5. Aunt Jane weighs 145 pounds. What is her weight in Newtons?

Answers

Answer 1

Answer:

i think its 644

Explanation:


Related Questions

If energy is conserved, then:
O A. initial (PE + KE) = final (PE + KE).
O B. the initial KE must be zero.
O C. the momentum does not change.
O D. PE(before) = KE(before).

Answers

Answer is A mark me brainliest

If energy is conserved, then initial (PE + KE) = final (PE + KE). So, the correct option is A.

What is meant by Law of Conservation of Energy ?

The law of conservation of energy states that, energy can neither be created nor be destroyed, but can be transformed from one form to another.

Here,

According to law of conservation of energy,

The total energy of an isolated system remains constant. That means, the total energy of the system in the initial state will be same as that in the final state.

The total mechanical energy is the sum of kinetic energy and potential energy.

TE = KE + PE

Therefore, the energy to be conserved in the system,

Initial TE = Final TE

So, Initial (KE + PE) = Final (KE + PE)

Hence,

If energy is conserved, then initial (PE + KE) = final (PE + KE).

To learn more about law of conservation of energy, click:

https://brainly.com/question/29775341

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What is the net force acting on a 52 kg object that has a velocity of 8.0 m/s and is moving in a circle of radius 1.6 m? a. 4000N b. 20880N c. 2500N d. 3500N

Answers

Σf = m a
Σf = m v^2 / r
Σf = 52 8^2 / 1.6
Σf = 2080 N
The net force acting of the object will be "2080 N".

Given values are:

Mass, m = 52 kgVelocity, v = 8.0 m/sRadius, r = 1.6 m

As we know the formula,

→ [tex]\Sigma f = ma[/tex]

or,

→ [tex]\Sigma f = \frac{mv^2}{r}[/tex]

By putting the values, we get

         [tex]= \frac{52\times 8^2}{1.6}[/tex]

         [tex]= \frac{52\times 64}{1.6}[/tex]

         [tex]= 2080 \ N[/tex]

Thus the response above is appropriate.

Learn more about force here:

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A 1500 kg car is being lifted by a hydraulic jack attached to a flat plate. Underneath the plate is a pipe with radius of 24 cm at the top and 2cm at the bottom. To generate an upward acceleration for the car of 1.0 m/s2, how much force must be applied to the small end of the pipe? The answer is NOT 102 N btw

Answers

Answer:

The force that must be exerted at the other end is 102.08 N/m₂

Explanation:

Using Pascal's formula; P₁ = P₂

P₁ = F₁/A₁ = m*g/(πr²)

m = 1500 kg; g = 9.8 m/s², r = 24 cm = 0.24 m

P₁ = (1500 * 9.8)/(22/7 * 0.24 * 0.24)

P₁ = 81235 N/m²

P₂ = F₂/A₂

A₂ = πr² where π = 22/7; r = 2.00 cm = 0.02m

A₂ = 22/7 * 0.02 * 0.02 = 0.00126 m²

P₂  = F₂ / 0.00126 m₂

substituting in the formula  P₁ = P₂

81235 N/m² =  F₂ / 0.00126 m²

F₂ = 81235 N/m² * 0.00126 m²

F₂ = 102.08 N/m₂

Explanation:

Answer:Why does water stop flowing out of the top hole even before the water level falls ... Calculate pressure on the bottom of the pool due to the water. 3. ... 2. If the ratio of A2 to A1 was 100, what force would be required to lift a 10000 N car? ... A cylinder with a radius of 11 cm and a height of 3.4 cm has a mass of 10.0 kg. a.

Explanation:

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