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3 years ago

What happens to gravity when someone jumps up?

Physics

1 answer:

Alla [95]3 years ago

5 0

Answer:

The direct answer to the question as written is as follows: nothing happens to gravity when someone jumps up - gravity continues exerting a force on the body of that particular someone proportional to (mass of someone) x (mass of Earth) / (distance squared). What you might be asking, however, is what is the net force acting on the body of someone jumping up. At the moment of someone jumping up there is an upward acceleration, i.e., an upward-directed force which counteracts the gravitational force - this is the net force ( a result of the jump force minus gravity). From that moment on, only gravity acts on the body. The someone moves upward gradually decelerating to the downward gravitational acceleration until they reaches the peak of the jump (zero velocity). Then, back to Earth.

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We now have an algebraic expression with only one variable, which can be solved. Once we have that, we can plug it back into one

Len [333]

Answer:

A=1

B=-2

Explanation:

Part A and B of the question wasn't given, however, I attached the relevant parts to solve this question as follows.

From part B as attached, it shows that the right option is C which is

2A+3B=-4

Substituting B with 3A-5 then we form the second equation as shown

2A+3(3A-5)=-4

By simplifying the above equation, we obtain

2A+9A-15=-4

Re-arranging, then

11A=-4+15

Finally

11A=11

A=1

To obtain B, we already know that 3A-5 so substituting the value of A into the above then we obtain

B=3(1)-5=-2

Therefore, required values are 1 and -2

3 0

3 years ago

A 0.683 kg mass moves in SHM at the end of a spring. It takes 1.41 s to move from the position with the spring fully extended to

dsp73

Answer:

Spring constant of the spring will be equal to 9.255 N /m

Explanation:

We have given mass m = 0.683 kg

Time taken to complete one oscillation is given T = 1.41 sec

We have to find the spring constant of the spring

From spring mass system time period is equal to $T=2\pi \sqrt{\frac{m}{K}}$ , here m is mass and K is spring constant

So $1.41=2\times 3.14\sqrt{\frac{0.683}{K}}$

$0.2245=\sqrt{\frac{0.683}{K}}$

Squaring both side

$0.0504=\frac{0.4664}{K}$

$K=9.255N/m$

So spring constant of the spring will be equal to 9.255 N /m

7 0

3 years ago

11) In which state of matter has the LEAST kinetic energy?

klio [65]

Solid particles hope this helped!

5 0

2 years ago

A) An automobile light has a 1.0-A current when it is connected to a 12-V battery. Determine the resistance of the light.

kirill [66]

Answer:

The resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

Explanation:

Given:

(A)

Current $I = 1$ A

Voltage $V = 12$ V

For finding the resistance,

$V = IR$

$R = \frac{V}{I}$

$R = \frac{12}{1}$

$R =$ 12Ω

(B)

For finding power delivered,

$P = I^{2} R$

$P = (1) ^{2} \times 12$

$P = 12$ Watt

(C)

For finding the potential difference,

$V = IR$

$V = 5 \times 10^{-3} \times 2$

$V = 10 \times 10^{-3}$

$V = 0.01$ V

Therefore, the resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

4 0

3 years ago

A very small sphere with positive charge 5.00uC is released from rest at a point 1.20cm from a very long line of uniform linear

JulijaS [17]

Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval (−L,0] for some large number L. The voltage V as a function of x on the interval (0,∞) is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element dx′, we have λ=dqdx′.V(x)=1/4πϵ0∫linedq/r=λ/4πϵ0∫−L0dx/x−x′=λ/4πϵ0(ln|x+L|−ln|x|)

5 0

3 years ago