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2 years ago

14

Which of the following has to happen for Earth's shadow to fall on the moon?

1 answer:

Ronch [10]2 years ago

6 0

The correct answer is a, lunar eclipse

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Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s starting from rest. Assuming that they have constan

elena55 [62]

Answer:

$10.07m/s^2$

Explanation:

Information we have:

velocities:

initial velocity: $v_{i}=0mph$ (starts from rest)

final velocity: $v_{f}=50mph$

time: $t=2.22s$

Since we need the answer in $m/s^2$ , we nees to convert the speed to meters per second:

$v_{f}=\frac{50miles}{1hour}(\frac{1hour}{3,600s} )(\frac{1609.34meters}{1mile} ) \\\\v_{f}=\frac{50*1609.34}{3600} m/s\\\\v_{f}=22.35m/s$

We find the acceleration with the following formula:

$a=\frac{v_{f}-v_{i}}{t}$

substituting the known values:

$a=\frac{22.35m/s-0m/s}{2.22s}\\ \\a=10.07m/s^2$

the acceleration is 10.07 $m/s^2$

8 0

2 years ago

Read 2 more answers

Which cell part are red blood cells missing

12345 [234]

They lack a cell nucleus.

3 0

3 years ago

Be sure to answer all parts. Assume the diameter of a neutral helium atom is 1.40 × 102 pm. Suppose that we could line up helium

ss7ja [257]

Answer:

The number of atoms are $1.86\times10^{8}$ .

Explanation:

Given that,

Diameter $D = 1.40\times10^{2}\ pm$

$D=1.40\times10^{2}\times10^{-12}\ m$

Distance = 2.60 cm

We calculate the number of atoms

Using formula of numbers of atoms

$Number\ of\ atoms =\dfrac{2.60\times10^{-2}}{1.40\times10^{2}\times10^{-12}}$

$Number\of\atoms =1.86\times10^{8}$

Hence, The number of atoms are $1.86\times10^{8}$ .

8 0

2 years ago

A meterstick is placed on a pivot point of 42.5cm and a 45g mass is hung at the 20cm mark. When released the meterstick remains

Vikki [24]

Hey i dont have an answer but i need the points for finals today. Thank you

6 0

3 years ago

Read 2 more answers

Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee

JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

Explanation:

Given that,

Mass of proton $m=1.67\times10^{-27}\ kg$

Speed $v= 9.00\times10^{7}\ m/s$

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2$

$K.E=6.76\times10^{-12}\ J$

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

$K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)$

$K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)$

$K.E=7.25\times10^{-12}\ m/s$

Hence, The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

7 0

3 years ago