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hodyreva [135]
3 years ago
7

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

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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)

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= (3.835  - 4.032 ) x 10⁴ k

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= - 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
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