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3 years ago

Q5: An ice skater moving at 12 m/s coasts

Physics

1 answer:

ExtremeBDS [4]3 years ago

8 0

u = 0.077

Explanation:

Work done by friction is

Wf = ∆KE + ∆PE

-umgx = ∆KE,. ∆PE =0 (level ice surface)

-umgx = KEf - KEi = -(1/2)mv^2

Solving for u,

u = v^2/2gx

= (12 m/s)^2/2(9.8 m/s^2)(95 m)

= 0.077

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A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

5 0

3 years ago

if a car that is 100 feet in front of you on route 70 west slams on the brakes while you are traveling 65 miles per hour (95 fee

fenix001 [56]

Answer:

t = 1.05 s

Explanation:

Given,

The distance between your vehicle and car, 100 ft

The constant speed of your vehicle, u = 95 ft/s

Since, the velocity is constant, a =0

If the car stopped suddenly, time left for you to hit the brake, t = ?

Using the second equation of motion,

S = ut + ½ at²

Substituting the given values in the equation

100 = 95 x t

t = 100/95

= 1.05 s

Hence, the time left for you to hit the brakes and stop before rear ending them, t = 1.05 s

5 0

3 years ago

Round your answers to one decimal place.this parallel circuit has two resistors at 15 and 40 ohms. what is the total resistance?

lubasha [3.4K]

1) The equivalent resistance of two resistors in parallel is given by:
$\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}$
so in our problem we have
$\frac{1}{R_{eq}} = \frac{1}{15 \Omega}+ \frac{1}{40 \Omega}=0.092 \Omega^{-1}$
and the equivalent resistance is
$R_{eq} = \frac{1}{0.092 \Omega^{-1}}=10.9 \Omega$

2) If we have a battery of 12 V connected to the circuit, the current in the circuit will be given by Ohm's law, therefore:
$I= \frac{V}{R_{eq}}= \frac{12 V}{10.9 \Omega}=1.1 A$

4 0

3 years ago

Read 2 more answers

Suppose you increase your walking speed from 4 m/s to 13 m/s in a period of 3 s. What is your acceleration?

iogann1982 [59]

The acceleration formula goes like this: a= (vf-vi)/t so it would be (13-4)/3 Thus the answer is 3m/s^2

7 0

3 years ago

an object is acted on by a drag force with a magnitude that is proportional to the speed. the object accelerates downward at 3.0

Nutka1998 [239]

The terminal speed of the object falling down is 66.67 m/s.

The terminal speed acquired by the body when,

Weight of the body = Drag force of the body

It is given,

Drag force is directly proportional to the speed,

So,

F = CV

Where F is drag force,

V is the speed,

C is the constant,

So, it can be written as C = F/V.

The weight of the body = mg

The weight of the body = 10m

M is the mass and g is the acceleration due to gravity,

The drag force when the speed is 20m/s.

Drag force = ma

a is the acceleration during the drag force which is given to be 3m/s²,

Drag force = 3m

Now we can write,

F₁/V₁ = F₂/V₂

F₁ is the drag force at 20m/s speed.

F₂ is the weight of the body and V₂ is the terminal speed,

Now, it can be written,

3m/20 = 10m/V₂

V₂ = 66.67 m/s.

So, the terminal speed is 66.67m/s.

To know more about terminal speed, visit,

brainly.com/question/14605362

#SPJ4

7 0

1 year ago