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3 years ago

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A 16.0 kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 25.0 N. Starting from rest, the

sled attains a speed of 1.00 m/s in 8.00 m. Find the coefficient of kinetic friction between the runners of the sled and the snow. You Answered

1 answer:

Fudgin [204]3 years ago

4 0

Answer:

$\mu_k = 0.15$

Explanation:

according to the kinematic equation

$v^{2} - u^{2} = 2aS$

Where

u is initial velocity = 0 m/s

a = acceleration

S is distance = 8.00 m

final velocity = 1.0 m/s

$a = \frac {v^{2}}{2S}$

$a = \frac {1{2}}{2*8.6}$

a = 0.058 m/s^2

from newton second law

Net force = ma

$f_{net} = ma$

F - f = ma

2 $5 - \mu_kN = ma$

$25 - \mu_kmg = ma$

$\frac {25 - ma}{mg} =\mu_k$

$\frac {25 - 16*0.058}{16*9.81} = 0.15$

$\mu_k = 0.15$

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Answer:

P = 6.63 kW

Explanation:

Given that,

Current, I = 260 A

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We need to find the power supplied to the starter motor. We know that,

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Put all the values,

P = 25.5 × 260

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or

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So, the power supplied to the motor is 6.63 kW.

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3 years ago

Read 2 more answers

An electric motor can drive grinding wheel at two different speeds. When set to high the angular speed is 2000 rpm. The wheel tu

shutvik [7]

a) The initial angular speed is 209.3 m/s

b) The angular acceleration is $-1.74 rad/s^2$

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

a)

The initial angular speed of the wheel is

$\omega_i = 2000 rpm$

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

$1 rev = 2\pi rad$

$1 min = 60 s$

We find:

$\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s$

b)

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

$\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s$

Now we can find the angular acceleration, given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_i = 209.3 rad/s$ is the initial angular speed

$\omega_f = 104.7 rad/s$ is the final angular speed

t = 60 s is the time interval

Substituting,

$\alpha = \frac{104.7-209.3}{60}=-1.74 rad/s^2$

c)

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

$\omega' = \omega_i + \alpha t$

where we have

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 40 s, we find

$\omega' = 209.3 + (-1.74)(40)=139.7 rad/s$

d)

The angular displacement of the wheel in a certain time interval t is given by

$\theta=\omega_i t + \frac{1}{2}\alpha t^2$

where

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 60 s, we find:

$\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad$

So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,

$\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev$

Learn more about circular motion:

brainly.com/question/2562955

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3 years ago

Why is our (a person's) gravitational pull NOT as strong as the Earth's gravitational pull

Bezzdna [24]

Answer:With gravity, two things with mass will want to move toward each other. However, we humans don't feel our gravity pulling on another person because it's not very big, but we do all feel the pull of Earth's gravity all the time - we're not all floating in the air, because that would be happening without Earth's gravity!

Explanation:

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3 years ago

You and your friend are stranded on a raft of mass 1000 kg which is not moving, you can neglect friction between the boat and th

Papessa [141]

Answer:

Mass of raft=m1= 1000kg

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Speed of paddle=v2=3 m/s

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Formula

V1= -(m2/m1)v2= -(40/1000)3= - 0.12m/s

Explanation:

Here the raft has much greater mass as compare the paddle and its velocity will always be negative due to recoil. 1 and 2 with masses and velocities are subscripts.

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3 years ago

Two objects (42.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h

Kazeer [188]

Answer:

a = 3.27 m/s²

T = 275 N

Explanation:

Given that:

Mass m₁ = 42.p0 kg

Mass m₂ = 21.0 kg

Consider both masses to be in a whole system, then:

The acceleration can be determined as:

$(m_1+m_2)a = g(m_1-m_2)$

Making acceleration the subject in the above formula;

$a =\dfrac{g(m_1-m_2)}{(m_1+m_2)}$

$a =\dfrac{9.8(42.0-21.0)}{(42.0+21.0)}$

$a =\dfrac{9.8(21.0)}{(63.0)}$

$a =\dfrac{205.8}{(63.0)}$

a = 3.27 m/s²

in the string, the tension is calculated using the formula:

$T = \dfrac{2m_1m_2g}{(m_1+m_2)}$

$T = \dfrac{2(42)(21)(9.81)}{(42+21)}$

$T = \dfrac{17304.84}{63}$

T = 274.68 N

T ≅ 275 N

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3 years ago