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3 years ago

What year was soccer added to the Olympic games ? a - 1996 b - 1904 c - 1900

Physics

2 answers:

vredina [299]3 years ago

5 0

Answer:

a - 1996

Explanation:

Thepotemich [5.8K]3 years ago

5 0

Answer:

1900

Explanation:

It's football but for the U.S terms, they call it Soccer, and it was in 1900

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Please can someone solve this question

Lorico [155]

Answer:both diagrams are attached.

Explanation:

3 0

2 years ago

A 10-ohm resistor has a constant current. If 1200 C of charge flow through it in 4 minutes what

Amanda [17]

Answer:

B 5.0 A .

Explanation:

Hello.

In this case, since we know the charge (1200 C), time (4 min =240 s) and resistance (10Ω) which is actually not needed here, we compute the current as follows:

$I=\frac{Q}{t}$

Then, for the given data, we obtain:

$I=\frac{1200C}{4min}*\frac{1min}{60s}\\\\I=5A$

Therefore, answer is B 5.0 A .

Best regards!

4 0

3 years ago

A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences

NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, $q=3\times 10^{-6}\ C$

Mass of charge particle, $m=2\times 10^{-6}\ C$

Speed, $v=5\times 10^{6}\ m/s$

Acceleration, $a=3\times 10^{4}\ m/s^2$

We need to find the minimum magnetic field that would produce such an acceleration. So,

$ma=qvB\ sin\theta$

For minimum magnetic field,

$ma=qvB$

$B=\dfrac{ma}{qv}$

$B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}$

B = 0.004 T

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0

3 years ago

A 20-turn coil of area 0.32 m2 is placed in a uniform magnetic field of 0.055 T so that the perpendicular to the plane of the co

makvit [3.9K]

Answer:

1.5 * 10^-2 Tm^2

Explanation:

Electric Flux = B.A cos(theta)

B = 0.055 T

A = 0.32 m^2

theta = 30

Electric Flux = (0.055 T).(0.32 m^2).Cos(30) = 0.0152 = 1.5 * 10^-2 Tm^2

5 0

3 years ago

Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the

ale4655 [162]

I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)

Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.

The 2.0-kg block feels

• the downward pull of its own weight, (2.0 kg) g

• the upward normal force of the surface, magnitude n₁

• kinetic friction, mag. f₁ = 0.30n₁, pointing in the negative horizontal direction

• the contact force of the larger block, mag. c₁, also pointing in the negative horizontal direction

• the applied force, mag. F, pointing in the positive horizontal direction

Meanwhile the 3.0-kg block feels

• its own weight, (3.0 kg) g, pointing downward

• normal force, mag. n₂, pointing upward

• kinetic friction, mag. f₂ = 0.30n₂, pointing in the negative horizontal direction

• contact force from the smaller block, mag. c₂, pointing in the positive horizontal direction (this is the force that is causing the larger block to move)

Notice the contact forces form an action-reaction pair, so that c₁ = c₂, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:

• net vertical force:

n₂ - (3.0 kg) g = 0 ==> n₂ = (3.0 kg) g ==> f₂ = 0.30 (3.0 kg) g

• net horizontal force:

c₂ - f₂ = 0 ==> c₂ = 0.30 (3.0 kg) g ≈ 8.8 N

4 0

3 years ago