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2 years ago

5

What is the slowest speed (in mph) that Mr. P can travel and still jump across the canyon? (I have attached a picture) Hint: The

answer is between 100-150 mph.

1 answer:

Alex17521 [72]2 years ago

7 0

might be 140mph, so that is a guess that i just made so plz let me know if im wrong or correct

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Can you access an instance variable from a static method? explain why or why not.

marusya05 [52]

<span>The reason a static method can't access instance variable is because static references the class not a specific instance of the class so there is no instance variable to access.</span>

5 0

3 years ago

(a) Two ions with masses of 4.39×10^−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic fiel

sammy [17]

Answer:

Part a)

$R_1 = 0.072 m$

Part b)

$R_2 = 0.036 m$

Part c)

d = 0.072 m

Explanation:

Part a)

As we know that the radius of the charge particle in constant magnetic field is given by

$R = \frac{mv}{qB}$

now for single ionized we have

$R_1 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(1.6 \times 10^{-19})(0.301)}$

$R_1 = 0.072 m$

Part b)

Similarly for doubly ionized ion we will have the same equation

$R = \frac{mv}{qB}$

$R_2 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(3.2 \times 10^{-19})(0.301)}$

$R_2 = 0.036 m$

Part c)

The distance between the two particles are half of the loop will be given as

$d = 2(R_1 - R_2)$

$d = 2(0.072 - 0.036)$

$d = 0.072 m$

6 0

2 years ago

Answer these questions please and you will get the brain list

Naddika [18.5K]

I am using the equation F=ma (force equals mass times acceleration) to solve these problems.

1. You are looking for force, and have mass and acceleration. You just plug in the values for mass and acceleration to get the force needed.

F=(15kg)(5m/s^2)

F=75N

2. Again, you are looking for force, and just need to plug in the values for mass and acceleration

F=(3kg)(2.4m/s^2)

F=7.2N

3. In this problem, you have force and mass, but need to find acceleration. To do this, you need to get acceleration alone on one side of the equation - divide each side by m. Your equation will now be F/m=a

a=(5N)/(3.7kg)

a=18.5m/s^2

I did not use significant figures. Let me know if you need to do that and need any help on that. Hope this helps!

7 0

2 years ago

Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the sp

kotegsom [21]

Answer:

m2 = 83.3 g

Explanation:

by conservation of momentum principle we have

$m_1v_{i1} + m_2v_{i2} = m_2v_{f2}$

as both sphere has same speed so $v_{i2} = v_{i1}$

$m_2 = \frac{m_1}[\frac{v_f2}{v_{f1}}+1}$

from conservation of kinetic energy principle we have

$\frac{1}{2}m_1v^{2}_{i1} + \frac{1}{2}m_2v^{2}_{i2} = \frac{1}{2}m_2v^{2}_f2$

$v_{f1} = \sqrt {\frac{(m_1+m_2) v^2_i1}{m_2}$

$v_{f1} = v_{i2}\sqrt {\frac{(m_1+m_2)}{m_2}$

$\frac{v_{f1}}{v_{i2}} =\sqrt {\frac{(m_1+m_2)}{m_2}$

substituting this value in above equation to get m2 value

$m_2 = \frac{m_1}{\sqrt {\frac{(m_1+m_2)}{m_2}+1}}$

solving for m2 we get

$m2 = \frac{m_1}{3}$

m_1 = 250 g

$=\frac{250}{3}$

m2 = 83.3 g

7 0

2 years ago

Given w= 4+,- 0.02 A=2.0+,-0.2 and 3.0+,-0.6 what is the value of w\ A y square

In-s [12.5K]

Explanation:

w = (4.52 ± 0.02) cm, x = ( 2.0 ± 0.2) cm, y = (3.0 ± 0.6) cm. Find z = x + y - w and its uncertainty.

z = x + y - w = 2.0 + 3.0 - 4.5 = 0.5 cm

Dz = Dx + Dy + Dw = 0.2 + 0.6 + 0.02 = 0.82 rounding to 0.8 cm

So z = (0.5 ± 0.8) cm

Solution with standard deviations, Eq. 1b, Dz = 0.633 cm

z = (0.5 ± 0.6) cm

Notice that we round the uncertainty to one significant figure and round the answer to match.

8 0

2 years ago