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3 years ago

g What is the separation between their paths in meters when they hit a target after traversing a semicircle

Physics

1 answer:

oee [108]3 years ago

8 0

Answer:

vm/B•q

Explanation:

The charge to mass ratio of any given particle, can be determined by using the formula

q/m = E/BB•R,

From the aforelisted formula,

Both E and B are the fields in the velocity selector.

B• is the magnetic field used to separate particles

R is the radius of the path of the particle.

Now, we rearrange by making R the subject of the formula, so that

R = Em/BB•q

In trying to find the velocity of the particles coming out of the velocity selector, we realize that

v = E/B, and when we substitute for this in the previous equation, we have

R = vm/B•q

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The square plates of a 3000-pF parallel-plate capacitor measure 40 mm by 40 mm and are separated by a dielectric that is 0.29 mm

Travka [436]

Answer:

The square plates of a 3000-pF parallel-plate capacitor measure 40 mm by 40 mm and are separated by a dielectric that is 0.29 mm thick and completely fills teh region between the plates.

Explanation:

5 0

2 years ago

Hi guys. Help with any of these questions! Pressure in physics year 7 :) <br> Thanks!

SVEN [57.7K]

Answer:

2 a) it is less dense than the water

2 b) it is more dense than the water

3 a ping pong ball is hollow and less dense than the water so it quickly bounces up to the surface of the water

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3 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the

Shalnov [3]

Heat required to raise the temperature of a given system is

$Q = ms\Delta T$

here we know that

m = mass

s = specific heat capacity

$\Delta T$ = change in temperature

now as we know that

mass of wood = 5 kg

mass of aluminium pan = 2 kg

change in temperature = 45 - 20 = 25 degree C

specific heat capacity of wood = 1700 J/kg C

specific heat capacity of aluminium = 900 J/kg C

now here we will find the total heat to raise the temperature of both

$Q = m_1s_1\Delta T_1 + m_2s_2\Delta T_2$

$Q = 5 * 1700 * 25 + 2 * 900 * 25$

$Q = 212500 + 45000$

$Q = 257500 J$

So heat required to raise the temperature of the system is 257500 J

4 0

3 years ago

Given the scope of Kayla’s education, training, and years of experience as a CMA, would this “favor” fall within the AAMA guidel

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Based on the information given, it can be inferred that the favor doesn't fall within the AAMA guidelines of her responsibilities.

From the information given, it should be noted that the guidelines of CMA as stipulated under the American Association of Medical Assistant prohibits the CMA from interpreting the medical data of the patient. Therefore, the favor that was asked by Dr. Hsu of Kayla is simply against the guidelines.

Even though the favor that was asked by Dr. Hsu was prohibited by AAMA, it should be noted that the final part of the favor about faxing the report to the internist would fall within AAMA guidelines.

In conclusion, the best way that Kayla can respond to Dr. Hsu is to decline doing the favor.

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6 0

2 years ago