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scZoUnD [109]
3 years ago
5

4. As a 500 N woman sits on the floor, the floor exerts a force on her of

Physics
1 answer:
monitta3 years ago
7 0

Answer: 1000

Explanation:

Its double her weight

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When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose anot
Softa [21]

Answer:

1.08 * 10^{14} J

Explanation:

Energy and mass are related by the famous equation developed by Albert Einstein:

E = mc^2

where m = mass and c = speed of light

This equation explains that an object with very small mass can produce a large amount of energy in reactions such as a nuclear reaction.

Hence, the energy produced by the explosion of a Plutonium bomb containing 3.6 grams of matter is:

E = 3.6 * 10^{-3} * (3 * 10^8)^2

E = 1.08 * 10^{14} J

3 0
3 years ago
Why is random sampling and random assignment in experimental research important?
Hunter-Best [27]

So the result is not biased or affected in some way

6 0
3 years ago
The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca
d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}

the elastic potencial energy is given by:

U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J

The work is defined as:

W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s

3 0
3 years ago
A 1.120 kg car is traveling with a speed of 40 m/s. find its energy
Aleonysh [2.5K]

Answer:

896 kJ

Explanation:

KInetic Energy = 1/2 m v^2

                         = 1/2 (1120)(40^2) = 896 000 J    or  896 kJ

4 0
2 years ago
What is the number of electrons that move past a point in a wire carrying 500 A of current in 4.0 minutes
mr Goodwill [35]
The current is defined as the amount of charge Q that passes through a given point of a wire in a time \Delta t:
I= \frac{Q}{\Delta t}
Since I=500 A and the time interval is
\Delta t=4.0 min=240 s
the charge is
Q=I \Delta t=(500 A)(240 s)=1.2 \cdot 10^5 C

One electron has a charge of q=1.6 \cdot 10^{-19}C, therefore the number of electrons that pass a point in the wire during 4 minutes is
N= \frac{Q}{q}= \frac{1.2 \cdot 10^5 C}{1.6 \cdot 10^{-19}C}=7.5 \cdot 10^{23} electrons
3 0
3 years ago
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