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3 years ago

A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find

Physics

1 answer:

pshichka [43]3 years ago

4 0

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0=4\pi \cdot 10^{-7} H/m$ is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

$I=1.41 A$ (current in the wire)

$B=5.61\mu T=5.61\cdot 10^{-6} T$ (strength of magnetic field)

Solving for r, we find the distance from the wire:

$r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m$

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You lift a 10-kg box to a height of 1m. How much work do you do on the box when you lift it from the ground? (g= 9.81 m/s2)

Andreyy89

Answer:

98.1 Joule

Explanation:

Solution,

⇒Mass(m)=10kg

⇒Weight(F)=Mg=10×9.81=98.1N

⇒Distance(d)=1m

Now,

Work done=F×d

=98.1×1

=98.1J

5 0

1 year ago

3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between

Natali5045456 [20]

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

$v^2 -u^2 = 2ad$ (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

$F=ma=-\mu mg$

where

m = 1000 kg is the mass of the car

$\mu = 0.80$ is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

$a=-\mu g$

And we can substitute it into eq.(1) to find d:

$v^2 -u^2 = 2(\-mu g) d\\d= \frac{v^2-u^2}{-2 \mu g}=\frac{0-(24 m/s)^2}{-2(0.80)(9.8 m/s^2)}=36.7 m \sim 37 m$

7 0

3 years ago

A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.

charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

7 0

2 years ago

A baseball thrown by a pitcher is hit by a batter. At the moment when the ball hits the bat, the force exerted on the bat by the

Slav-nsk [51]

Newton's Laws of motion describe the motion of an object based on the applied force

The correct option that gives the relationships between the forces is the the option;

$\underline{F_{ball}}$ on bat = $\underline{F_{bat}}$ on ball

(Either option A or B without the minus symbol before $F_{bat}$ , likely typographical error)

Reason:

Force exerted on the bat by the ball = $F_{ball}$

Force exerted on the ball by the bat = $F_{bat}$

Given that the batter hits the ball with the bat, the force exerted by the bat on the ball, $F_{bat}$ , is reacted to by the force of the ball acting on the bat, $F_{ball}$

According to Newton's third law of motion, action and reaction are equal and opposite. Therefore, at the moment when the ball hits the bat, we have;

$\underline{F_{ball}}$ on bat = $\underline{F_{bat}}$ on ball

Where the force of the bat is high, the ball is accelerated to travel at high speed

Learn more Newton's Laws of motion here:

brainly.com/question/24522313

8 0

2 years ago

(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro

Kipish [7]

Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:

$v = \frac{m + M}{m} \cdot \sqrt{2gh}$

where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum.

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:

$\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}$ (1)

where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively.

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}$

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}$

$\frac{v_{2}}{v_{1}} = 1.41$

Therefore, the second projectile was 1.41 times faster than the first.

I hope it helps you!

8 0

2 years ago