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Trava [24]
2 years ago
6

What is a vector quantity?​

Physics
1 answer:
Nat2105 [25]2 years ago
7 0

Answer:

A quantity that has magnitude and direction. It's usually represented by an arrow whose direction is the same direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude

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. A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is orient
Dmitriy789 [7]

Answer:

The magnetic flux through surface is 2.22 \times 10^{-3} Wb

Explanation:

Given :

Magnitude of magnetic field B = 0.078 T

Radius of circle r = 0.10 m

Angle between field and surface normal \theta = 25°

From the formula of flux,

\phi = B.A

\phi = BA\cos \theta

Where \theta = angle between magnetic field line and surface normal, A = area of circular surface.

A = \pi r^{2}

A = 3.14 \times (0.10) ^{2}

A = 0.0314 m^{2}

Magnetic flux is given by,

\phi = 0.078 \times 0.0314 \times \cos 25

\phi = 2.22 \times 10^{-3} Wb

Therefore, the magnetic flux through surface is 2.22 \times 10^{-3} Wb

6 0
3 years ago
You serve a volleyball with a mass of 1.4 kg. The ball leaves with a speed of 13 m/s. Calculate KE
NemiM [27]

Answer:

118.3 J

Explanation:

Givens:

m = 1.4 kg

V = 13 m/s

Formula for kinetic energy:

KE = (1/2)*(m)*(v)^2

KE = .5*(1.4 kg)*(13 m/s)^2

KE 118.3 J

J = Joules

7 0
3 years ago
What is the most effective means of establishing awareness of hazards in commercial, industrial, and storage facilities with lar
coldgirl [10]

Answer:

C: Contacting the facilities.

7 0
3 years ago
(Astronomy)
olya-2409 [2.1K]

Answer:c ITs C ^w^

Explanation:

4 0
2 years ago
An archer shoots an arrow with a mass of 45.0 grams from bow pulled
Sladkaya [172]

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

3 0
2 years ago
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