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anzhelika [568]
2 years ago
12

a football player starts from rest and speeds up to a final velocity of 12 m/s. the speeding up takes a total of 6 seconds. what

is the football players acceleration
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
8 0

Answer:

2 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 12 m/s

Time (t) = 6 s

Acceleration (a) =?

The acceleration of the player can be obtained as follow:

v = u + at

12 = 0 + (a × 6)

12 = 6a

Divide both side by 6

a = 12 / 6

a = 2 m/s²

Thus, the acceleration of the player is 2 m/s²

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Romashka-Z-Leto [24]

Answer:

If the canoe heads upstream the speed is zero. And directly across the river is  8.48 [km/h] towards southeast

Explanation:

When the canoe moves upstream, it is moving in the opposite direction of the normal river current. Since the velocities are vector (magnitude and direction) we can sum each vector:

Vr = velocity of the river = 6[km/h}

Vc = velocity of the canoe = -6 [km/h]

We take the direction of the river as positive, therefore other velocity in the opposite direction will be negative.

Vt = Vr + Vc = 6 - 6 = 0 [km/h]

For the second question, we need to make a sketch of the canoe and we are watching this movement at a high elevation. So let's say that the canoe is located in point 0 where it is located one of the river's borders.

So we are having one movement to the right (x-direction). And the movement of the river to the south ( - y-direction).

Since the velocities are vector we can sum each vector, so using the Pythagoras theorem we have:

Vt = \sqrt{(6)^{2} +(-6)^{2} } \\Vt=8.48[km/h]

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3 years ago
PLEASE HELP TIMED
Igoryamba

Radio waves have longer wavelengths and lower frequencies than microwaves.

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3 years ago
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A uniform magnetic field of magnitude 0.23 T is directed perpendicular to the plane of a rectangular loop having dimensions 7.8
Arlecino [84]

Answer:

0.0025116weber/m²

Explanation:

Magnetic field density (B) is the ratio of the magnetic flux (¶) through the loop to its cross sectional area (A).

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Dimension of the rectangular loop = 7.8 cm by 14 cm

Area of the rectangular loop perpendicular to the field B = 7.8cm×14cm

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Converting this value to m²

Area of the loop = 109.2 × 10^-4

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Answer:

Cruising at 35,000 feet in an airliner, straight toward the east,

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