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3 years ago

Can Someone Help Me With This?

Physics

1 answer:

Stels [109]3 years ago

4 0

$\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} a,b=\stackrel{bases}{parallel~sides}\\ h=height\\[-0.5em] \hrulefill\\ a=3\\ b = 9\\ A=84 \end{cases}\implies 84=\cfrac{h(3+9)}{2}\implies 84=\cfrac{h(12)}{2} \\\\\\ 84=6h\implies \cfrac{84}{6}=h\implies 14=h$

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Find the ratio of the final speed of the electron to the final speed of the hydrogen ion, assuming non-relativistic speeds. Take

KiRa [710]

Answer:

$\frac{V_{e}}{V_{h}}=0.428*10^{2}$

Explanation:

From conservation of energy states that

$K_{i}+v_{i}=v_{f}+K_{f}\\ as\\K_{i}=0\\K_{f}=1/2mv^{2}\\ v_{i}=qv\\v_{f}=0\\So\\qv=1/2mv^{2}\\ v=\sqrt{\frac{2qv}{m} }\\ Velocity_{electron}=\sqrt{\frac{2qv}{m_{e}} }\\Velocity_{hydrogen}=\sqrt{\frac{2qv}{m_{h}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{\frac{2qv}{m_{e}}}{\frac{2qv}{m_{h}}}}\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{m_{h}}{m_{e}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{1.67*10^{-27} }{9.11*10^{-31} } }\\\frac{V_{e}}{V_{h}}=0.428*10^{2}$

5 0

3 years ago

how do Swati and Banks adjust their body position during a skydiving jump so they can fall at the same rate

professor190 [17]

While skydiving, its not just freely falling under Earth's gravity. Additional force called drag acts against the gravity which slows down the rate of fall. Drag is caused by the air molecules which pushes against the body as it falls through them. This is actually a significant amount of force which slows down the rate of fall of the body. Drag depends on the contact surface area and weight. More the surface area in contact, more would be the drag. The sitting position of the skydiver would experience less drag than the chest down position because of the less contact surface area of the body with the air molecules while in the former case. No two persons have identical body shape and weight. Hence, the rate of fall can be made nearly equal but not exactly equal. This is would be possible when they are having same body position.

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2 years ago

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 2.5 x 10-15 m before

Tasya [4]

Explanation:

It is given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, $d=2.5\times 10^{-15}\ m$

Speed of light, $c=3\times 10^{8}\ m\s$

Let t is the time interval required for the strong interaction to occur. The speed is given by :

$c=\dfrac{d}{t}$

$t=\dfrac{d}{c}$

$t=\dfrac{2.5\times 10^{-15}\ m}{3\times 10^{8}\ m/s}$

$t=8.33\times 10^{-24}\ s$

So, the time interval required for the strong interaction to occur is $8.33\times 10^{-24}\ s$ . Hence, this is the required solution.

8 0

3 years ago

a small table has a mass of 4kg, stands on four legs, each leg having an area of 0.001 m2. what is the pressure exerted by the t

Gemiola [76]

Answer:

P = 10 kPa

Explanation:

Given that,

The mass of a small table, m = 4 kg

The area of each leg = 0.001 m²

We need to find the pressure exerted by the table on the floor. Pressure is equal to the force per unit area. So

$P=\dfrac{mg}{4\times A}\\\\P=\dfrac{4\times 10}{4\times 0.001}\\P=10000\ Pa\\\\or\\\\P=10\ kPa$

So, the required pressure is 10 kPa.

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2 years ago

Please help with science! Describe the magnetic field in the vicinity of a bar magnet.

Katen [24]

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3 years ago