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2 years ago

9

Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is

the coefficient of kinetic friction?

1 answer:

GarryVolchara [31]2 years ago

6 0

that would be given by

[email protected]

@ representing coefficient of kinetic friction.

thus 19.5/51.7 = 0.377

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James gently releases a ball at the top of a slope, but does not push the ball. The ball rolls down the slope. Which force cause

Rzqust [24]

D. Gravitational force.

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2 years ago

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. A little car has a maximum acceleration of 2.57 m/s2. What is the new maximum acceleration of the little car if it tows anothe

valkas [14]

Answer:

$a'=1.285\ m/s^2$

Explanation:

Let m be the mass of a little car and m' be the mass of another car.

We know that,

Force = mass × acceleration

ATQ,

m × a = 2m × a'

a = 2 × a'

$a'=\dfrac{a}{2}\\\\a'=\dfrac{2.57}{2}\\\\a'=1.285\ m/s^2$

So, the acceleration of another little car is equal to $1.285\ m/s^2$ .

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2 years ago

Noise-canceling headphones are an application of destructive interference. Each side of the headphones uses a microphone to pick

jeka57 [31]

Answer:

2.5 ms

Explanation:

v = Speed of sound in air = 343 m/s

f = Frequency = 200 Hz

Wavelength is given by

$\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{343}{200}\\\Rightarrow \lambda=1.715\ m$

In the case of destructive interference, path difference is given by

$x=\dfrac{\lambda}{2}\\\Rightarrow x=\dfrac{1.715}{2}\\\Rightarrow x=0.8575\ m$

Delay is givenn by

$t=\dfrac{x}{v}\\\Rightarrow t=\dfrac{0.8575}{343}\\\Rightarrow t=0.0025\ s$

The minimum headphone delay, that will cancel this noise is 2.5 ms

6 0

2 years ago

For a metal that has an electrical conductivity of 7.1 x 107 (Ω-m)-1, do the following: (a) Calculate the resistance (in Ω) of a

jonny [76]

Answer:

(a) 0.0178 Ω

(b) 3.4 A

(c) 6.4 x 10⁵ A/m²

(d) 9.01 x 10⁻³ V/m

Explanation:

(a)

σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹

d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m

Area of cross-section of the wire is given as

A = (0.25) π d²

A = (0.25) (3.14) (2.6 x 10⁻³)²

A = 5.3 x 10⁻⁶ m²

L = length of the wire = 6.7 m

Resistance of the wire is given as

$R=\frac{L}{A\sigma }$

$R=\frac{6.7}{(5.3\times10^{-6})(7.1\times10^{7}) }$

R = 0.0178 Ω

(b)

V = potential drop across the ends of wire = 0.060 volts

i = current flowing in the wire

Using ohm's law, current flowing is given as

$i = \frac{V}{R}$

$i = \frac{0.060}{0.0178}$

i = 3.4 A

(c)

Current density is given as

$J = \frac{i}{A}$

$J = \frac{3.4}{5.3\times10^{-6}}$

J = 6.4 x 10⁵ A/m²

(d)

Magnitude of electric field is given as

$E = \frac{J}{\sigma }$

$E = \frac{6.4 \times 10^{5}}{ 7.1 \times 10^{7}}$

E = 9.01 x 10⁻³ V/m

5 0

2 years ago

Block 1 of mass m1 slides along a frictionless floor and into a one-dimensional elastic collision with stationary block 2 of mas

Elan Coil [88]

<span>Answers: (a) 2.0 m/s (b) 4 m/s

Method:

(a) By conservation of momentum, the velocity of the center of mass is unchanged, i.e., 2.0 m/s.

(b) The velocity of the center of mass = (m1v1+m2v2) / (m1+m2)

Since the second mass is initially at rest, vcom = m1v1 / (m1+m2)

Therefore, the initial v1 = vcom (m1+m2) / m1 = 2.0 m/s x 6 = 12 m/s

Since the second mass is initially at rest, v2f = v1i (2m1 /m1+m2 ) = 12 m/s (2/6) = 4 m/s </span>

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2 years ago