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3 years ago

13

What is the magnetic force (in newtons) on a particle traveling in a 1.5 T magnetic field if q = 7.5 microcoulombs and v = 1.75

× 106 m/s at a 45° angle to the magnetic field? Answer with three signficant digits.

1 answer:

Alla [95]3 years ago

3 0

Given:

B(Magnetic field): 1.5 T

q= 7.5 microcoulombs

v= 1.75 x 10 ∧6 m/s

The angle ∅ between B and v is 45 °.

Now we know that F= qvB sin ∅

Substituting these values we get:

F= 7.5 x 10∧-6 x 1.75 x 10∧6 x 1.5 x sin 45

F= 16.752 N

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3 years ago

Read 2 more answers

a crowbar having the lenght of 1.75 is used to balance a load of 500N if the distance between the fulcrum and the load is 0.5m f

Lemur [1.5K]

The effort applied = 200 N

<h3>Further explanation</h3>

Equilibrium :

$\tt F_1.d_1=F_2.d_2$

Total length = 1.75

The distance between the fulcrum and the load is 0.5 m ⇒ d₁=0.5 m

The distance between the fulcrum and the force applied :

$\tt 1.75-0.5=1.25~m\rightarrow d_2=1.25~m$

A load of 500N ⇒ F₁=500 N

The force applied :

$\tt 500\times 0.5=F_2\times 1.25\\\\F_2=\dfrac{500\times 0.5}{1.25}=200~N$

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3 years ago

A 0.106-A current is charging a capacitor that has square plates 4.60 cm on each side. The plate separation is 4.00 mm.

nikdorinn [45]

Answer:

a

$\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

b

$I = 0.106 \ A$

Explanation:

From the question we are told that

The current is $I = 0.106 \ A$

The length of one side of the square $a = 4.60 \ cm = 0.046 \ m$

The separation between the plate is $d = 4.0 mm = 0.004 \ m$

Generally electric flux is mathematically represented as

$\phi_E = \frac{Q}{\epsilon_o}$

differentiating both sides with respect to t is

$\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} * \frac{d Q}{ dt}$

=> $\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} *I$

Here $\epsilon_o$ is the permitivity of free space with value

$\epsilon _o = 8.85*10^{-12} C/(V \cdot m)$

=> $\frac{d \phi_{E}}{dt} = \frac{0.106}{8.85*10^{-12}}$

=> $\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

Generally the displacement current between the plates in A

$I = 8.85*10^{-12} * 1.1977 *10^{10}$

=> $I = 0.106 \ A$

3 0

3 years ago

A 7.0kg skydiver is descending with a constant velocity

Vikentia [17]

Answer:

The air resistance on the skydiver is 68.6 N

Explanation:

When the skydiver is falling down, there are two forces acting on him:

- The force of gravity, of magnitude $mg$ , in the downward direction (where m is the mass of the skydiver and g is the acceleration due to gravity)

- The air resistance, $R$ , in the upward direction

So the net force on the skydiver is:

$F=mg-R$

where

m = 7.0 kg is the mass

$g=9.8 m/s^2$

According to Newton's second law of motion, the net force on a body is equal to the product between its mass and its acceleration (a):

$F=ma$

In this problem, however, the skydiver is moving with constant velocity, so his acceleration is zero:

$a=0$

Therefore the net force is zero:

$F=0$

And so, we have:

$mg-R=0$

And so we can find the magnitude of the air resistance, which is equal to the force of gravity:

$R=mg=(7.0)(9.8)=68.6 N$

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4 years ago