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2 years ago

Why is nuclear energy an important discussion in today's world?

Physics

1 answer:

aksik [14]2 years ago

8 0

Answer:

Nuclear energy is not limited to the generation of electricity, but may equally well be used for such important tasks as desalination, production of hydrogen, space heating and process-heat applications in industry as well as for extraction of carbon from CO2 to combine with hydrogen to create synthetic liquid fuels.

Explanation:

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In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce

noname [10]

Answer:

14.7 m/s

Explanation:

a = acceleration experienced by driver's head = 50 g = 50 x 9.8 m/s² = 490 m/s²

v₀ = initial speed of the driver = 0 m/s

v = final speed of the driver after 30 ms

t = time interval for which the acceleration is experienced = 30 ms = 0.030 s

Using the equation

v = v₀ + a t

Inserting the values

v = 0 + (490) (0.030)

v = 14.7 m/s

6 0

2 years ago

In an isolated system, the total heat given off by warmer substances equals the total heat energy gained by cooler substances. N

galina1969 [7]

Answer:

The temperature of the cooler substance was close to the room temperature. Therefore, the system experienced less change

Explanation:

Generally, in a closed system containing two bodies at different temperatures, there is a flow of heat energy from the body at a higher temperature to the body at a lower temperature. The effect is more significant when there is a large temperature difference between the bodies. However, if the temperature difference is small or insignificant, the change will be less.

3 0

1 year ago

A balloon of mass M is floating motionless in the air. A person of mass less than M is on a rope ladder hanging from the balloon

Alex73 [517]

Answer:

Down with a speed less than v

Explanation:

Let the person's mass be represented by m

The mass of the balloon = M

Total momentum is conserved thus F = external force = 0

when the person starts to climb the ladder, external force is still equal to zero.

We know that change is p = F * change in t

in this equation, F is the external force = 0

Hence change in p = 0

This means that total momentum is conserved thus F = external force = 0

(owing to the fact that exterior forces on the system are balanced)

Since all external force was zero before they began to climb the ladder, as they climb all external forces will still = 0

Thus, As the person starts moving

mv + MV = 0

or mv = MV

V = mv/M

Since M > m, m/M will definitely be a number that is less than 1,

Hence, V = (a number less than 1)v.

This means that V < v, or the balloon moves down a speed V which is less than v.

4 0

2 years ago

A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a

Yanka [14]

Answer:

Diameter of the cylinder will be $d=2.998\times 10^4m$

Explanation:

We have given young's modulus of steel $E=207GPa=207\times 10^9Pa$

Change in length $\Delta l=0.38mm$

Length of rod $l=500mm$

Load F = 11100 KN

Strain is given by $strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}$

We know that young's modulus $E=\frac{stress}{strain}$

So $207\times 10^9=\frac{stress}{7.6\times 10^{-4}}$

$stress=1573.2\times 10^{-5}N/m^2$

We know that stress $=\frac{force}{artea }$

So $1573.2\times 10^{-5}=\frac{11100\times 1000}{area}$

$area=7.055\times 10^{8}m^2$

So $\frac{\pi }{4}d^2=7.055\times 10^{8}$

$d=2.998\times 10^4m$

6 0

2 years ago

An electron in a vacuum chamber is fired with a speed of 9800 km/s toward a large, uniformly charged plate 75 cm away. The elect

melisa1 [442]

Answer:

The plate's surface charge density is $-8.056\times10^{-9}\ C/m^2$

Explanation:

Given that,

Speed = 9800 km/s

Distance d= 75 cm

Distance d' =15 cm

Suppose we determine the plate's surface charge density?

We need to calculate the surface charge density

Using work energy theorem

$W=\Delta K.E$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{i}^2$

Here, final velocity is zero

$W=0-\dfrac{1}{2}mv_{i}^2$ ...(I)

We know that,

$W=-Fd$

$W=-E\times e\times d$

$W=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d$ ...(II)

From equation (I) and (II)

$-\dfrac{1}{2}mv_{i}^2=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d$

Charge is negative for electron

$\lambda=\dfrac{mv^2\epsilon_{0}}{(-e)d}$

Put the value into the formula

$\lambda=-\dfrac{9.1\times10^{-31}\times(9800\times10^{3})^2\times8.85\times10^{-12}}{1.6\times10^{-19}\times(75-15)\times10^{-2}}$

$\lambda=-8.056\times10^{-9}\ C/m^2$

Hence, The plate's surface charge density is $-8.056\times10^{-9}\ C/m^2$

3 0

2 years ago