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2 years ago

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Physics

2 answers:

Alex17521 [72]2 years ago

8 0

(8.3 min)/(1 AU) = (T)/(1.52 AU)

(8.3 min)x(1.52 AU) = (T x 1 AU)

T = (8.3 min x 1.52 AU) / (1 AU)

T = 12.62 minutes

solmaris [256]2 years ago

3 0

Answer:

d=1.49×1011m

Explanation:

Velocity is defined as the rate of travel, and can be found using the distance formula.

velocity=distancetime

Rearranging this formula we can solve for distance given velocity and time of travel.

d=vt

We are given velocity and time, and so can solve for distance, but if we plug in the values given;

d=(3.00×108m/s)(8.3minutes)

We can see that the units do not match up. Since seconds are the SI unit for time, we will need to convert 8.3 minutes to seconds.

t=(8.3minutes)(60seconds/minute)=(498s)

Now our units work out and we can solve for distance.

= 15.85

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What is the speed of a bobsled whose distance-time graph indicates that it traveled 119m in 29s?

Assoli18 [71]

Do 112m /29s which it will be 3.862 which if you round it, it will be 3.86 m/s

7 0

3 years ago

Read 2 more answers

A 250–g piece of gold is at 19 °C. 5.192 kJ of energy is added to it by heat. The specific heat of gold is 129 J/(kg·°C). Calcul

algol [13]

Answer:

A. DT is given by Q= MCs DT

m = mass of the substances

Cs= is it's specific heat capacity

Ck= Q

Mk ×DTk

=250 × 9 × 5

129

=Dt = 180.1085271

answer is 180degree C.

Explanation:

B. = 25×10 ×100

1.082

=2500

1.082

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2 years ago

The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an

9966 [12]

Answer:

$-4.941*10^8J.$

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

$\Delta U = \Delta_{perogee}-\Delta_{Apogee}$

Where,

$U= \frac{-GmM_e}{r}$

Replacing

$\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}$

$\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})$

Our values are given by,

$m = 85.5Kg$

$M_e = 5.97*10^{24}Kg$

$r_A = 7330Km$

$r_p = 6610Km$

$G = 6.67*10^{-11}Nm^2/Kg^2$

Replacing at the equation,

$\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})$

$\Delta U = -4.941*10^8J$

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was $-4.941*10^8J.$

4 0

2 years ago

A horizontal spring with spring constant 200N/m is compressed by 15cm and used to launch a 2kg box across a frictionless horizon

kobusy [5.1K]

Answer:

Explanation:

Given that,

Spring constant k=200N/m

Compression x = 15cm = 0.15m

Attached mass m =2kg

Coefficient of kinetic friction uk= 0.2

The energy in the spring is given as

U =½kx²

U = ½ × 200 × 0.15²

U = 2.25J

Force in the spring is given by Hooke's law

F = ke

F = 200×0.15

F = 30N

The weight of body which is equal to the normal is give as

W = mg

W = 2 × 9.81

W = 19.62N

W = N = 19.62 Newton's 2nd Law

From law of friction,

Fr = uk•N

Fr = 0.2 × 19.62

Fr = 3.924

Using newton second law again

Fnet = F - Fr

Fnet = 30 - 3.924

Fnet = 26.076

Work done by net force is given as

W = Fnet × d

W = 26.076d

Then, the work done by this net force is equal to the energy in the spring

W = U

26.076d = 2.25

d = 2.25/26.076

d = 0.0863m

Which is 8.63cm

So the box will slide 8.63cm before stopping

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