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3 years ago

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Suppose that you changed the area of the bottom surface of the friction cart without changing its mass, by replacing the Teflon

slab with one that was smaller but thicker. The contact area would shrink, but the normal force would be the same as before. Would this change the friction force on the sliding cart

1 answer:

insens350 [35]3 years ago

4 0

Answer:

in this case the weight of the vehicle does not change , consequently the friction force should not change

Explanation:

The friction force is a macroscopic manifestation of the interactions of the molecules between the two surfaces, this force in the case of solid is expressed by the relation

fr = μ N

W-N= 0

N = W

as in this case the weight of the vehicle does not change nor does the Normal one, consequently the friction force should not change

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An aeroplane accelerates from rest to 65 m/s for take-off. It travels for

Virty [35]

Answer:

1.76m/s²

Explanation:

Given parameters:

Initial velocity = 0m/s

Final velocity = 65m/s

Distance traveled = 1200m

Unknown:

Acceleration = ?

Solution:

This is linear velocity and we apply the appropriate motion equation to solve this problem.

V² = U² + 2as

S is the distance

u is the initial velocity

V is the final velocity

a is the acceleration

Now, insert the parameters and solve;

65² = 0² + 2 x a x 1200

4225 = 2400a

a = 1.76m/s²

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3 years ago

Which one of the following accurately describes the force of gravity?

elena55 [62]

Choice ' C ' is a true statement.
The other choices aren't.

8 0

3 years ago

A motorcycle of mass 100 kilograms travels around a flat, circular track of radius 10 meters with a constant speed of 20 meters

JulijaS [17]

Answer:

100/10 = 10 , 10 × 10 = 100÷20 = 5

I'm pretty sure its wrong

8 0

3 years ago

Read 2 more answers

How fast is a wave traveling if it has a wavelength of 7 meters and a frequency of 11 Hz?

pashok25 [27]

Answer:

$\huge{ \boxed{ \bold{ \sf{77 \: m/s}}}}$

☯ Question :

How fast is a wave travelling if it has a wavelength of 7 meters and a frequency of 11 Hz?

☯ $\underbrace{ \sf{Required \: Answer : }}$

☥ Given :

Wavelength ( λ ) = 7 meters
Frequency ( f ) = 11 Hz

☥ To find :

Speed of sound ( v ) = ?

☄ We know ,

$\boxed{ \sf{v = f \times λ}}$

where ,

v = speed of sound
f = frequency
λ = wavelength

Now, substitute the values and solve for v.

➺ $\sf{v = 11 \times 7}$

➺ $\boxed{ \sf{v = 77 \: m/s}}$

-------------------------------------------------------------------

✑ Additional Info :

Frequency : The number of complete vibrations made by a particle of a body in one second is called it's frequency. It is denoted by the letter f . The SI unit of frequency is hertz ( Hz ).

Wavelength : The distance between two consecutive compressions or rarefactions of a sound wave is called wavelength of that wave. It is denoted by λ ( lambda ) and it's SI unit is m.

Speed of a sound wave : The distance covered by a sound wave in one second is called speed of sound wave. It depends on the product of wavelength and frequency of the wave.

Hope I helped!

Have a wonderful time! ツ

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7 0

3 years ago

A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu

Montano1993 [528]

Answer:

The angle of separation is $\Delta \theta = 0.93 ^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta _ i = 56^o$

The refractive index of violet light in diamond is $n_v = 2.46$

The refractive index of red light in diamond is $n_r = 2.41$

The wavelength of violet light is $\lambda _v = 400nm = 400*10^{-9}m$

The wavelength of red light is $\lambda _r = 700nm = 700*10^{-9}m$

Snell's Law can be represented mathematically as

$\frac{sin \theta_i}{sin \theta_r} = n$

Where $\theta_r$ is the angle of refraction

=> $sin \theta_r = \frac{sin \theta_i}{n}$

Now considering violet light

$sin \theta_r__{v}} = \frac{sin \theta_i}{n_v}$

substituting values

$sin \theta_r__{v}} = \frac{sin (56)}{2.46}$

$sin \theta_r__{v}} = 0.337$

$\theta_r__{v}} = sin ^{-1} (0.337)$

$\theta_r__{v}} = 19.69^o$

Now considering red light

$sin \theta_r__{R}} = \frac{sin \theta_i}{n_r}$

substituting values

$sin \theta_r__{R}} = \frac{sin (56)}{2.41}$

$sin \theta_r__{R}} = 0.344$

$\theta_r__{R}} = sin ^{-1} (0.344)$

$\theta_r__{R}} = 20.12^o$

The angle of separation between the red light and the violet light is mathematically evaluated as

$\Delta \theta = \theta_r__{R}} - \theta_r__{V}}$

substituting values

$\Delta \theta =20.12 - 19.69$

$\Delta \theta = 0.93 ^o$

6 0

3 years ago