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Whitepunk [10]
3 years ago
12

When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose anot

her bomb is tested, and 1.5 grams of matter are converted into energy. How many joules of energy are released by the explosion?
Physics
1 answer:
gayaneshka [121]3 years ago
7 0

Answer:

The value is E =  1.35 *10^{14} \ J

Explanation:

From the question we are told that

    The mass of matter converted to energy on first test is  m  =  1 \  g  = 0.001 \  kg

    The mass of matter converted to energy on second test m_1 =  1.5 \  g = 1.5 *10^{-3} \ kg

    Generally the amount of energy that was released by  the explosion is  mathematically  represented as  

         E =  m * c^2

=>       E =  1.5 *10^{-3}  * [ 3.0 *10^{8}]^2

=>       E =  1.35 *10^{14} \ J

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Can someone give me an example or explanation of calculating energy from voltage? Many thanks!
antoniya [11.8K]
Expression to calculate energy from voltage: E= V*Q where E= energy, V= voltage, and Q= charge

Additional help:
-To find the Voltage ( V )
[ V = I x R ] V (volts) = I (amps) x R (Ω)

-To find the Current ( I )
[ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)

-To find the Resistance ( R )
[ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)

I hope that helps to some extent-
7 0
2 years ago
An ice skater has a moment of inertia of 5.0 kgm2 when her arms are outstretched. at this time she is spinning at 3.0 revolution
givi [52]
With arms outstretched,
Moment of inertia is I = 5.0 kg-m².
Rotational speed is ω = (3 rev/s)*(2π rad/rev) = 6π rad/s
The torque required is
T = Iω = (5.0 kg-m²)*(6π rad/s) = 30π 

Assume that the same torque drives the rotational motion at a moment of inertia of 2.0 kg-m².
If u = new rotational speed (rad/s), then
T = 2u = 30π
u = 15π rad/s
   = (15π rad/s)*(1 rev/2π rad)
   = 7.5 rev/s

Answer: 7.5  revolutions per second.

7 0
3 years ago
1. A car is stopped at a red light. When the light turns green, the car starts
Lubov Fominskaja [6]

Answer:

Because of inertia (Newton's First Law of motion)

Explanation:

According to Newton's First Law of motion:

"An object at rest (or in motion at constant velocity) will tend to stay at rest (or tend to keep moving with same velocity) unless acted upon an unbalanced force"

In this problem, the object we are analyzing is the coffee cup.

At the beginning, the cup is at rest, together with the car.

Later, the car starts moving when the light turns green.

If we apply Newton's First Law of motion to the cup, we see that the coffee cup tends to keep its state of rest: for this reason, as the car moves forward, the coffee in the cup will spill backward, into the rear seat. This property of an object to mantain its state of motion is also called as inertia.

8 0
3 years ago
Institutions that anticipate and recognize significant external forces and modify their strategies and operations will prosper a
vesna_86 [32]

Answer:True

Explanation: External factors are factors outside of an organisation or an institution that has the capacity of either adversely or positively affect the institution or the Organisation.

For an institution to prosper and perform optimally in an economy.or a country, such Institution must put into cognisance the possible external threats that are capable of affecting it, when Organisations or Institution put strategies in place to control or mitigate such externalities,the institution or Organisation will sure Prosper.

5 0
3 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
2 years ago
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