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2 years ago

What are the products of a fusion reaction? Check all that apply.

Physics

1 answer:

Free_Kalibri [48]2 years ago

8 0

Energy, heavier atoms atoms, neutron and a proton are the products of a fusion reaction.

<h3>What are the products of a fusion reaction?</h3>

Nuclear fusion is a type of reaction in which two or more atomic nuclei fuse or combine to form heavy nuclei and subatomic particles such as neutrons or protons and release of energy.

So we can conclude that Energy, heavier atoms, neutron and a proton are the products of a fusion reaction.

Learn more about reaction here: brainly.com/question/11231920

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Hiw many times lager then a centimeter is a dekagram

Digiron [165]

Answer: 1 000 times

Explanation: 1 000 centimetre is equal to 1 dekacemetre.

Using conversion will apply in converting every number.

6 0

3 years ago

An airplane flies with a constant speed of 720 km/hr. How long will it take to travel a distance of 1440 km? Please show your wo

choli [55]

Answer:

2 h

Explanation:

Velocity =Distance/time ⇒ time = distance/speed

= 1440/720

= 2 h

6 0

3 years ago

6kg of human blood at a temperature of 65degees celcius is mixed with 4kg of human blood ata temperature of 20 degrees celcius .

Gennadij [26K]

Answer:

$T_f=47^{\circ}$

Explanation: Two samples of blood that have different masses and temperatures and are mixed, we have to find the final temperature of the mixture. the final temperature can be found using the following formula:

$T_f=\frac{(m_1\cdot T_1+m_2T_2)}{(m_1+m_2)}\Rightarrow(1)$

(1) Formula basically tells us that the product of mass and temperature remains constant throughout, so the addition of two products of the two separate blood samples would be equal to the product of final temperature and the total mass of the mixture. Mathematically this means that:

$\mleft(m_1+m_2\mright)T_f=(m_1\cdot T_1)+(m_2T_2)$

Using (1) and plugging in the corresponding values, we get the answer as follows:

$\begin{gathered} T_f=\frac{(m_1\cdot T_1+m_2T_2)}{(m_1+m_2)}\Rightarrow(1) \\ m_1=6\operatorname{kg} \\ m_2=4\operatorname{kg} \\ T_1=65^{\circ} \\ T_2=20^{\circ} \\ \therefore\rightarrow \\ T_f=\frac{(6kg\cdot65^{\circ}+4\operatorname{kg}\cdot20^{\circ})}{(6kg+4\operatorname{kg})}=\frac{(390+80)}{10}=\frac{470}{10}=47^{\circ} \\ \therefore\rightarrow \\ T_f=47^{\circ} \end{gathered}$

4 0

2 years ago

For fully developed laminar pipe flow in a circular pipe, the velocity profile is u(r) = 2(1-r2 /R2 ) in m/s, where R is the inn

sdas [7]

Answer:

a) $v_{max} = 2\ \textup{m/s}$

b) $v_{avg} = 1\ \textup{m/s}$

c) Q = 1.256 × 10⁻³ m³/s

Explanation:

Given:

The velocity profile as:

$u(r) = 2(1-\frac{r^2}{R^2} )$

Now, the maximum velocity of the flow is obtained at the center of the pipe

i.e r = 0

thus,

$v_{max}=u(0) = 2(1-\frac{0^2}{R^2} )$

$v_{max} = 2\ \textup{m/s}$

Now,

$v_{avg} = \frac{v_{max}}{2}\ \textup{m/s}$

$v_{avg} = \frac{2}}{2}\ \textup{m/s}$

$v_{avg} = 1\ \textup{m/s}$

Now, the flow rate is given as:

Q = Area of cross-section of pipe × $v_{avg}$

Q = $\frac{\pi D^2}{4}\times v_{avg}$

Q = $\frac{\pi 0.04^2}{4}\times 1$

Q = 1.256 × 10⁻³ m³/s

7 0

3 years ago

A car is initially moving at 35 km/h along a straight highway. To pass another car, it speeds up to 135 km/h in 10.5 seconds at

Aleks04 [339]

Acceleration = (velocity final-velocity initial)/ time
where
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
= 37.5 m/s
velocity initial = 35 km/hr x 1hr /3600 s x 1000 m/1 km
= 9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units = (2.646m/s^2)/(9.8m/s^2)
= 0.27 units

6 0

4 years ago