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aev [14]
3 years ago
8

Atoms that emit particles and energy from their nuclei are

Physics
1 answer:
saw5 [17]3 years ago
4 0
Atoms that emit particles and energy from their nuclei are (Radioactive).
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Substitution solve for x &amp; y<br>-4x - 2y equals 14<br>-10x+7y=-25​
den301095 [7]

Answer:

-4×-2y=14 (1)

-10×+7y=-25 (2)

multiplying eq 1 by 7 and eq 2 by 2 and add eq. 1 and 2

-28×-14y=98

-20×+14y=-50

___________

-28×=48

×=48/-28

×=-12/7

now

-4×-2y=14

-4*-12/7-2y=14

48/7-2y=14

-2y=14-48/7

-2y=(98-48)/7

-2y=50/7

y=-50/14

y=-25/7

8 0
3 years ago
Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass
Alecsey [184]

Answer:

a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

Explanation:

a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

p₁ = p₂

37.7 kgm/s = 40.3v

v = 37.7/40.3 = 0.94 m/s

So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

7 0
3 years ago
A flutist assembles her flute in a room where the speed of sound is 342m/s . When she plays the note A, it is in perfect tune wi
USPshnik [31]

Answer:

a.3Hz

b.0.0034m

Explanation:

First, we know the flute is an open pipe, because open pipe as both end open and a close organ pipe as only one end close.

The formula relating the length and he frequency is giving as

f=\frac{nv}{2l}\\.

a.we first determine the length of the flute at the fundamental frequency i.e when <em>n</em>=1 and when the speed is in the 342m/s

Hence from

f=\frac{nv}{2l}\\\\l=\frac{342}{2*440}\\ l=0.389m\\.

since the value of the length will remain constant, we now use the value to determine the frequency when the air becomes hotter and the speed becomes 345m/s.

f=\frac{nv}{2l} \\f=\frac{345}{2*0.389}\\f=443.4Hz

Hence the require beat is

B=/f_{1}-f_{2}/\\B=/440-443/\\B=3Hz.

b. since the length is dependent also on the speed and frequency, we determine the new length when she plays with a fundamental frequency when the speed of sound is 345m/s

using the formula

L_{new}=\frac{v}{2f}\\\\L_{new}=\frac{345}{2*440}\\\\L_{new}=0.39204

Now to determine the extension,

L_{extend}=L_{new}-L_{old}\\L_{extend}=0.39204- 0.38864\\L_{extend}=0.0034m\\

4 0
3 years ago
A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr
Musya8 [376]

Answer:

162500000.  

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

I=\dfrac{q}{t}

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

n=\dfrac{260\times 10^{-3}\ t}{1.6\times 10^{-9}}

n=162500000 t

\dfrac{n}{t}=16250000

The number of electron passe per second will be 162500000.

4 0
2 years ago
What is the BEST example of a purebred
ki77a [65]

no se   nada pvodebrhtre

8 0
3 years ago
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