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4 years ago

Change in Energy Quick Check

Physics

1 answer:

nika2105 [10]4 years ago

4 0

Answer:

The force of gravity on the object decreases. (FALSE)

The potential energy of the object decreases. (TRUE)

The acceleration due to gravity decreases. (FALSE)

The kinetic energy of the object decrease (FALSE)

Explanation:

FORCE OF GRAVITY:

Force of gravity on the object is the weight of object, which depends upon the mass and value of acceleration due to gravity (W = mg). Since, the value mass is constant on Earth and acceleration due to gravity is also constant on earth.

Therefore, force of gravity remains same on the object. The statement is false.

POTENTIAL ENERGY:

Potential energy of object depends upon the mass, value of acceleration due to gravity, and change in height of object (P.E = mgΔh). Since, the value mass is constant on Earth and acceleration due to gravity increases as the object moves towards the surface of earth. But, the height of object is decreasing.

Therefore, potential energy of object decreases. The statement is true.

ACCELERATION DUE TO GRAVITY:

Acceleration due to gravity depends upon the altitude (gh = g[1 - 2h/Re]). Since, the height of object is decreasing.

Therefore, acceleration due to gravity increases. The statement is false.

But, this change is not significant.

KINETIC ENERGY:

Kinetic Energy of object, which depends upon the mass and velocity of the object (K.E = mv²/2). Since, the value mass is constant on Earth and velocity increases as the object moves towards the surface of earth.

Therefore, kinetic energy of the object also increases. The statement is false.

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romanna [79]

Answer:

B if I am correct

Explanation:

I am actually studying in my classroom for this. Glad I saw your question.

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4 years ago

I am starting a PowerPoint presentation about Cat-Eye Syndrome. I don't know how to start my sentence. How can I start?

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3 years ago

Gbenga needs to get glasses to correct his farsightedness. His eyes currently cannot focus on objects that are within 2 ft (or 6

Scilla [17]

Answer:

Focal Length = 38.61cm, Power = 2.59 Diopter, Converging lens.

Explanation:

When an object is placed 25cm from Gbenga's eye, the glasses lens must produce an image 61cm away (Gbenga's eye near point).

An image 61cm from the eye will be (61cm - 1.6cm) from the glasses.

i.e. $d_{i}=61cm-1.6cm=59.4cm$

and $d_{o} = 25cm - 1.6cm = 23.4cm$

note $d_{i}$ will be negative because the image is formed on the same side as the object.

finally, $d_{i}=-59.4cm\\d_{o}=23.4cm$

the formula for finding the focal length $f$ is given as

$f=\frac{d_{i}d_{o} }{d_{o}+d_{i} }$

$f=\frac{-59.4*23.4}{23.4-59.4} \\$

$f=\frac{-1389.96}{-36}$

$f=38.61cm$

The focal length is positive which indicates converging lens

power $p=\frac{1}{f}$

but $f\\$ must be in metres

Therefore, $f=38.61cm=0.3861m$

$p=\frac{1}{0.3861}$

$p=2.59 Diopter$

3 0

4 years ago

A 1.10 μF capacitor is connected in series with a 1.92 μF capacitor. The 1.10 μF capacitor carries a charge of +10.1 μC on one p

miss Akunina [59]

Answer:

(a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

Explanation:

Given that,

Charge = 10.1 μC

Capacitor C₁ = 1.10 μF

Capacitor C₂ = 1.92 μF

Capacitor C₃ = 1.10 μF

Potential V₁ = 51.5 V

Let V₁ and V₂ be the potentials on the two plates of the capacitor.

(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor

Using formula of potential difference

$V_{1}=\dfrac{Q}{C_{1}}$

Put the value into the formula

$V_{1}=\dfrac{10.1 \times10^{-6}}{1.10\times10^{-6}}$

$V_{1}=9.18\ V$

The potential on the second plate

$V_{2}=V-V_{1}$

$V_{2}=51.5 -9.18$

$V_{2}=42.32\ v$

(b). We need to calculate the equivalent capacitance of the two capacitors

Using formula of equivalent capacitance

$C=\dfrac{C_{1}\timesC_{2}}{C_{1}+C_{2}}$

Put the value into the formula

$C=\dfrac{1.10\times10^{-6}\times1.92\times10^{-6}}{(1.10+1.92)\times10^{-6}}$

$C=6.99\times10^{-7}\ F$

$C=0.69\ \mu F$

Hence, (a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

7 0

3 years ago

Please help, very confused!

PtichkaEL [24]

D. the last choice because the info above tells u so

4 0

3 years ago

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