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3 years ago

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A block is sliding down an inclined plane that makes an angle of 65o with the horizontal. If the coefficient of kinetic friction

between the block and plane is 0.17, what is the magnitude of the block’s acceleration down the plane?

1 answer:

Nezavi [6.7K]3 years ago

4 0

Answer:

8.2 m/s²

Explanation:

m = mass of the block

μ = Coefficient of kinetic friction = 0.17

$F_{n}$ = Normal force on the block by the ramp

$f_{k}$ = kinetic frictional force

Force equation perpendicular to ramp surface is given as

$F_{n} = mg Cos65$

Kinetic frictional force is given as

$f_{k} = \mu F_{n}$

$f_{k} = \mu mg Cos65$

Force equation parallel to ramp surface is given as

$mg Sin65 - f_{k} = ma$

$mg Sin65 - \mu mg Cos65 = ma$

$g Sin65 - \mu g Cos65 = a$

$(9.8) Sin65 - (0.17) (9.8) Cos65 = a$

$a = 8.2$ m/s²

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In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

sergij07 [2.7K]

Answer:

A) $\Phi=83.84\times 10^{-9}$

B) $\Phi=0 Wb$

C) $emf=5.4090\times 10^{-4}V$

Explanation:

Given that:

no. of turns i the coil, $n=200$

area of the coil, $a=13.1 \times 10^{-4}\,m^2$

time interval of rotation, $t=3.1\times 10^{-2}\,s$

intensity of magnetic field, $B=6.4\times 10^{-5}\,T$

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

$\Phi=B.a\,cos \theta$ ..................................(1)

$\theta$ is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}$

$\Phi=83.84\times 10^{-9} Wb$

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴ $\theta=90^{\circ}$

From eq. (1)

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}$

$\Phi=0 Wb$

(C)

According to the Faraday's Law we have:

$emf=n\frac{B.a}{t}$

$emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}$

$emf=5.4090\times 10^{-4}V$

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3 years ago

An object is placed 50.0 cm in front of a convex mirror. where can be the image located if the focal length is 40 cm from the mi

sergiy2304 [10]

Answer:

Image will form at distance 22.22 cm behind the mirror

Explanation:

As we know that the mirror formula is given as

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

now we know that

object distance from mirror is

$d_o = -50 cm$

Focal length of the mirror is given as

$f = 40 cm$

now we have

$\frac{1}{-50} + \frac{1}{d_i} = \frac{1}{40}$

$\frac{1}{d_i} = \frac{1}{50} + \frac{1}{40}$

$d_i = 22.22 cm$

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2 years ago

It was a children versus grown-ups competition at school. One event required the adult to throw a basketball as far as he could.

aleksklad [387]

Based on Newton's Law, the event is not fair because:

<span>No, because a basketball is bigger than a baseball, and objects that are bigger accelerate slower.

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3 years ago

Read 2 more answers

There are 2.2 pounds in a kilogram. if a boys mass is 40kg, what is his height in pounds?

sasho [114]

The simplest way to do this is to set up equivalent fractions, like this-

$\frac{1}{2.2}$ = $\frac{40}{x}$

Solve for x by using cross multiplication.

40*2.2= 88
1*x=88
x=88

Therefore, the boy weighs 88lbs.

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2 years ago

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10) A soccer player kicks a soccer ball (m = 0.42 kg) accelerating from rest to 32.5m/s in 0.21s. Determine the force that sends

MaRussiya [10]

Explanation:

(10) Mass of a soccer player, m = 0.42 kg

Initial speed, u = 0

Final speed, v = 32.5 m/s

Time, t = 0.21 s

We need to find the force that sends soccer ball towards the goal.

Force, F = ma

$F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.42 \times (32.5-0)}{0.21}\\\\F=65\ N$

So, 65 N of force soccer ball sends towards the goal.

(11) Mass of the satellite, m = 72,000 kg

Initial speed, u = 0 m/s

Final speed, v = 0.63 m/s

Time, t = 1296 s

We need to find the force is exerted by the rocket on the satellite.

Force, F = ma

$F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{72,000\times (0.63-0)}{1296}\\\\F=35\ N$

So, 35 N of the force is exerted by the rocket on the satellite.

Hence, this is the required solution.

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2 years ago