А A van accelerates from Amst to zomst in 8s. How far does it<br>travel in<br>this time?

A 2-column table with 5 rows. The first column titled metal has entries aluminum, cork, iron, lead, wax. The second column title

astraxan [27]

Answer:Cork and wax

Explanation:

9 0

2 years ago

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The core of a certain reflected reactor consist of a cylinder 10 ft high 10 ft in diameter The measured maximum-to-average flux

Westkost [7]

Answer:

The maximum power density in the reactor is 37.562 KW/L.

Explanation:

Given that,

Height = 10 ft = 3.048 m

Diameter = 10 ft = 3.048 m

Flux = 1.5

Power = 835 MW

We need to calculate the volume of cylinder

Using formula of volume

$V =\pi r^2 h$

Put the value into the formula

$V=\pi\times(1.524)^2\times 3.048$

$V= 22.23\m^3$

$V = 22.23\times10^{3}\ Liter$

We need to calculate the maximum power density in the reactor

Using formula of power density

$P=\dfrac{E}{V}$

Where, P = power density

E = energy

V = volume

Put the value into the formula

$P=\dfrac{835\times10^{6}}{22.23\times10^{3}}$

$P=37561.85 = 37.562\times KW/L$

Hence, The maximum power density in the reactor is 37.562 KW/L.

6 0

3 years ago

A 0.75μF capacitor is charged to 70 V . It is then connected in series with a 55Ω resistor and a 140 Ω resistor and allowed to d

Ipatiy [6.2K]

Answer:

Explanation:

Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.

Energy dissipates in 55Ω resistor is given by V²/R

Since the 55ohms and 140ohms l discharge the capacitor fully, the voltage will be zero volts and this voltage will be shared by the resistor in ratio.

So for 55ohms, using voltage divider rule

V=R1/(R1+R2) ×Vt

V=55/(55+140) ×70

V=19.74Volts is across the 55ohms resistor.

Then, energy loss will be

E=V²/R

E=19.74²/55

E=7.09J

7.09J of heat is dissipated by the 55ohms resistor

6 0

3 years ago

Qual a capacidade térmica de um objeto que, ao receber 10000 cal de energia, tem sua temperatura elevada de 25°C para 75°C?

Stolb23 [73]

Answer:

$200 cal/^{\circ}C$

Explanation:

When heat energy is supplied to an object, the temperature of the object increases according to the equation:

$Q=C\Delta T$

where

Q is the heat supplied

C is the heat capacity of the object

$\Delta T$ is the change in temperature

In this problem we have:

$Q=10,000 cal$ is the energy supplied

$\Delta T=75C-25C=50C$ is the change in temperature of the object

Therefore, the heat capacity of the object is:

$C=\frac{Q}{\Delta T}=\frac{10,000}{50}=200 cal/^{\circ}C$

6 0

2 years ago

Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0

2 years ago

А A van accelerates from Amst to zomst in 8s. How far does ittravel inthis time?​

А A van accelerates from Amst to zomst in 8s. How far does ittravel inthis time?