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4 years ago

5

Light traveling though air strikes the surface of the four different materials shown. Which material reflects light but does not

refract it?

1 answer:

dangina [55]4 years ago

8 0

The spoons .................

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It takes brooke 10minutes to run 1 mile what is her speed in miles per minutes

hjlf

If you divide miles by minutes, the answer will have units of
miles per minute, which is exactly what you want.

(1 mile) / (10 minutes) = 1/10 mile/minute = 0.1 mile per minute

6 0

4 years ago

An electron moving with a velocity v⃗ = 5.0 × 107 m/s i^ enters a region of space where perpendicular electric and a magnetic fi

stiv31 [10]

Answer:

Magnetic field, $B=2\times 10^{-4}\ T$

Explanation:

Given that,

Velocity of electron, $v=5\times 10^7\ m/s$

It enters a region of space where perpendicular electric and a magnetic fields are present.

Magnitude of electric field, $E=10^4\ V/m$

We need to find the magnetic field will allow the electron to go through the region without being deflected.

Magnetic force on the electron, $F_m=qvB\ sin\theta$ .......(1)

Electric force on the electron, F = q E........(2)

From equation (1) and (2) we get:

$qvB\ sin\theta=qE$

$B=\dfrac{E}{v}$

$B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}$

B = 0.0002 T

or

$B=2\times 10^{-4}\ T$

Hence, this is the required solution.

3 0

3 years ago

A helium-filled balloon occupies a volume of 15 cubic meters at sea level. the balloon is released and raises to a point in the

Elena L [17]

According to Boyle’s law, For a fixed amount of an ideal gas kept at a fixed temperature, P (pressure) and V (volume) are inversely proportional.

Therefore,

$P_{1} V_{1} =P_{2} V_{2}$

Given $P_{1} = 1 atm$ , $V_{1} = 15 \ cubic \ meter$ and $P_{2} = 0.75\ atm$ .

Thus,

$V_{2} = \frac{P_{1}\times V_{1} }{ P_{2} } = \frac{1\times 15}{0.75} =20 m^3$

5 0

3 years ago

An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce

Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

$f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Frequency of pendulum is given by

$f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

Given in the question

$f_p=\dfrac{1}{2}f_s$

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m$

The unstretched length of the spring is 2.67 m

6 0

3 years ago

What form of energy is released into the atmosphere by the earth's surface

Crank

Answer:

Thermal Energy (Heat)

8 0

3 years ago

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