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Anuta_ua [19.1K]

2 years ago

5

Determine the orbits period of the moon when the distance between the earth and the moon is 3.82 x 10 to the power of 8

2 answers:

777dan777 [17]2 years ago

6 0

Answer:

3820000000

Explanation:

Gennadij [26K]2 years ago

3 0

This statement is infact true

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A 1.5-kilogram cart initially moves at 2.0 m/s. It is brought to rest by a constant net force in 0.3 seconds. What is the magnit

ser-zykov [4K]

We know, F = m.a
Here, m = 1.5 Kg
a = v/t = 2/0.3 = 6.67 m/s

Substitute their values,
F = 1.5 * 6.67
F = 10 N

In short, Your Answer would be 10 Newtons

Hope this helps!

5 0

3 years ago

A flare is launched from a boat. The height, , in meters, of the flare above the water is approximately modelled by the function

Tpy6a [65]

Answer:

10 seconds

Explanation:

If the height is modeled by the function $h(t)=-15t^{2} +150t$ , then the seconds it takes to reach the water (when the height equals 0) is modeled by the following.

$0=-15t^{2} +150t\\0=-15t*(t-10)\\\\t=0\\t=10$

6 0

2 years ago

When a star begins to run out of fuel, what two types of stars can it become

Nikolay [14]

Answer:

A red giant or red super giant then a white dwarf

Explanation:

This also depends on what the star classification was to begin with.

7 0

2 years ago

An organism lives in the darkest aquatic ecosystem, where light cannot reach. Which ecosystem is this organism most likely in?

bazaltina [42]

The deep sea vents at the very bottom of the ocean. The thermal vent communities as otherwise stated.

5 0

2 years ago

Read 2 more answers

A certain elastic conducting material is stretched into a circular loop of 1.6 cm radius. It is placed with its plane perpendicu

inysia [295]

Answer:

Induced emf in the loop is 0.0603 volts.

Explanation:

It is given that,

Radius of the circular loop, r = 1.6 cm = 0.016 m

Magnetic field, B = 0.8 T

When released, the radius of the loop starts to shrink at an instantaneous rate of 75.0 cm/s, $\dfrac{dr}{dt}=75\ cm/s=0.75\ m/s$

We need to find the magnitude of induced emf at that instant. Induced emf is given by :

$\epsilon=\dfrac{-d\phi}{dt}$

Where

$\phi$ is the magnetic flux, $\phi=B\times A$

$\epsilon=\dfrac{-d(BA)}{dt}$ , A is the area of cross section

$\epsilon=-\dfrac{-d(B(\pi r^2))}{dt}$

$\epsilon=-2\pi r B(\dfrac{dr}{dt})$

$\epsilon=-2\pi \times 0.016\times 0.8 \times 0.75\ m/s$

$\epsilon=0.0603\ V$

So, the induced emf in the loop is 0.0603 volts. Hence, this is the required solution.

3 0

3 years ago