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3 years ago

A Savior In Physics Please Help In This Question!!!!!!!!!!!!!!!!

Physics

1 answer:

Ipatiy [6.2K]3 years ago

5 0

Hello there!

To find the speed when you have the kinetic energy and the mass, you just start with the kinetic energy formula then solve for speed.

Kinetic energy = $\frac{1}{2} mv^2$

V = $\sqrt{ \frac{2*K.E{} }m$

$V = \sqrt{ \frac{2*1.09*10 ^{-3} }{9.34*10 ^{-2} } }$

V = $\frac{0.00218}{0.0934}$

V = 0.02334

Thus, the speed is 0.02 m/s

I hope this helps!

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Which pair of quantities includes one quantity that increases as the other decreases during simple harmonic motion?

11111nata11111 [884]

Kinetic energy and potential energy pair is the quantity in which one will increase then other will decrease

As we know that sum of kinetic energy and potential energy will always remain conserved

So here we will have

$KE + PE = constant$

so here as we move away from mean position the kinetic energy will decrease while at the same time potential energy will increase.

So the pair of potential energy and kinetic energy will satisfy the above condition

7 0

3 years ago

Read 2 more answers

2. Two cars A and B are heading forward. The velocity of A and B are 30 m/s and 20 m/s

KengaRu [80]

Answer:

A.) 27000 kgm/s

18000 kgm/s

B.) Va = 22 m/s

C.) 19800 kgm/s

25200 kgm/s

Explanation: Given that the velocity of A and B are 30 m/s and 20 m/s. And of the same mass M = 9 × 10^5g

M = 9×10^5/1000 = 900 kg

A.) Initial momentum of A

Mu = 900 × 30 = 27000 kgm/s

Initial momentum of B

Mu = 900 × 20 = 18000 kgm/s

B.) if they have an accident and then the velocity of the B is 28 m/s, find out velocity of A.

Momentum before impact = momentum after impact

Given that Vb = 28 m/s

27000 + 18000 = 900Va + 900 × 28

45000 = 900Va + 25200

900Va = 45000 - 25200

900Va = 19800

Va = 19800/900

Va = 22 m/s

C.) Momentum of A after impact

MV = 900 × 22 = 19800 kgm/s

Momentum of B after impact

MV = 900 × 28 = 25200 kgm/s

8 0

3 years ago

Which of the following is a valuable part of the scientific method?

yulyashka [42]

Trial and error

scientific laws and theories are proven by experimental data and large bodies of evidence.

7 0

3 years ago

What is the energy released in this B- nuclear reaction 2K-> 2Ca0,e? (The atomic mass of 42 K is 41.962403 u and that of 42Ca

Katyanochek1 [597]

<u>Answer:</u> The energy released in the given nuclear reaction is 3.526 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

$_{19}^{42}\textrm{K}\rightarrow _{20}^{42}\textrm{Ca}+_{-1}^{0}\textrm{e}$

We are given:

Mass of $_{19}^{42}\textrm{K}$ = 41.962403 u

Mass of $_{20}^{42}\textrm{Ca}$ = 41.958618 u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(41.962403-41.958618)=0.003785u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.003785u)\times c^2$

$E=(0.003785u)\times (931.5MeV)$ (Conversion factor: $1u=931.5MeV/c^2$ )

$E=3.526MeV$

Hence, the energy released in the given nuclear reaction is 3.526 MeV.

7 0

3 years ago

Consider a sound wave moving through the air modeled with the equation s(x, t) = 5.00 nm cos(60.00 m−1x − 18.00 ✕ 103 s−1t). Wha

GaryK [48]

Answer:

Shortest time = 58.18 × 10^(-6) s

Explanation:

We are given;

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t))

Let us set x = 0 as origin.

Now, for us to find the time difference, we need to solve 2 equations which are;

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t1))

And

s(x,t) = 5.00 nm cos((60.00 m^(−1)x) − (18.00 X 10³ s^(−1)t2))

Now, since the wave starts from maxima at time at t = 0, the required time would be the difference (t2 - t1)

Thus, the solutions are;

t1 = (1/(18 × 10³)) cos^(-1) (2.5/5)

And

t2 = (1/(18 × 10³)) cos^(-1) (-2.5/5)

Angle of the cos function is in radians, thus;

t1 = 58.18 × 10^(-6) s

t2 = 116.36 × 10^(-6) s

So,

Required time = t2 - t1 = (116.36 × 10^(-6) s) - (58.18 × 10^(-6) s) = 58.18 × 10^(-6) s

4 0

3 years ago