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4 years ago

12

(50 POINTS) Set the ball and the barrier about two feet apart on the table. Grasp both ends of the ruler. Push the ball using th

e center of the ruler so the ball rolls, hits the barrier, and rebounds from it. Describe the motion of the ball as it moves.

2 answers:

padilas [110]4 years ago

8 0

bounces off barrier back to rluler ....

nataly862011 [7]4 years ago

6 0

Answer:

Can you please Identify the force exerted on the ball from the time the ruler touches it?

Explanation:

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At t = 0 the components of a radio-controlled car's velocity are vx = 0.5 m/s and vy = 1.2 m/s. Find the components of the car's

FinnZ [79.3K]

Answer:

x-component of velocity = 5.7 m/s

y-component of velocity = -1.4 m/s

Explanation:

Use first equation of motion to find components of velocity at a given time:

$v = u + at$

where, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

Given:

$u_x= 0.5 m/s\\u_y=1.2 m/s\\a_x=2 m/s^2\\a_y=-1m/s^2\\t = (2.6-0)s =2.6 s$

$v_x=u_x+a_xt\\\Rightarrow v_x=0.5+2\times 2.6\\\Rightarrow v_x=5.7 m/s$

$v_y=u_y+a_yt\\\Rightarrow v_y=1.2-1\times 2.6\\\Rightarrow v_y=-1.4 m/s$

5 0

3 years ago

A solenoid having an inductance of 5.41 μH is connected in series with a 0.949 kΩ resistor. (a) If a 16.0 V battery is connected

vekshin1

Answer:

(A) $9.14\times 10^{-9}sec$

(B) $6.20\times 10^{-3}A$

Explanation:

We have given inductance $L=5.41\mu H=5.41\times 10^{-6}H$

Resistance $R=0.949kohm=0.949\times 10^3ohm$

Time constant of RL circuit is equal to $\tau =\frac{L}{R}$

$\tau =\frac{5.41\times 10^{-6}}{0.949\times 10^3}=5.70\times 10^{-9}sec$

Battery voltage e = 16 volt

(a) It is given current becomes 79.9% of its final value

Current in RL circuit is given by

$i=i_0(1-e^{\frac{-t}{\tau }})$

According to question

$0.799i_0=i_0(1-e^{\frac{-t}{\tau }})$

$e^{\frac{-t}{\tau }}=0.201$

${\frac{-t}{\tau }}=ln0.201$

${\frac{-t}{5.7\times 10^{-9} }}=-1.6044$

$t=9.14\times 10^{-9}sec$

(b) Current at $t=\tau sec$

$i=i_0(1-e^{\frac{-t}{\tau }})$

$i=\frac{16}{0.949\times 10^3}(1-e^{\frac{-\tau }{\tau }})$

$i=6.20\times 10^{-3}A$

3 0

3 years ago

Fish conservation zones extend as much as_____ from a nations’ coast.

Ann [662]

200 miles from a nation's coast

7 0

3 years ago

Read 2 more answers

A particle with charge −5 µC is located on

Nataly [62]

Answer:

36.25 N

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the magnitude of the two charges

r is the separation between the two charges

Moreover:

- The force is repulsive if the two charges have same sign

- The force is attractive if the two charges have opposite sign

In this problem, we have 3 charges:

$q_1=-5\mu C = -5\cdot 10^{-6}C$ is the charge located at $x=+10 cm = +0.10 m$

$q_2=+6\mu C=+6\cdot 10^{-6}C$ is the charge located at $x=-8 cm =-0.08 m$

$q_3=+2\mu C=+2\cdot 10^{-6}C$ is the charge located at $x=-2 cm=-0.02 m$

The force between charge 1 and charge 3 is:

$F_{13}=\frac{kq_1 q_3}{(x_1-x_3)^2}=\frac{(9\cdot 10^9)(5\cdot 10^{-6})(2\cdot 10^{-6})}{(0.10-(-0.02))^2}=6.25 N$

And since the two charges have opposite sign, the force is attractive, so the force on charge 3 is to the right (towards charge 1).

The force between charge 2 and charge 3 is:

$F_{23}=\frac{kq_2 q_3}{(x_2-x_3)^2}=\frac{(9\cdot 10^9)(6\cdot 10^{-6})(2\cdot 10^{-6})}{(-0.08-(-0.02))^2}=30.0 N$

And since the two charges have same sign, the force is repulsive, so the force on charge 3 is to the right (away from charge 2).

So the two forces on charge 3 have same direction (to the right), so the net force is the sum of the two forces:

$F=F_{13}+F_{23}=6.25+30.0=36.25 N$

8 0

4 years ago

In a photoelectric effect experiment, it is observed that violet light does not eject electrons from a particular metal. Next, r

liubo4ka [24]

Answer:

No ejection of photo electron takes place.

Explanation:

When a photon of suitable energy falls on cathode, then the photoelectrons is emitted from the cathode. This phenomenon is called photo electric effect.

The minimum energy required to just eject an electron is called work function.

The photo electric equation is

E = W + KE

where, E is the incident energy, W is the work function and KE is the kinetic energy.

W = h f

where. h is the Plank's constant and f is the threshold frequency.

Now, when the violet light is falling, no electrons is ejected. When the red light is falling, whose frequency is less than the violet light, then again no photo electron is ejected from the metal surface.

6 0

3 years ago