This is an image of a satellite traveling around Earth. Explain what are the two forces that are keeping the satelite around Earth without flying off or hitting the ground.​

Answer:

i think it's C thx correct me if wrong

Answer:120.3676

Explanation: using the molecular calculator and molar mass of MgSO4. hope this helps!

Answer:

P=mv

Explanation:

m = 2000kg

v = 10m/s

2000×10=20000

Answer: 2000kgm/s

The momentum will be =  2 * [tex]10^{4}[/tex] kg m/s

Momentum is a property of a moving body that the body has by virtue of its mass and motion and that is equal to the product of the body's mass and velocity

momentum = mass * velocity

mass = 2000 kg

velocity = 10 m/s

momentum = 2000 * 10 = 2 * [tex]10^{4}[/tex] kg m/s

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Answer:

λ/2 = length of pipe

The pipe is open at both ends having wavelength of A-N-A or antinode-node-antinode which is 1/2 wavelength

λ = 331 m/s / (65.4 / s) = 5.06 m     wavelength of sound

L = 5.06 m / 2 = 2.53 m      length of pipe

whats the restraunt called

Based on the scenario, Remy has likely been influenced by framing effect.

In Psychology, framing effect can be defined as a cognitive bias wherein a person's choice from a group of options is primarily influenced by the positive or negative connotations pertaining to the options.

In this scenario, Remy did a research on the barbecue restaurant and finds that there are a number of both negative and positive reviews about it. However, he convinced himself by relying on the positive reviews and decides to go to the restaurant.

In conclusion, we can deduce that Remy has likely been influenced by framing effect.

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Answer:

I believe the answer is speed up.

Explanation:

this is because when water heats up the molecules move father apart from each other they speed up, eventually causing the water to boll

Answer:

Explanation:

a) the acceleration of the puck on the rough ice.

a = μg = 0.550(9.81) = 5.3955 = 5.40 m/s²

 (comes from μ = F/N = ma/mg = a/g)

b) the distance from the end boards the puck is when it comes to a stop.

v² = u² + 2as

0² = 12.0² + 2(-5.40)s

s = 13.3 ft

so distance from the boards is

15.7 - 13.3 = 2.4 m

by the way...that's some VERY rough ice...more like sand.

Answer:

They are...if I'm correct Chemically combined, sorry if I'm wrong.

Answer:

Explanation:

Speed is total distance traveled over time taken to do so.

If AB is measured in kilometers, time (t) for the whole trip is

t = AB/60 + AB/40  

t = 2AB/120 + 3AB/120

t = 5AB/120 hrs

Average speed is distance over time

s = 2AB / (5AB/120)

s = 2(120)/5

s = 48 km/hr

s = 48(1000 m/km / 3600 s/hr) = 13.333333.... 13 m/s

Velocity is displacement over time.

As displacement is zero, velocity is zero

v = 0 km/hr = 0 m/s

Pretty harsh reporting answers just because they are wrong.

Answer:

0.02976°C

Explanation:

Heat Supplied = mcø

ø = 500/(4*4200)

ø = 0.02976 °C

The density of the liquid will be equal to   [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]

The density of an object is defined as the ratio of the mass of an object to the volume of the object.

Volume of tube = 2^2 * pi * 30 = 377 cm^3

Volume of tube submerged = 25* 377 / 30 = 314 cm^3

Buoyancy = weight of liquid displaced

Volume of liquid displaced = 314 cm^3

Mass of tube and lead = 250 + 30 = 280 g

Now from the mass density by definition

[tex]\rho = \dfrac{m}{v}[/tex]

[tex]m=\rho \times v[/tex]

Mass of liquid displaced = Mass being supported

[tex]314 \times \rho = 280 g[/tex]

[tex]\rho= \dfrac{280}{ 314 } = .892 \frac{g}{cm^3}[/tex]

Thus the density of the liquid will be equal to   [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]

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The density of the liquid is 1.67 g/[tex]cm^3[/tex].

The volume of the tube is

30 * 4 * 3.14 * 0.25 = 94.2 [tex]cm^3[/tex].

The mass of the lead pellets and the plastic tube is

30 + 250 = 280 g.

The volume of the lead pellets is

250 / 11.34 = 22 [tex]cm^3[/tex].

The volume of the liquid that the tube displaces is

94.2 - 22 = 72 [tex]cm^3[/tex].

The density of the liquid is

280 / 72 = 1.67 g/ [tex]cm^3[/tex].

Therefore, the density of the liquid is 1.67 g/ [tex]cm^3[/tex].

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Answer:

Your parents or anyone in your ancestry.

Explanation:

Answer:

Dynamic stretching

Explanation:

Dynamic stretching is a form of stretching beneficial in sports utilizing momentum from form.

7.4 s

Explanation:

Given:

[tex]v_0 = 73\:\text{m/s}[/tex]

[tex]v = 0[/tex]

[tex]a = -9.8\:\text{m/s}^2[/tex]

[tex]t = ?[/tex]

To solve the time it takes for the object to come to a stop, we are going to use the equation below:

[tex]v = v_0 + at \Rightarrow t = \dfrac{v - v_0}{a}[/tex]

Using the given values above, we get

[tex]t = \dfrac{0 - 73\:\text{m/s}}{-9.8\:\text{m/s}^2}[/tex]

[tex]\;\;\;\;= 7.4\:\text{s}[/tex]

Answer:

a = 0.01m/s²

Explanation:

V_f = V_0+a*t

V_f = Velocity final

V_0 = Velocity initial

a = acceleration

t = time

a = (V_f-V_0)/t

a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))

a = 0.01m/s²

Answer:

Explanation:

E = mgh

mg = E/h

mg = 60/0.75 = 80 N

Answer:

Let’s substitute in our givens.

(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)

I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.

Note that I have considered the bullet’s velocity to be in the positive direction,

The answer is vf = 0.280 m/s

Answer:

Explanation:

30 = e^5k

ln30 = lne^5k

ln30 = 5k

k = ln30/5

k = 0.68023947...

round to your heart's content.

Answer: An answer on what? I’ll never ignore you!

Explanation:

Answer:

an answer on what?

Explanation: Im here to help!!

Answer:

[tex]a[/tex]

Explanation:

is a corect anser

Answer:

Explanation:

momentum is mass times velocity

p = mv

so take the momentum of the truck in question 17 and divide by the mass of this car

v = p/m = p / 1400

Gases have heavier molecules. Since all gases have the same average kinetic energy at the same temperature, lighter molecules move faster and heavier molecules move slower on average.

Answer:

P = 2 pi (L / g)^1/2

P2 / P1 = (8 / 2)^1/2 = 2

The period would be twice as long or 5.6 sec.

Newton's third law allows to find the result for the value of the reaction force during the collision is:

Newton's third law stable that the forces appear in pairs or ea that when two bodies interact, the interaction forces appear in the two bodies simultaneously, in general they are called action and reaction forces.

These furas are of the same magnitude, but in the opposite direction, each one applied to one of the bodies.

They indicate that the most lighter body collides with the one with the greatest mass with a force of F = 90 N. If we call this the action, the larger body must react with a force of equal magnitude on the lighter body.

Consequently, the reaction force is F = 90 N directed towards the lighter body.

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Hi there!

We can use the following:

W = ΔKE = F · d

Find the work done on the cart:

W = 200 · 10 = 2000 J

Now, this is equal to the change in kinetic energy of the object. Its initial kinetic energy is 0 J since it starts from rest, so:

2000J = KEf - KEi

KE is given as:

[tex]KE = \frac{1}{2}mv^2[/tex]

2000J = 1/2(55)v²

4000 = 55v²

√(4000/55) = 8.53 m/s

The acceleration due to gravity would be 5.95 m/s²

A force is known to be a push or pull and it is the change in momentum per time. It can be expressed by using the relation.

From the parameters given:

By replacing the given values into the above equation, we can determine the acceleration.

∴

625 N = 105 kg × acceleration.

[tex]\mathbf{acceleration = \dfrac{625 \ N}{105 \ kg}}[/tex]

acceleration = 5.95 N/kg

acceleration = 5.95 m/s²

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Hi there!

Recall Newton's Law of Universal Gravitation:

[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]

Where:

Fg = Force of gravity (N)

G = Gravitational Constant

m1, m2 = masses of objects (kg)

r = distance between objects (m)

Plug in the given values stated in the problem:

[tex]F_g = (6.673*10^{-11})\frac{50 * 50}{25^2} = \boxed{2.669 * 10^{-10} N}[/tex]

The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

Let suppose that the wheel rotates at constant angular speed ([tex]\omega[/tex]), in radians per second, the tangential speed of a wad of chewing gum to the rim of the wheel ([tex]v[/tex]), in centimeters per second, is:

[tex]v = 2\pi\cdot r\cdot f[/tex] (1)

Where:

If we know that [tex]f = 5.13\,hz[/tex] and [tex]r = 41.5\,cm[/tex], then the tangential speed of the chewing gum is:

[tex]v = 2\pi\cdot (41.5\,cm)\cdot (5.13\,hz)[/tex]

[tex]v \approx 1337.659\,\frac{cm}{s}[/tex]

The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

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