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3 years ago

What would happen to Earth if it stopped in its orbit around the sun?

Physics

2 answers:

erica [24]3 years ago

6 0

It would crash into the sun

Sloan [31]3 years ago

4 0

Answer:

it would cit would crash into the sun

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A bicyclist, initially at rest, begins pedaling and gaining speed steadily for 4.90s during which she covers 25.0m.

emmasim [6.3K]

The bicyclist accelerates with magnitude a such that

25.0 m = 1/2 a (4.90 s)²

Solve for a :

a = (25.0 m) / (1/2 (4.90 s)²) ≈ 2.08 m/s²

Then her final speed is v such that

v ² - 0² = 2a (25.0 m)

Solve for v :

v = √(2 (2.08 m/s²) / (25.0 m)) ≈ 10.2 m/s

Convert to mph. If you know that 1 m ≈ 3.28 ft, then

(10.2 m/s) • (3.28 ft/m) • (1/5280 mi/ft) • (3600 s/h) ≈ 22.8 mi/h

8 0

2 years ago

If the force applied to an object is not greater than the starting friction, what will happen to the object?

bonufazy [111]

Answer:

Explanation:

the object will not move as the force exerted is not sufficient enough to overcome its force of friction

3 0

3 years ago

Read 2 more answers

A horse has an acceleration of 2 m/s2. If it starts from rest, how fast is it going after 1.7 seconds?

fgiga [73]

(2 m/s²) · (1.7 s) = 3.4 m/s

6 0

3 years ago

Read 2 more answers

Help again please !!!

rusak2 [61]

D I think .... don’t be mad if I’m wrong

5 0

3 years ago

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

$\rho_m = 846lb/ft^3$

$g = 32.17405ft/s^2$

$h_1 = 1in = \frac{1}{12} ft$

For the air the defined properties would be

$\rho_a = 0.0075lb/ft^3$

$g = 32.17405ft/s^2$

$h_2 = ?$

We have for equilibrium that

$\text{Pressure change in Air}=\text{Pressure change in Mercury}$

$\rho_m g h_1 = \rho_a g h_2$

Replacing,

$(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)$

Rearranging to find $h_2$

$h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}$

$h = 9400ft$

Therefore the elevation of the mountain top is 9400ft

7 0

3 years ago