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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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In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

$\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63$ seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

$D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m$ .

Now, for the maximum height, H, applying the equation of motion as

$v^2=u^2+2as$

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, $0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H$

$\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}$

$\Rightarrow H = 3.25 m$ .

Hence, the maximum height covered is 3.25 m.

8 0

3 years ago

A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several

mamaluj [8]

Answer:

The charge on the dust particle is $q_d = 6.94 *10^{-13} \ C$

Explanation:

From the question we are told that

The length is $l = 2.0 \ m$

The width is $w = 4.0 \ m$

The charge is $q = -10\mu C= -10*10^{-6} \ C$

The mass suspended in mid-air is $m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg$

Generally the electric field on the carpet is mathematically represented as

$E = \frac{q}{ 2 * A * \epsilon _o}$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}$

$E = -70621.5 \ N/C$

Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles

$F__{E}} = F__{G}}$

=> $q_d * E = m * g$

=> $q_d = \frac{m * g}{E}$

=> $q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}$

=> $q_d = 6.94 *10^{-13} \ C$

4 0

4 years ago

the Earth exerts a gravitational force of 500 newtons on an object what is the mass of the object in kilograms

MAXImum [283]

Use Newton's second law F = mass * acceleration
In your problem F = 500, and we know gravity is working on it so use a = 9.81
Substitute into the equation
500 = m * 9.81
m = 50.97 kg

7 0

3 years ago

The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a

Degger [83]

Answer:

147.456077993 Hz

Explanation:

$f_0$ = Frequency of the sonar = 22 kHz

$v_w$ = Velocity of the whale = 4.95 m/s

v = Speed of sound in water = 1482 m/s

The difference in frequency is given by

$\Delta f=f_0\times\dfrac{2v_{w}}{v-v_w}\\\Rightarrow \Delta f=22000\times\dfrac{2\times 4.95}{1482-4.95}\\\Rightarrow \Delta f=147.456077993\ Hz$

The difference in frequency is 147.456077993 Hz

6 0

3 years ago

Since the Wagon is being pulled down hill is it increasing (C)

Ket [755]

<em>Since the wagon is being pulled down hill with a constant velocity, all the forces of the wagon would be (C) increasing.</em>
<em>You are correct! **</em>

4 0

4 years ago