3. A crane can lift a 5000 N object to height of 30 m in 120 seconds. The power at which the crane is

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst

Archy [21]

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

$E = \frac{1}{2}mv^{2}$

So, $v = \sqrt{\frac{2E}{m}}$ .... (1)

Now,

$r = \frac{mv}{Bq}$

Substitute the value of v from equation (1), we get

$r = \frac{\sqrt{2mE}}{Bq}$

Let the radius of the alpha particle is r2.

For proton

So, $r_{1} = \frac{\sqrt{2m_{1}E}}{Bq_{1}}$ ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, $r_{2} = \frac{\sqrt{2m_{2}E}}{Bq_{2}}$ ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

$\frac{r_{1}}{r_{2}}=\frac{q_{2}}{q_{1}}\times \sqrt{\frac{m_{1}}{m_{2}}}$

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

$\frac{r_{1}}{r_{2}}=\frac{2q}}{q}}\times \sqrt{\frac{m}}{4m}}=1$

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

8 0

3 years ago

Read 2 more answers

A basketball player standing under the hoop shoots the ball straight up with an initial velocity of v0Part (a) What is the maxim

Mama L [17]

Answer:

Maximum height will be $h=\sqrt{\frac{v_0^2}{2g}}$

Explanation:

We have given initial velocity through which basketball is thrown $=v_0$

Acceleration due to gravity $=g\ m/sec^2$

At maximum height velocity will be zero

So final velocity v = 0 m /sec

According to third equation of motion $v^2=u^2+2gh$

$0^2=v_0^2-2gh$ ( Negative sign is due to upward acceleration )

$h=\sqrt{\frac{v_0^2}{2g}}$

6 0

3 years ago

The_______ of an object is given relative to an origin

Setler79 [48]

The slope of an object is given relative to an origin.

8 0

3 years ago

A 6.0-cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge p

Rufina [12.5K]

Answer:

0.0072 m³/s

Explanation:

Using Bernoulli's law

P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂ since the pipe is horizontal

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

flow rate is constant

A₁v₁ = A₂v₂

A₁ = πr₁² = π (0.06/2)² = 0.0028278 m²

A₂ = πr₂² = π (0.0225)² = 0.00159 m²

v₁ = (A₂ / A₁)v₂

v₁ = (0.00159 m²/ 0.0028278 m²) v₂ = 0.562 v₂

substitute v₁ into the Bernoulli's equation

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

500 ( 1 - 0.3161 ) v₂² = (31.0 - 24 ) × 10³ Pa

341.924 v₂² = 7000

v₂² = 20.472

v₂ = √ 20.472 = 4.525 m/s

volume follow rate = 0.00159 m² × 4.525 m/s = 0.0072 m³/s

8 0

3 years ago

Speed and Motion you went from the starting line to the finish line at different rates. If you repeated the activity while carry

Phantasy [73]

Answer:

It will cause kinetic energy to increase.

Explanation:

Given that Speed and Motion you went from the starting line to the finish line at different rates.

If you repeated the activity while carrying weights but keeping your times the same, the weight carried will add up to the mass of the body.

And since Kinetic energy K.E = 1/2mv^2

Increase in the mass of the body will definitely make the kinetic energy of the body to increase.

Since the time is the same, that means the speed V is the same.

Weight W = mg

m = W/g

The new kinetic energy will be:

K.E = 1/2(M + m)v^2

This means that there will be increase in kinetic energy.

3 0

3 years ago