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s344n2d4d5 [400]
3 years ago
11

A circular loop of wire with radius 10.0 cm is located in the xy-plane in a region of uniform magnetic field. A field of 2 T is

directed in the z-direction, which is upward. (a) What is the magnetic flux through the loop
Physics
1 answer:
kati45 [8]3 years ago
4 0

Answer:

\phi=628.3

Explanation:

From the question we are told that

Radius r=10.0 cm

Magnetic fieldB=2T

Generally the equation for area of circular path is mathematically given by

 Area=\pi r^2

 A=\pi 10^2

 A=314.15m^2

Generally the equation for Magnetic flux is mathematically given by

 \phi=BA

 \phi=2*314.15

 \phi=628.3

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Un libro del peso di 12 N è in equilibrio su un tavolo. Sapendo che il coefficiente di attrito statico vale 0,5, la forza di att
Tanzania [10]

Answer:

60

Explanation:

Translation -

A book weighing 12 N is balanced on a table. Knowing that the static friction coefficient is 0.5, how much is the friction force worth?

Friction force is

f = u * n

f = 0.5 * 12N

f = 60

4 0
4 years ago
Julie is cycling at a speed of 3.4 meters/second. If the combined mass of the bicycle and Julie is 30 kilograms, what is the kin
nexus9112 [7]

Answer:

A

Explanation:

KE = 1/2 mv^2

=1/2(30kg)( 3.4 m/s)^2

=173.4 joules

=1.7×10^2 joules

6 0
3 years ago
Please help me with questions 1, 2 and 3. <br> i need a step by step explanation
kifflom [539]

Answer:

1) d

2) 5 m/s

3) 100

Explanation:

The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:

(i) x=\frac{1}{2}at^2+v_0t+x_0

The equation for velocity v and a constant acceleration a is:

(ii) v=at+v_0

1) Solve equation (ii) for acceleration a and plug the result in equation (i)

(iii) a = \frac{v -v_0}{t}

(iv) x = \frac{v-v_0}{2t}t^2+v_0t + x_0

Simplify equation (iv) and use the given values v = 0, x₀ = 0:

(v) x=-\frac{v_0}{2}t + v_0t= \frac{v_0}{2}t

2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v:v=at+v_0=0.2\frac{m}{s^2} * 10s+3\frac{m}{s}=2\frac{m}{s}+3\frac{m}{s}=5\frac{m}{s}

3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}

5 0
3 years ago
What is the value of acceleration due to gravity at the pole, equator and the centre of the earth
fgiga [73]

Answer:

In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator.

7 0
2 years ago
A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 222 kg and it
inn [45]

Answer:

The buoyant force is 3778.8 N in upward.

Explanation:

Given that,

Mass of balloon = 222 Kg

Volume = 328 m³

Density of air = 1.20 kg/m³

Density of helium = 0.179 kg/m³

We need to calculate the buoyant force acting

Using formula of buoyant force

F_{b}=\rho_{air}\times V_{b}\times g

Where, \rho_{air} = density of air

V = Volume of balloon

g = acceleration due to gravity

Put the value into the formula

F_{b}=1.20\times321\times9.81

F_{b}=3778.8\ N

This buoyant force is in upward direction.

Hence, The buoyant force is 3778.8 N in upward.

4 0
3 years ago
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