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kozerog [31]
3 years ago
8

Can someone explain to me #4.

Physics
1 answer:
nalin [4]3 years ago
4 0
A) We balance the masses: 4(1.00728) vs 4.0015 + 2(0.00055)4.02912 vs. 4.0026This shows a "reduced mass" of 4.02912 - 4.0026 = 0.02652 amu. This is also equivalent to 0.02652/6.02E23 = 4.41E-26 g = 4.41E-29 kg.
b) Using E = mc^2, where c is the speed of light, multiplying 4.41E-29 kg by (3E8 m/s)^2 gives 3.96E-12 J of energy.
c) Since in the original equation, there is only 1 helium atom, we multiply the energy result in b) by 9.21E19 to get 3.65E8 J of energy, or 365 MJ of energy.
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A baseball player friend of yours wants to determine his pitching speed. you have him stand on a ledge and throw the ball horizo
zhenek [66]

Answer:

The pitching speed of your friend is 33.20 m/s

Explanation:

<em>Lets explain how to solve the problem</em>

Your friend throw the ball horizontally that means the vertical initial

component of velocity is zero (u_{y}=0).

The ball is thrown from a height 4 meters above the ground.

The height h=u_{y}t+\frac{1}{2}gt^{2}

<u><em>Remember:</em></u> the height is negative value because its below the point of

thrown (initial position)

h = -4 m , u_{y}=0 and g = -9.8 m/s²(downward)

<em>Substitute these values in the rule above</em>

⇒ 4=0-\frac{1}{2}(9.8)t^{2}

⇒ -4 = -4.9t² (multiply both sides by -1)

⇒ 4 = 4.9t² (divide both sides by 4.9)

⇒ 0.81633 = t² (take √ for both sides)

⇒ <em>t = 0.9035</em>

Then the time of the ball to land on the ground is 0.9035 seconds

The range of the ball on the ground is 30 m

The range R=u_{x}*t, where u_{x} is the horizontal

component of the initial velocity

R = 30 meters and t = 0.9035

⇒ 30=u_{x}(0.9035) (divide both sides by 0.9035)

⇒ u_{x}=33.20 m/s

<em>The pitching speed of your friend is 33.20 m/s </em>

4 0
3 years ago
The specific heat capacity of sea water is 4100 J/Kg°C and the boiling point of 100.6 °C. (i) Calculate the energy required to r
maw [93]

Answer:

334.314 (kJ)

Explanation:

1) the formula for the required energy is: Q=c*m(Bp-t), where c - 4100 J/kg*C; m - 0.9 kg; Bp - 100.6 C; t - 10 C.

2) according to the formula above:

Q=4100*0.9*(100.6-10)=41*9*906=334314 (J).

6 0
3 years ago
Which of the following calculations is equal to 1?
lutik1710 [3]

Answer:

we need to see the answers but probably 1 or -1

5 0
3 years ago
The newly formed xenon nucleus is left in an excited state. Thus, when it decays to a state of lower energy a gamma ray is emitt
nevsk [136]

Answer:3.87*10^-4

Explanation:

What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca

We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.

C=f*lambda

3*10^8=f*3.44*10^-12

F=0.87*10^20 hz

Then with the frequency, find the energy emitted using equation

E=hf E = freq*Plank's constant

E=.87*10^20*6.62*10^-34

E=575.94*10^(-16)

With this energy, convert into MeV from joules.

With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.

Plugging and computing all necessary numbers gives you

3.87*10^-4 u.

6 0
3 years ago
a car headed north at 15 m/s accelerates for 4.25 s to reach a velocity of 28.3 m/s. What is the acceleration of the car?
Liono4ka [1.6K]

<u>Answer:</u>

The acceleration of the car is 3.13 m/s^2

<u>Explanation:</u>

In the question it is given that car initially heads north with a velocity 15 m/s. It then accelerates for 4.25 s and in the end its velocity is 28.3 m/s.

initial velocity u = 15 m/s

time t=4.25 s

final velocity v=28.3 m/s

The equation of acceleration is

a= \frac{(v-u)}{t}

= \frac{(28.3-15)}{4.25} =  \frac {13.3}{4.25} =3.13m/s^2

The value of acceleration is positive, here since the car is speeding up. If it was slowing down the value of acceleration would be negative.

7 0
3 years ago
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