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3 years ago

Which type of nuclear waste can be disposed of in a landfill?

Physics

1 answer:

marysya [2.9K]3 years ago

8 0

Low-level nuclear waste can be disposed of in a landfill.

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A particle leaves the origin with a speed of 3.6 times 106 m/s at 34 degrees to the positive x axis. It moves in a uniform elect

gogolik [260]

Answer:

1.2X10^3NC

Explanation:

Pls see attached file

3 0

3 years ago

A square plate of edge length 9.0 cm and negligible thickness has a total charge of 6.90 10-6 C. Estimate the magnitude E of the

SVETLANKA909090 [29]

Answer:

E= 4.35*10^6 N/C

Explanation:

Let's find the area charge density of the plate

α= 6.9*10^-6/9*10^-2 = 7.7*10^-5C/m2

Now we can calculate the electric field just of the plate

E =α/2e =7.7*10^-5/2*8.85*10^-12 = 4.35*10^6 N/C

7 0

3 years ago

Calculate the acceleration of a car if it's velocity increases from 15 m/s to 75m/s in 5 seconds.

Andreas93 [3]

initial velocity=u=15m/s
Final velocity=v=75m/s
Time=t=5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{75-15}{5}$

$\\ \sf\longmapsto Acceleration=\dfrac{60}{5}$

$\\ \sf\longmapsto Acceleration=12m/s^2$

7 0

3 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

4 years ago

I need c only please answer it

Elodia [21]

It will be moving at high speeds

5 0

3 years ago