Using this equation: LaTeX: v\:=\frac{\Delta x}{\Delta t}v = Δ x Δ t, solve for LaTeX: \Delta tΔ t Group of answer choices LaTeX: \Delta t\:=\frac{v}{\Delta x} Δ t = v Δ x Δ t = v Δ x LaTeX: \Delta t\:=\Delta x\:.\:v Δ t = Δ x . V Δ t = Δ x . V LaTeX: \Delta t\:=\frac{\Delta x}{v} Δ t = Δ x v LaTeX: \Delta t\:=\Delta x-v

Answers

Answer 1

Answer:

Answer:

$\Delta t = (\Delta x)/(v) \n\n$

Explanation:

Given the expression $v\:=(\Delta x)/(\Delta t)$ , we are to find the value of Δt from the expression. To find Δt, we will make Δt the subject of the formula from the given expression as shown;

$v\:=(\Delta x)/(\Delta t)\n\ncross\ multiply\n\nv*\Delta t = \Delta x\n\nDivide \ both\ sides \ of\ the \ resulting \ equation \ by \ v\n\n(v*\Delta t)/(v) = (\Delta x)/(v) \n\n\Delta t = (\Delta x)/(v) \n\n$

Hence the equivalent expression for $\Delta t = (\Delta x)/(v) \n\n$ .

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Can any kind soul help me please

Answers

Explanation:

Hey there!

Given;

Mass of the Earth (M) = 6.0*10^24kg.

Radius of Earth (R) = 6400km = 6400*1000= 6.4*10^6.

Gravitational constant (G) = 6.67*10^-11Nm^2/kg^2.

Height from surface of Earth (h) = 350km= 3.5*10^5m

Gravity (g) = ?

We have;

$g = \frac{G.M}{ {(R + h)}^(2) }$

Where "G" is gravitational constant, "M" is mass of earth, "g" is acceleration due to gravity, "h" height from surface of Earth.

Keep all values.

$g = \frac{6.67 * {10}^( - 11) * 6.0 * {10}^(24) }{ (6.4 * {10}^(6) + 3.5 * {10}^(5) )^2}$

$or \: g = \frac{40.002* {10}^(13) }{ (6.4* {10}^(6) + 3.5*10^5)^2}$

g=8.7m/s^2

Therefore,gravityfromdistance350kmabove theEarth'ssurfaceis8.7m/s^2.

Hopeithelps...

Calculate the equivalent resultant force for the following loading system and determinethe location of the resultant force intersects the member BC from point C.StDOID175 lbA) 416 lb, 0.1 ftB) 416 lb, 10.9 ftC) 325 lb, 2.29 ftD) 260 lb. 8.71 ftNeo

Answers

Okay so first u will multiply then divide then add all

If there are 50 grams of U-238 on day zero of radioactive decay, how much will there be after 4.5 billion years

Answers

We use the formula of the half-life to calculate for the remaining U-238 after 4.5 billion years. The formula is expressed as A = A₀ (1/2)^(t/h) where A is the final amount, A₀ is the initial amount of the substance, t is the time and h is the half-life of the substance wherein for U-238 h is equal to 4.47 billion years.

A = A₀ (1/2)^(t/h)
A = 50 (1/2)^(4.5 / 4.47)
A = 24.88 g

25 if using usatsyprep

If a 78.2 kg kangaroo weighs 6742 N on the planet katar, calculate katar's radius

Answers

Answer:

about 7500 km

Explanation:

The mass of planet Katar is 7.27×10^25 kg. The force on the kangaroo is given by ...

F = GmM/r^2

Solving for r, we have ...

r = √(GmM/F) = √(6.67×10^-11 × 78.2 × 7.27×10^25 / 6742) ≈ √(5.62×10^13)

This is in meters, so the radius in km is about ...

r ≈ 7500 km

The process in which a substance changes from a gaseous state to the liquid state is A. critical temperature.
B. absolute zero.
C. evaporation.
D. condensation.

Answers

This process is called D. Condensation.

You can easily notice this happen on Air conditioning devices, where the cold gaseous air is turned into water that drips from the air conditioning unit due to condensation.

By definition we have to:

In physics, condensation is a change of state by which a substance passes from a gaseous state to a liquid state.

For example, in a refrigeration cycle, the refrigerant leaves the compressor in the gaseous state, from here, it enters the condenser, a state change called condensation occurs and the refrigerant leaves in liquid form. This process is useful in air conditioners.

Answer:

The process in which a substance changes from a gaseous state to the liquid state is:

D. condensation.

1. A substance which allows visible light to pass through freely is ______. A. opaque B. reflective C. transparent D. translucent2. A substance can be opaque to certain wavelengths of electromagnetic radiation, but transparent to others. A. True B. False
3. _______ is when either the vertical or the horizontal component of light waves is eliminated.

A. Refraction
B. Scattering
C. Transmission
D. Polarization

Answers

1. Transparent
2. True
3. Polarisation

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