such as electrons & ions, moving through an electrical conductor.

The average velocity attained bycharged partical ,such as electrons,in a material due to electric fields

Answer:

if buoyant force is equal is equal to the weight of body then our body float. if density of our body is less than density of liqiud then our body float in water.

Explanation:

tilting it will raise the height of its center of gravity.

Answer:

The velocity at that time would be [tex](-67.975\; \rm s)[/tex]. The velocity of this particle is continuously changing.

Explanation:

Differentiate the expression for position [tex]x[/tex] with respect to time [tex]t[/tex] to find an expression for velocity.

[tex]\begin{aligned}v(t) &= \frac{d}{dt}[x(t)] \\ &= \frac{d}{dt} \left[ 7.4 + 9.2\, t - 2.1\, t^{3}\right]\\ &= 9.2 - 6.3\, t^{2}\end{aligned}[/tex].

Hence, at [tex]t = 3.5\; \rm s[/tex], velocity would be [tex]v(3.5) = 9.2 - 6.3 \times (3.5)^{2} = -67.975\; \rm m[/tex].

Since velocity [tex]v(t)[/tex] changes with time [tex]t[/tex], the velocity of this particle is continuously changing.

Answer:

I dont know answer Sorry For that thank u

Answer:

ccccccccc or bbbbbbb or aaaaaaaa or ddddddd