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2 years ago

Glenn shoots an arrow at a 30.0 degree angle. It has a velocity of 65.0 m/s How far will the arrow travel?

Physics

1 answer:

Burka [1]2 years ago

3 0

I just shot my shot at this little 52 ms far

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How many neutrons does element X have if its atomic number is 23 and its mass number is 90?

givi [52]

Answer:

Explanation:

- The atomic number (Z) of an atom is equal to the number of protons in the nucleus

- The mass number (A) of an atom is equal to the sum of protons and neutrons in the nucleus

Therefore, calling p the number of protons and n the number of neutrons, for element X we have:

Z = p = 23

A = p + n = 90

Substituting p=23 into the second equation, we find the number of neutrons:

n = 90 - p = 90 - 23 = 67

4 0

3 years ago

Read 2 more answers

30.42

CaHeK987 [17]

Answer: Try C

Explanation:

It's the only one that makes since.

7 0

3 years ago

Which of the following wire gage sizes is the thickest? <br> A. 8<br> B. 10<br> C. 14<br> D. 0

maks197457 [2]

A because it's the smaller the thicker you just can't have 0 gage

5 0

3 years ago

The density of mobile electrons in copper metal is 8.4 1028 m-3. Suppose that i = 4.6 1018 electrons/s are drifting through a co

Vesna [10]

Answer:

The time is 106.7 minute.

Explanation:

Given that,

Density $= 8.4\times10^{28}\ m^3$

Current $i = 4.6\times10^{18}\ electron/s$

Diameter of wire = 1.2 mm

Length = 31 cm

We need to calculate the drift velocity

Using formula of drift velocity

$v_{d}=\dfrac{I}{neA}$

$v_{d}=\dfrac{Ne}{tne\times\pi r^2}$

Put the value into the formula

$v_{d}=\dfrac{4.6\times10^{18}}{8.4\times10^{28}\times\pi\times(0.6\times10^{-3})^2}$

$v_{d}=4.842\times10^{-5}\ m/s$

We need to calculate the time

Using formula for time

$v_{d}=\dfrac{l}{t}$

$t=\dfrac{l}{v_{d}}$

Where, l = length

$v_{d}$ = drift velocity

Put the value into the formula

$t=\dfrac{31\times10^{-2}}{4.842\times10^{-5}}$

$t=6402.31\ sec$

$t=106.7\ minute$

Hence, The time is 106.7 minute.

7 0

3 years ago

A solar cooker, really a concave mirror pointed at the Sun, focuses the Sun's rays 13.8 cm in front of the mirror. Part A What i

shepuryov [24]

From the concept of optics on a curvature of a spherical mirror, the proportion for which the focal length is equivalent to half the radius of curvature is fulfilled. Mathematically this is

$f = \frac{R}{2}$

Here,

f = Focal Length

R = Radius

Rearranging to find the radius we have,

$R = 2f$

Replacing with our values,

R = 2(13.8cm)

R = 27.6cm

Therefore the radius of the spherical surface from which the mirror was made is 27.6cm

6 0

3 years ago