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Svet_ta [14]
3 years ago
10

Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, excep

t when the symbol is needed to make your meaning clear. For example, 1⋅105 is not the same as 1105. When you need to be explicit, type * (Shift + 8) to insert the multiplication operator. You will see a multiplication dot (⋅) appear in the answer box. Do not use the symbol ×. For example, for the expression ma, typing m ⋅ a would be correct, but mxa would be incorrect.Enter the expression ma .
Physics
2 answers:
lidiya [134]3 years ago
8 0

Answer:

ma= ma

m⋅a = m⋅a

And equivalently:

am=ma

a⋅m = m⋅a

Explanation:

Question

Assuming this question "Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, except when the symbol is  needed to make your meaning clear. For example, 1*10^5 is not the same as 110^5 . When you need to be explicit, type * (Shift + 8) to insert the multiplication operator. You will see a multiplication dot (⋅) appear in the answer box. Do not use the symbol x. For example, for the expression ma,

typing m⋅a would be correct, but mxa would be incorrect".

Solution to the problem

For this case we want to write a expression for ma, and based on the previous info we can write:

ma= ma

m⋅a = m⋅a

And equivalently:

am=ma

a⋅m = m⋅a

But is not correct do this:

mxa=mxa

axm = mxa

lei3 years ago
0 0

Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, except when the symbol is needed to make your meaning clear. For example, 1⋅105 is not the same as 1105. When you need to be explicit, type * (Shift + 8) to insert the multiplication operator. You will see a multiplication dot (⋅) appear in the answer box. Do not use the symbol ×. For example, for the expression ma, typing m ⋅ a would be correct, but mxa would be incorrect.Enter the expression ma .

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A scientist is studying a shock wave from an earthquake. What kind of wave is being studying?
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3 0
3 years ago
A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels
Crank

Answer:

v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }

Explanation:

The average velocity is total displacement divided by time:

v_{avg} =\dfrac{D_{tot}}{t}

And in the case of vertical v_{avg}

v_{avg}=\dfrac{y_{tot}}{t}

where y_{tot} is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height H with initial velocity v_0 is given by:

y=H+v_0t-\dfrac{1}{2} gt^2

The time it takes for the rock to reach maximum height is when y'(t)=0, and it is

t=\frac{v_0}{g}

The vertical distance it would have traveled in that time is

y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2

y_{max}=\dfrac{2gH+v_0^2}{2g}

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

y_{down}=\dfrac{2gH+v_0^2}{2g}+H

Therefore, the total displacement throughout the rock's journey is

y_{tot}=y_{max}+y_{down}

y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H

\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

t=\dfrac{v_0}{g},

and it should take the rock the same amount of time to return to the roof, and it takes another t_0 to go from the roof of the building to the ground; therefore,

t_{tot}=2\dfrac{v_0}{g}+t_0

where t_0 is the time it takes the rock to go from the roof of the building to the ground, and it is given by

H=v_0t_0+\dfrac{1}{2}gt_0^2

we solve for t_0 using the quadratic formula and take the positive value to get:

t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

Therefore the total time is

t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH}  }{g}}

Now the average velocity is

v_{avg}=\dfrac{y_{tot}}{t}

v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }

\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }

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3 years ago
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