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Yanka [14]
3 years ago
9

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

l direction with variable speed and from a variable vertical position and a fixed horizontal position x=0.
The robot is calibrated by adjusting the speed at which the sphere is launched and the height of the robot’s sphere launcher. Depending on where the spheres land on the ground, students earn points based on the accuracy of the robot. The robot is calibrated so that when the spheres are launched from a vertical position y=H and speed v0, they consistently land on the ground on a target that is at a position x=D. Positive directions for vector quantities are indicated in the figure.

When the students arrive at the competition, it is determined that the height of the sphere launcher can no longer be adjusted due to a mechanical malfunction. Therefore, the spheres must be launched at a vertical position of y=H2. However, the spheres may be launched at speed v0 or 2v0.
Question: In a clear response that may also contain diagrams and/or equations, describe which speed, v0 or 2v0, the students should launch the sphere at so that they earn the maximum number of points in the competition.

Physics
1 answer:
jeyben [28]3 years ago
7 0

Answer:

V_0

Explanation:

Given that, the range covered by the sphere, M, when released by the robot from the height, H, with the horizontal speed V_0 is D as shown in the figure.

The initial velocity in the vertical direction is 0.

Let g be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. V_0 will remain constant throughout the projectile motion.

So, if the time of flight is t, then

D=V_0t\; \cdots (i)

Now, from the equation of motion

s=ut+\frac 1 2 at^2\;\cdots (ii)

Where s is the displacement is the direction of force, u is the initial velocity, a is the constant acceleration and t is time.

Here, s= -H, u=0, and a=-g (negative sign is for taking the sigh convention positive in +y direction as shown in the figure.)

So, from equation (ii),

-H=0\times t +\frac 1 2 (-g)t^2

\Rightarrow H=\frac 1 2 gt^2

\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)

Similarly, for the launched height 2H, the new time of flight, t', is

t'=\sqrt {\frac {4H}{g}}

From equation (iii), we have

\Rightarrow t'=\sqrt 2 t\;\cdots (iv)

Now, the spheres may be launched at speed V_0 or 2V_0.

Let, the distance covered in the x-direction be D_1 for V_0 and D_2 for 2V_0, we have

D_1=V_0t'

D_1=V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_1=\sqrt 2 D [from equation (i)]

\Rightarrow D_1=1.41 D (approximately)

This is in the 3 points range as given in the figure.

Similarly, D_2=2V_0t'

D_2=2V_0\times \sqrt 2 t [from equation (iv)]

\Rightarrow D_2=2\sqrt 2 D [from equation (i)]

\Rightarrow D_2=2.82 D (approximately)

This is out of range, so there is no point for 2V_0.

Hence, students must choose the speed V_0 to launch the sphere to get the maximum number of points.

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In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
Daniel [21]

Answer:

v_2=-133.17m/s, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before (p_i) and after (p_f) the explosion. We will take the east direction as positive.

Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

Which means (since we want v_2 and M=m_1+m_2):

v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

5 0
3 years ago
Two conducting spheres are each given a charge Q. The radius of the larger sphere is three times greater than that of the smalle
padilas [110]

Answer:

Explanation:

If E₀ is the electric field outside the smaller sphere and r is the radius of larger sphere.

E₀ = kQ/r²

The radius of the larger sphere is 3r and the charge on both sphere is same then the electric field outside the larger sphere is given as

E = kQ/(3r)² = kQ/9r² = 1/9 (kQ/r²)= 1/9 x E₀

hence the correct option is e.

5 0
3 years ago
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
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Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
Please explain the method of making electricity from non renewable sources like coal?
Leno4ka [110]
The coal is  urned to heat up water. this produces steam. the steam turns a turbine that turns a generator which provided energy yhat can be transferred into electrisity
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What solutes do not dissolve well in water?
hammer [34]

Answer:

Many substances do not dissolve in water and that is because they are non-polar and do not interact well with water molecules. A common example is oil and water. Oil contains molecules that are non-polar, thus they do not dissolve in water.

Explanation:

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