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3 years ago

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An ice cube of mass 50.0 g can slide without friction up and down a 25.0 degree slope. The ice cube is pressed against a spring

at the bottom of the slope, compressing the spring 0.100 m . The spring constant is 25.0 N/m . When the ice cube is released, how far will it travel up the slope before reversing direction? Express your answer numerically in meters to three significant figures. d = nothing m

1 answer:

lozanna [386]3 years ago

8 0

Answer:

0.6 m

Explanation:

When a spring is compressed it stores potential energy. This energy is:

Ep = 1/2 * k * x^2

Being x the distance it compressed/stretched.

When the spring bounces the ice cube back it will transfer that energy to the cube, it will raise up the slope, reaching a high point where it will have a speed of zero and a potential energy equal to what the spring gave it.

The potential energy of the ice cube is:

Ep = m * g * h

This is vertical height and is related to the distance up the slope by:

sin(a) = h/d

h = sin(a) * d

Replacing:

Ep = m * g * sin(a) * d

Equating both potential energies:

1/2 * k * x^2 = m * g * sin(a) * d

d = (1/2 * k * x^2) / (m * g * sin(a))

d= (1/2 * 25 * 0.1^2) / (0.05 * 9.81 * sin(25)) = 0.6 m

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In the figure, a proton is projected horizontally midway between two parallel plates that are separated by 0.50 cm, and are 5.60

noname [10]

a) Minimum speed of the proton: $6.05\cdot 10^6 m/s$

b) Angle of the velocity: $\theta=-5.1^{\circ}$

Explanation:

a)

The proton experiences a vertical force due to the electric field, given by:

$F=qE$

where

$q=1.6\cdot 10^{-19}C$ is the proton charge

$E=610,000 N/C$ is the magnitude of the electric field

The vertical acceleration of the proton is therefore

$a=\frac{qE}{m}$

where

$m=1.67\cdot 10^{-27}kg$ is its mass

Therefore, the vertical position of the proton at time t is

$y(t)=\frac{1}{2}at^2=\frac{1}{2}\frac{qE}{m}t^2$

where we assumed that the initial vertical velocity is zero (because the proton is fired horizontally) and the initial vertical position, halfway between the two plates, is the origin.

The horizontal motion of the proton instead is uniform, so the horizontal position is given by

$x(t)=v_0 t$

where $v_0$ is the initial speed. This equation can be rewritten as

$t=\frac{x(t)}{v_0}$

And substituting into the eq. for y,

$y(t) = \frac{1}{2}\frac{qE}{m} \frac{x^2}{v_0^2}$

Solving for the initial speed,

$v_0 = \sqrt{\frac{qEx^2}{2my}}$

The proton just misses one of the plate when

x = 5.60 cm = 0.056 m (length of the plates)

y = 0.25 cm = 0.0025 m (half the distance between the plates)

Therefore, we find the initial speed:

$v=\sqrt{\frac{(1.6\cdot 10^{-19})(610,000)(0.056)^2}{2(1.67\cdot 10^{-27})(0.0025)}}=6.05\cdot 10^6 m/s$

b)

In order to find the angle, we just need to analyze the horizontal and vertical component of the final velocity of the proton.

The horizontal velocity is constant so it is:

$v_x = v_0 = 6.05\cdot 10^6 m/s$

The vertical velocity is given by:

$v_y^2 - u_y^2 = 2ay$

where:

$u_y=0$ (initial vertical velocity is zero)

$a=\frac{qE}{m}$ (acceleration)

y = 0.0025 m (vertical displacement)

Solving for $v_y$ ,

$v_y = \sqrt{2ay}=\sqrt{2\frac{qEy}{m}}=5.4\cdot 10^5 m/s$

Therefore, the final angle of the velocity with respect to the horizontal is:

$\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.4\cdot 10^5}{6.05\cdot 10^6})=5.1^{\circ}$

And since the electric field is downward (the proton just misses the lower field), it means that the angle is below the horizontal:

$\theta=-5.1^{\circ}$

Learn more about electric fields:

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3 years ago

A tabletop gamer has designed a game that requires three dice to be thrown onto a tray with a measurement grid. To add an extra

Airida [17]

Answer:

The resulting coordinate is $(x,y) = (-0.170 \ m , -0.038 \ m )$

Explanation:

From the question we are told that

The mass of the first dice is $m_1 = 11.10 \ g$

The mass of the second dice is $m_2 = 15.10 \ g$

The mass of the third dice is $m_3 = 18.90 \ g$

The coordinate of the first dice is $(x_1, y_1) = (0.3150\ m , -0.4990 \ m )$

The coordinate of the second dice is $(x_2,y_2) = (-0.4050 \ m , 0.4850 \ m )$

The coordinate of the third dice is $(x_3 , y_3) = (-0.1150 \ m , -0.1850\ m )$

Generally the resulting coordinate of the center of mass of the dice in the x-axis is mathematically evaluated as

$x\ cm = \frac{m_1 * x_1 + m_2 * x_2 + m_3 * x_3}{m_1 + m_2 +m_3 }$

i.e the summation of the moments about their x-axis divided by the magnitude of their masses

substituting values

$x\ cm = \frac{11.0 * 0.3150 + 15.10 * (-0.4050) + 18.90 * (-0.1150)}{11.100 + 15.10 +18.90 }$

$x\ cm = -0.170 \ m$

Generally the resulting coordinate of the center of mass of the dice in the y-axis is mathematically evaluated as

$y\ cm = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3}{m_1 + m_2 +m_3 }$

$y\ cm = \frac{ 11.10 * (-0.4990) + (15.10) * (0.4850) + (18.90) * (-0.1850)}{ 11.10 + 15.10 +18.90 }$

$y\ cm = -0.038 \ m$

Thus the resulting coordinate is $(x,y) = (-0.170 \ m , -0.038 \ m )$

4 0

3 years ago

20.0 -kg cannonball is fired from a cannon with muzzle speed of 1000m/s at an angle of 37.0° with the horizontal. A second ball

Dovator [93]

The mechanical energy for the first and the second ball is

$10 ^{7} \: joules.$

Mass of the first ball = 20 kg

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 37°

The maximum height reached by the first ball is,

$= \frac{ u {}^{2} _{1}sin {}^{2} θ}{2g}$

$= \frac{ {1000}^{2} sin {}^{2}37°}{2 \times 9.8}$

$= 18478.69 \: m$

The maximum height of the first cannonball is 17478.69 m.

The initial speed at which a cannonball is fired from a cannon =1000 m/s

The angle made by the cannonball while being fired from the cannon = 90 °

$= \frac{ u {}^{2} _{2}sin {}^{2} θ}{2g}$

$= \frac{ 1000{}^{2}sin^{2} 90°}{2 \times 9.8}[tex] = 51020.41 \: m$

For the first ball, total mechanical energy= Potential energy at maximum height + kinetic energy at the maximum height

So, the total mechanical energy is,

= mgh \: + \frac{1}{2}mv {}^{2} _{x}[/tex]

$= 20 \times 9.8 \times 18478.64 \times \frac{20}{2} (1000 \: cos37 °)$

$= 10 ^{7} The potential energy at the maximum height, = m _{2}gh$

$= 20 \times 9.8 \times 51020.41$

$= 10 ^{7} \:J$

Therefore, the total mechanical energy for the first and the

$\:second \: cannonball \: is \: 10 ^{7} \:joules.$

To know about energy, refer to the below link:

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5 0

2 years ago

Light can make things happen when it is ______. Fill in the missing word. COrrect answer gets brainliest!

MatroZZZ [7]

Answer:

Light can make things happen when it is shining.

(or glowing)

8 0

3 years ago

Read 2 more answers

A 3.4kg aluminium ball has an apparent mass of 2.10kg when submerged in a particular liquid. Calculate the density of the liquid

Shalnov [3]

Answer:

1083.3kg/m ³

Explanation:

Given parameters:

Mass of aluminium ball = 3.4kg

Apparent mass of ball = 2.1kg

Unknown:

Density of the liquid = ?

Solution:

Density is is the mass per unit volume of any give substance;

Density = $\frac{mass}{volume}$

Now, we must understand that the apparent weight of the aluminium is the part of its weight supported by the fluid;

Pl = $\frac{Ml}{Vl}$

Pl = density of liquid

Ml = mass of liquid

Vl = volume of liquid

Mass of liquid = Mal - Mapparent

Mal = mass of aluminium

The volume of liquid displaced is the same as the volume of the aluminium according to Archimedes's principle;

Ml = Pl x Vl

Val = Vl

Val volume of aluminium

Vl = volume of liquid

****

Pal = $\frac{Mal}{Val}$

Val = $\frac{Mal}{Pal}$

Val = volume of aluminium

Mal = mass of aluminium

Pal = density of aluminium

*****

Since the Val = Vl

Ml = Pl x $\frac{Mal}{Pal}$

Since

Ml = Mal - Mapparent

Mal - Mapparent = Pl x $\frac{Mal}{Pal}$

Pal = density of aluminium = 2712 kg/m ³

3.4 - 2.1 = Pl x $\frac{3.4}{2712}$

1.3 = Pl x 0.00123

Pl = 1083.3kg/m ³

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3 years ago