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3 years ago

A mass m = 4.6 kg hangs on the end of a massless rope L = 2.16 m long. The pendulum is held horizontal and released from rest.

1 answer:

4 0

ind the velocity my using the conservation of energy mgh=1/2mv^2. At the bottom, the mass is turning in a circle with tension pointing up towards the center of the circle and weight pointing down. Use the centripetal force requirement mv^2/r=T-W and solve for T. Hope this helps.

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

jeyben [28]

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.

7 0

3 years ago

Suppose the Sun moved twice the distance from Earth. Describe immediate and future changes to our planet. Justify your reasoning

Reptile [31]

Our year would now be 2.8 times longer, we would also be receiving only 1/4 of the energy from the sun that we currently do. This means that we’d now be out beyond the orbit of Mars and right at the edge of the asteroid belt, and things would rapidly get very cold with temperatures expected to drop by around 50 degrees Celsius on average, and that’s with our current atmospheric composition which would not be stable in the new conditions. And also, any living thing on earth would die.

5 0

2 years ago

A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.

marishachu [46]

Answer:

a) the magnitude of the force is

F= Q( $\frac{kqs}{r^3}$ ) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b) the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = $\frac{kq}{r^{2} }$

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = $\frac{kp}{r^{3} }$ , where p = q × s

E(r) = $\frac{kqs}{r^{3} }$ where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q( $\frac{kqs}{r^3}$ )

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

8 0

3 years ago

What is the power output of a pump which can raise 60kg of water height to height of 10m every minute

sattari [20]

Answer:

<h3>Power = Work Done/time</h3>

=> Power = 60×10×10/60

=> Power = 6000/60

=> Power = 100 Watt

Hence the power output of a pump is 100 Watts.

8 0

2 years ago

8. Define resistance and resistivity and also give the relation between them. Explain the

frozen [14]

Answer

I hope it's helps you

6 0

3 years ago