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larisa [96]
3 years ago
7

The current through a wire that is 0.80 m long is 5.0 A. The wire is at right angles to a 0.60 T magnetic field. What is the mag

nitude of the force on the wire?
0.24 N

2.4 N

24 N

240 N
Physics
1 answer:
Oksanka [162]3 years ago
3 0

Answer: B

F = 2.4 N

Explanation: Given that

Magnetic field B = 0.60 T

Current I = 5 A

Length L = 0.8 m

The force on an individual charge moving at the drift velocity Vd is given by F = qVdB sin θ.

Since current = charge/time

I = q/t

q = It

Also, recall that V = L/t. Therefore

F = I × t × L/t × B × sinØ

F = ILBsin θ

Since the wire is at right angles to the magnetic field

Sin θ = sin 90 = 1 then,

F = 5 × 0.8 × 0.6 = 2.4N

The force on the wire is 2.4 N

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The position-time equation for a cheetah chasing an antelope is:
pochemuha

Answer:

x = 1.6 + 1.7 t^2      omitting signs

a) at t = 0     x = 1.6 m

b) V = d x / d t = 3.4 t

at t = 0     V = 0

c) A = d^2 x / d t^2 = 3.4     (at t = 0  A = 3.4 m/s^2)

d)  x = 1.6 + 1.7 * (4.4)^2 = 34.5    (position at 4.4 sec = 34.5 m)

4 0
2 years ago
Which of the following is a scientific law?
Varvara68 [4.7K]

Answer:

B. For a gas in a closed container at a constant temperature, the product of the pressure and the volume remains constant.

Explanation:

The rest are societal laws, as they are telling you something you should avoid or follow.

Hope this helps :)

7 0
3 years ago
Read 2 more answers
Two objects gravitationally attract with a force of 100 N. If the mass of one object is doubled and the mass of the other object
barxatty [35]

Hello!

Recall the equation for gravitational force:
F_g = \frac{Gm_1m_2}{r^2}

Fg = Force of gravity (N)
G = Gravitational constant

m1, m2 = masses of objects (kg)
r = distance between the objects' center of masses (m)

There is a DIRECT relationship between mass and gravitational force.

We are given:
F_g = 100N

If we were to double one mass and triple another, according to the equation:
F'_g = \frac{G(2m_1)(3m_2)}{r^2} = 6(\frac{G(m_1)(m_2)}{r^2}) = 6F_g

Thus:
6 * F_g = 6 * 100 = \boxed{600N}

5 0
2 years ago
Stored energy in an object due to its position is
Temka [501]
Is called potential energy
3 0
3 years ago
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Honeybees acquire a charge while flying due to friction with the air. A 120 mg bee with a charge of + 23 pC experiences an elect
Alex Ar [27]

Answer:

  • \frac{F}{W}=1.95\times10^{-6}
  • 51304447 \frac{N}{C}
  • Upward

Explanation:

The weight of the bee is:

W=mg=(120\times10^{-6}kg)(9.81\frac{m}{s^{2}})=1.18\times10^{-3}N

with m the mass and g the gravity acceleration.

Electric force of the bee is related with the electric field of earth by:

F_{e}=qE=(23\times10^{-12}C)(100\frac{N}{C})=2.3\times10^{-9}

with q the charge, E the electric field and Fe the electric force.

So:

\frac{F}{W}=\frac{2.3\times10^{-9}}{1.18\times10^{-3}}=1.95\times10^{-6}

Because Newton's first law we should make the net force on it equals cero:

F+F_e+W=0

F=-(F_e+W)=-(2.3\times10^{-9} +1.18\times10^{-3})=-1.1800023\times10^{-3}

with W the weight, Fe the electric force on the bee due earth's electric field and F the force.

So, the applied electric field should be:

E_a=\frac{-1.1800023\times10^{-3}}{23\times10^{-12}}=-51304447 \frac{N}{C}

The negative sign indicates that the electric field should be opposite to earth's electric field, so it should be upward.

7 0
3 years ago
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