7. A runner increases his velocity from 0 m/s to 20 mis in 2 What is the runner's average acceleration?

2 decaliters + 800 deciliters = __________ liters?

Stolb23 [73]

Answer is 100 liters

8 0

3 years ago

Read 2 more answers

Water vapor enters the atmosphere through

Blizzard [7]

Answer:

Heat from the Sun causes water to evaporate from the surface of lakes and oceans. This turns the liquid water into water vapor in the atmosphere. Plants, too, help water get into the atmosphere through a process called transpiration! ... Water can also get into the atmosphere from snow and ice.

Most water vapor enters the atmosphere via evaporation and transpiration. Evaporation occurs when a single water molecule on a liquid water surface gains enough kinetic energy (often by solar radiation) to break the bond which holds the molecules together. Really Hopes this helps!

Explanation:

6 0

3 years ago

How is velocity different from speed? Based on distance and/or direction

Mnenie [13.5K]

Answer:

Speed is solved with time and distance but has no direction

Average velocity is solved with Δx/Δt and has a direction

6 0

2 years ago

A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in

Leno4ka [110]

Answer:

The electric field is $35\cos(90\pi t)\ mV/m$

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length $n= 1.65\times10^{3}$

Current $I = 6.00\sin 90\pi t$

We need to calculate the induced emf

$\epsilon =\mu_{0}nA\dfrac{dI}{dt}$

Where, n = number of turns per unit length

A = area of cross section

$\dfrac{dI}{dt}$ =rate of current

Formula of electric field is defined as,

$E=\dfrac{\epsilon}{2\pi r}$

Where, r = radius

Put the value of emf in equation (I)

$E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}$ ....(II)

We need to calculate the rate of current

$I=6.00\sin 90\pi t$ ....(III)

On differentiating equation (III)

$\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)$

Now, put the value of rate of current in equation (II)

$E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}$