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1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra

UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

$F_r = mgsin\theta$

$F_r = \mu N$

$F_r = \mu mg cos\theta$

Matching the two equation we have,

$\mu N = \mu mg cos\theta$

$\mu = tan\theta$

Applying energy conservation,

$\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1$

$h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})$

Frictionless surface

$\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2$

$\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2$

$h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})$

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between $h_1$ and $h_2$ is given by

$\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}$

$\frac{h_1}{h_2} = \frac{1}{2}$

8 0

2 years ago

Gold and silicon are mutually insoluble in the solid state and form a eutectic system with a eutectic temperature of 636 k and a

kupik [55]

Yupp its c because my dad farted

3 0

2 years ago

Kate is researching air pollution and finds some information on ozone. She knows that ozone is a good thing as part of the ozone

jek_recluse [69]

"At ground level, ozone contributes to smog" so it is also an air pollutant.

Option: A

<u>Explanation</u>:

ozone is naturally present in stratosphere and acts as shield against harmful ultraviolet radiations. But it acts a pollutant contributing to global warming when it is present in lower level atmosphere particularly troposphere. In this level it combines with primary pollutants that is "nitrogen oxides" and "volatile organic" compounds to form secondary pollutant which absorbs outgoing radiation and contributes in raising the temperature. It has harmful impacts on vegetation as well as human health.

7 0

2 years ago

Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless sur

d1i1m1o1n [39]

Answer:

The speed of bullet and wooden bock coupled together, V = 22.22 m/s

Explanation:

Given that,

Mass of the bullet, m = 0.04 Kg

Mass of the wooden block, M = 0.5 Kg

The initial velocity of the bullet, u = 300 m/s

The initial velocity of the wooden block, U = 0 m/s

The final velocity of the bullet and wooden bock coupled together, V = 0 m/s

According to the conservation of linear momentum, the total momentum of the body after impact is equal to the total momentum before impact.

Therefore,

mV + MV = mu + MU

V(m+M) = mu

V = mu/(m+M)

Substituting the values in the above equation,

V = 0.04 Kg x 300 m/s / (0.04 Kg+ 0.5 Kg)

= 22.22 m/s

Hence, the speed of bullet and wooden bock coupled together, V = 22.22 m/s

8 0

2 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

2 years ago