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4 years ago

57. Estimate Potential Energy A boulder with a

Physics

1 answer:

Oksanka [162]4 years ago

3 0

Answer: 4.9 x 10^6 joules

Explanation:

Given that:

mass of boulder (m) = 2,500 kg

Height of ledge above canyon floor (h) = 200 m

Gravita-tional potential energy of the boulder (GPE) = ?

Since potential energy is the energy possessed by a body at rest, and it depends on the mass of the object (m), gravitational acceleration (g), and height (h).

GPE = mgh

GPE = 2500kg x 9.8m/s2 x 200m

GPE = 4900000J

Place result in standard form

GPE = 4.9 x 10^6J

Thus, the gravita-tional potential energy of the boulder-Earth system relative to the canyon floor is 4.9 x 10^6 joules

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A cyclotron is used to produce a beam of high-energy deuterons that then collide with a target to produce radioactive isotopes f

serg [7]

Answer:

19080667.0818 m/s

0.637294 m

$2.1875\times 10^{15}$

Explanation:

m = Mass of deuterons = $3.34\times 10^{-27}\ kg$

v = Velocity

K = Kinetic energy = 3.8 MeV

d = Diameter

B = Magnetic field = 1.25 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

t = Time = 1 s

i = Current = 350 μA

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 3.8\times 10^6\times 1.6\times 10^{-19}}{3.34\times 10^{-27}}}\\\Rightarrow v=19080667.0818\ m/s$

The speed of the deuterons when they exit is 19080667.0818 m/s

In this system the centripetal and magnetic force will balance each other

$\dfrac{mv^2}{r}=qvB\\\Rightarrow \dfrac{mv^2}{\dfrac{d}{2}}=qvB\\\Rightarrow d=\dfrac{2mv}{qB}\\\Rightarrow d=\dfrac{2\times 3.34\times 10^{-27}\times 19080667.0818}{1.6\times 10^{-19}\times 1.25}\\\Rightarrow d=0.637294\ m$

The diameter is 0.637294 m

Current is given by

$i=\dfrac{nq}{t}\\\Rightarrow n=\dfrac{it}{q}\\\Rightarrow n=\dfrac{350\times 10^{-6}\times 1}{1.6\times 10^{-19}}\\\Rightarrow n=2.1875\times 10^{15}$

The number of deuterons is $2.1875\times 10^{15}$

8 0

3 years ago

A car drives 215km east and then 45km north. What is the magnitude of the cars displacement? Round you answer to nearest whole n

kifflom [539]

Answer: $219.65\ \text{km}$

Explanation:

Given

Car drives 215 km east and then 45 km North

Displacement is East direction is

$\Rightarrow \vec{r_1}=215\hat{i}$

Now, the displacement from that to 45 km North is given by

$\Rightarrow \vec{r_{21}}=45\hat{j}$

Net displacement is $\vec{r_2}$

$\Rightarrow \vec{r_2}=\vec{r_1}+\vec{r_{21}}\\\Rightarrow \vec{r_2}=215\hat{i}+45\hat{j}$

Magnitude of the displacement is

$\Rightarrow \left | r_{2} \right |=\sqrt{\left ( 215 \right )^2+\left ( 45 \right )^2}\\\Rightarrow \left | r_{2} \right |=\sqrt{48250}\\\Rightarrow \left | r_{2} \right |=219.65\ \text{km}$

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