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3 years ago

If it takes 0.5 seconds to travel 5 Millimeters? What's the mph?

Physics

1 answer:

Angelina_Jolie [31]3 years ago

3 0

I think it would me 5/0.5 = 10mph

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An object is dropped onto the moon (gm = 5 ft/s2). How long does it take to fall from an elevation of 250 ft.?

anyanavicka [17]

Answer:

10 seconds

Explanation:

x = x₀ + v₀ t + ½ at²

250 = 0 + (0) t + ½ (5) t²

250 = 2.5 t²

t² = 100

t = 10

It takes 10 seconds to land from a height of 250 ft.

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3 years ago

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What effect does the sun have on the earth. written response

maw [93]

It had a gravitational force that acts so that it pulls Earth in an orbital, which explains temperature variations throughout the years.

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3 years ago

The United States, England, and Russia were members of the:

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Allied powers
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3 years ago

a ship travels a port p and travels 30 km due north. then it changes course and travels 20 km in a direction 30° east of north

liq [111]

When we represent what is given to us on a coordinate plane, we have a figure as shown in the attachment.

To find the distance between P and R, we have to find the Net Displacement of the ship (brown arrow in the figure).

For that, we use the rules for Vector addition.

We see that the first displacement $D_{1}$ = 30 km (blue arrow) is along the y-axis, but the second part of the ship's journey $D_{2}$ = 20 km (red arrow) is at an angle with reference to y-axis.

So, we first find the components of the red arrow along X and Y.

Component of $D_{2}$ along X-axis is given by $D_{2x} = D_{2} Sin 30$ = 10 km

Component of $D_{2}$ along Y-axis is given by $D_{2y} = D_{2} Cos 30$ = 17.32 km

We now add all the vectors along X and along Y separately.

Net Displacement along X $D_{netX} = 10 km$

Net Displacement along Y $D_{netY} = 30 + 17.32$ = 47.32 km

Now that we have the components of the net displacement along X and Y, we make use of Pythagorean Theorem to calculate the $D_{net}$

$D_{net} = \sqrt{D_{netX} ^{2} + D_{netY} ^{2}}$

Therefore, [tex]D_{net} = 48.37 km.

Hence, the distance between the ports P and R is 48 km.

6 0

3 years ago

An electron moves with velocity v⃗ =(5.9i−6.4j)×104m/s in a magnetic field B⃗ =(−0.63i+0.65j)T. Determine the z-component of the

HACTEHA [7]

Answer:

Explanation:

Force on the electron = q ( v x B )

q = - 1.6 x 10⁻¹⁹

v = (5.9i−6.4j)×10⁴

B = (−0.63i+0.65j)

v x B = (5.9i−6.4j)×10⁴ x (−0.63i+0.65j)

= (3.835 - 4.032 ) x 10⁴ k

= - 1970 k

Force on the electron = q ( v x B )

= - 1.6 x 10⁻¹⁹ x -1970 k

= 3.152 x 10⁻¹⁶ k

z-component of the force on the electron

Fz = 3.152 x 10⁻¹⁶ N

7 0

2 years ago