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3 years ago

Student Exploration: Nuclear Decay. Has anyone done a Gizmos lab on this?

Physics

1 answer:

LUCKY_DIMON [66]3 years ago

6 0

Answer:NOOPE

Explanation:IM THE MYSTERY MAN WHOOOOSHHHHHHHHHHHHHHH ???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

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Identify one type of electromagnetic energy that has a shorter wave length than green light

11111nata11111 [884]

When you look at the acronym ROY G. BIV, you get the wavelengths from longest to shortest. Since Green is in the middle of ROY G. BIV, anything to the right of it would have a shorter wavelength. You can choose in the visible range Blue, Indigo, or Violet.

5 0

4 years ago

Two people, one with mass m1 and the other with mass m2, stand on a stationary sled with mass M on a frozen lake. Assume that th

lozanna [386]

Answer:

Part a)

Velocity of sled

$v = \frac{m_1 s}{m_1 + m_2 + M}$

velocity of first man who jump off

$v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}$

Part b)

Velocity of sled

$v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s$

Also the speed of second person is given as

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}$

Part c)

change in kinetic energy of sled + two people is given as

$KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

Explanation:

As we know that here we we consider both people + sled as a system then there is no external force on it

So here we can use momentum conservation

since both people + sled is at rest initially so initial total momentum is zero

now when first people will jump with relative velocity "s" then let say the sled + other people will move off with speed v

so by momentum conservation we have

$0 = m_1(v - s) + (m_2 + M)v$

$v = \frac{m_1 s}{m_1 + m_2 + M}$

so velocity of the sled + other person is

$v = \frac{m_1 s}{m_1 + m_2 + M}$

velocity of first man who jump off

$v_1 = \frac{m_1 s}{m_1 + m_2 + M} - s$

$v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}$

Part b)

now when other man also jump off with same relative velocity

so let say the sled is now moving with speed vf

so by momentum conservation we have

$(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) = m_2(v_f - s) + Mv_f$

$(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) + m_2s = (m_2 + M)v_f$

Now we have

$v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s$

Also the speed of second person is given as

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s - s$

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}$

Part c)

change in kinetic energy of sled + two people is given as

$KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

here we know all values of speed as we found it in part a) and part b)

4 0

3 years ago

How many neutrons are in calcium

vladimir1956 [14]

Answer:

I believe there are 20 Neutrons in Calcium.

Explanation:

6 0

4 years ago

Touching your finger to your nose, action force?

blsea [12.9K]

Yes i think since action force means moving in one direction

5 0

4 years ago

To stop a car, first you require a certain reaction time to begin braking; then the car slows under the constant braking deceler

kirza4 [7]

Answer:

a.) 2.66 seconds for 79.2 km/h and 1.82 seconds for 47.8 km/h

b.)-4.13 m/s2 for 79.2 km/h and -3.63 m/s2 for 47.8 km/h

Explanation:

Now for (a)

time = velocity/distance

for velocity = 79.2 km/h and distance 0.0586 km

t = (0.0586/79.2)*3600

t = 2.66 seconds (Please note that multiplication with 3600 is to convert hours into seconds)

for velocity 47.8 km/h and distance 0.0242 km

t = (0.0242/47.8)*3600

t = 1.82 seconds

Now for (b)

2as = vf2-vi2

so,

2a(58.6) = -(79.2*1000/3600)^2 (Please note that multiplication and division with 1000 and 3600 respectively is to convert speed unit from km/h to m/s)

a = -4.13 m/s2 for 58.6 m

and

2a(24.2) = -(47.8*1000/3600)^2

a = -3.63 m/s2 for 24.2 m.

Please note that "-" sign express the deceleration.

5 0

3 years ago