Ask question

Ask question

All categories

3 years ago

11

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 $5 - \mu_kN = ma$

$25 - \mu_kmg = ma$

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

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

$\mu_k = 0.15$

You might be interested in

What Type of relationship exists between the Temperature of a Star and the Wavelength of a Star?

Debora [2.8K]

This is known as Wien's Law:
The relationship is:
wavelength = 0.0029/temperature

It is an inversely proportional relationship.

3 0

3 years ago

Read 2 more answers

three point charges are arranged in a line. charge q3=+5.00 nC and is located at the origin. charge q2=-3.00 nC and is located a

valentinak56 [21]

Answer:

$q_1=+0.375\ {10}^{-9}$

Explanation:

Electrostatic Forces

The force exerted between two point charges $q_1$ and $q_2$ separated a distance d is given by Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{d^2}$

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

$\displaystyle q_3=+5\ 10^{-9}\ c$

$\displaystyle q_2=-3\ 10^{-9}\ c$

The distance between $q_3$ and $q_2$ is

$\displaystyle d_2=4cm=0.04\ m$

The distance between $q_3$ and $q_1$ is

$\displaystyle d_1=2cm=0.02\ m$

We must find the value of $q_1$ such that

$\displaystyle |F_3|=0$

Applying Coulomb's formula for $q_1$ is

$\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}$

Now for $q_2$

$\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}$

If the total force on $q_3$ is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{k\ q_1\ q_3}{d_1^2}=\frac{k\ q_2\ q_3}{d_2^2}$

Simplfying and solving for $q_1$

$\displaystyle q_1=\frac{q_2\ d_1^2}{d_2^2}$

$\displaystyle q_1=\frac{3.10^{-9}\ 0.02^2}{0.04^2}$

$\boxed{\displaystyle q_1=+0.375\ {10}^{-9}}$

5 0

3 years ago

A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 t. What minimum current in the rod is nee

slamgirl [31]

Answer:

$i = 4.9 A$

Explanation:

Force on a current carrying rod due to magnetic field is given as

$F = iLB$

here we know that

$i =$ current in the rod

$B = 0.10 T$

$L = 1.0 m$

now magnetic force is balanced by the weight of the rod

so we will have

$iLB = mg$

$i(1.0)(0.10) = 0.05 \times 9.8$

$i = 4.9 A$

8 0

3 years ago

A car is traveling at 16 m/s^2

Leya [2.2K]

Really? wow that's pretty cool

4 0

3 years ago

An airplane has wings, each with area A, designed so that air flows over the top of the wing at 265 m/s and underneath the wing

exis [7]

Answer

given,

Pressure on the top wing = 265 m/s

speed of underneath wings = 234 m/s

mass of the airplane = 7.2 × 10³ kg

density of air = 1.29 kg/m³

using Bernoulli's equation

$P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2$

$\Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)$

$\Delta P =\dfrac{1}{2}\times 1.29\times (265^2-234^2)$

$\Delta P =9977.5 Pa$

Applying newtons second law

2 Δ P x A - mg = 0

$A =\dfrac{mg}{2\Delta P}$

$A =\dfrac{7.2\times 10^3 \times 9.8}{2\times 9977.5}$

A = 3.53 m²

7 0

3 years ago