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3 years ago

A change in kinetic of an object is equal to the ___

Physics

1 answer:

podryga [215]3 years ago

4 0

Answer:C..net work done on the object.

Explanation:

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(a) The Sun orbits the Milky Way galaxy once each 2.60 x 108y , with a roughly circular orbit averaging 3.00 x 104 light years i

PilotLPTM [1.2K]

Answer:

Part a)

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

$v = 2.18 \times 10^5 m/s$

Explanation:

Time period of sun is given as

$T = 2.60 \times 10^8 years$

$T = 2.60 \times 10^8 (365 \times 24 \times 3600) s$

$T = 8.2 \times 10^{15} s$

Now the radius of the orbit of sun is given as

$R = 3.00 \times 10^4 Ly$

$R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m$

$R = 2.84 \times 10^20 m$

Part a)

centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = \frac{4\pi^2}{T^2} R$

$a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})$

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

orbital speed is given as

$v = \frac{2\pi R}{T}$

$v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}$

$v = 2.18 \times 10^5 m/s$

5 0

4 years ago

Aluminum has a density of

agasfer [191]

<h3><u>Volume is 0.1848 m³</u></h3><h3 />

Explanation:

<h2>Given:</h2>

m = 49.9 kg

ρ = 270 kg/m³

<h2>Required:</h2>

volume

<h2>Equation:</h2>

$ρ \:= \:\frac{m}{v}$

where: ρ - density

m - mass

v - volume

<h2>Solution:</h2>

Substitute the value of ρ and m

$ρ \:= \:\frac{m}{v}$

$270\: kg/m³\:= \:\frac{49.9\:kg}{v}$

$(v)\:270\: kg/m³\:= \:49.9\:kg$

$v\:= \:\frac{49.9\:kg}{270\: kg/m³}$

$v\:= \:0.1848\:m³$

<h2>Final Answer:</h2><h3><u>Volume is 0.1848 m³</u></h3>

3 0

3 years ago

Read 2 more answers

The perception of an image first, followed by noticing individual pieces of the

juin [17]

Answer:

The answer is Top-Down processing

Explanation:

I had this question on a apex quiz and i got it correct.

7 0

3 years ago

What is 7 to the power of 5

LekaFEV [45]

7⁵ = 7 × 7× 7× 7× 7

= 16807

5 0

3 years ago

Read 2 more answers

A Foucault pendulum consists of a brass sphere with a diameter of 32.0 cm suspended from a steel cable 10.0 m long (both measure

Sergio039 [100]

Answer:

T = 44.35 °C

Explanation:

d = 32cm

R = 16 cm

Lsteel = 10m

T1 = 20° C

Space = 0.3cm

The space between the sphere and the floor is represented by δL(total) after the temperature increases.

As the temperature increases, both will expand.

So,

0.3 x 10^(-2) = δL(steel) + δR(brass)

= {L(o) x α(steel) x δT} + {R(o) x {α(brass) x δT}

= {10 x 1.2 x 10^(-5) x (T-20)} + {0.16 x 2 x 10^(-5) x (T-20)}

= 12.32 x 10^(-5) x(T-20)

Therefore (T-20) = (0.3 x 10^(-2)) / {12.32 x 10^(-5)}

T = 20 + 24. 35 = 44.35 °C

8 0

3 years ago