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3 years ago

8

what would you expect to happen to the acceleration if all friction were removed from the ramp, making the net force even higher

than 600n

2 answers:

Kruka [31]3 years ago

6 0

Answer:

If friction were removed from the ramp, the acceleration of the skateboarder will increase.

Explanation:

Gnoma [55]3 years ago

5 0

Answer:

Increase

Explanation:

The acceleration will increase more if all the friction were removed from ramp.

Acceleration is the rate of change of velocity with time.

Frictional force is a force that opposes motion of a body.

A large friction will reduce the velocity of a body significantly and bring it to stop, a state of no acceleration.
When friction is removed, the acceleration of a body will increase significantly since the velocity will change rapidly.
At each turn of time, the velocity keeps increasing due to absence of the inhibiting force of friction.

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A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?

forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

$Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}$

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

$e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm$

Therefore, the extension of the wire is 0.362 mm.

8 0

3 years ago

Find the angle of refraction of a ray of light that enters a diamond (n=2.419) from air at an angle of 18.0° to the normal. ONLY

Tpy6a [65]

Answer:

n

Explanation:

hb

3 0

3 years ago

The free-fall acceleration on Mars is 3.7 m/s^2. What length of pendulum has a period of 1.0 s on Earth? What length of pendulum

NemiM [27]

Answer:

Explanation:

Given

Free fall acceleration on mars $g_{m}=3.7\ m/s^2$

Time Period of pendulum on earth $T=1\ s$

Time period of Pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

for earth

$1=2\pi\cdot \sqrt{\frac{L}{9.8}}$

$L=\frac{9.8}{(2\pi )^2}$

$L=0.498\ m$

(b)For same time period on mars length is given by

$L'=\frac{g_m}{(2\pi )^2}$

$L'=\frac{3.7}{39.48}$

$L'=0.0936\ m$

$L'=9.36\ cm$

3 0

3 years ago

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago

A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.

Kruka [31]

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

$a=\dfrac{v-u}{t}$ .............(1)

Now, the second equation of motion is :

$s=ut+\dfrac{1}{2}at^2$

Put the value of a in above equation as :

$s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2$

$s=\dfrac{t(u+v)}{2}$

$u=\dfrac{2s}{t}-v$

$u=\dfrac{2\times 47}{8.6}-2.3$

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

8 0

3 years ago