Plz, Help I just need this question to be answered so I can be done with this worksheet.

A light bulb is a kind of lever is the statement true or false

Anon25 [30]

Answer:

False

Explanation:

It would be a screw

8 0

3 years ago

Another droplet of the same mass falls 8.4 cm from rest in 0.250 s, again moving through a vacuum. Find the charge carried by th

Andrej [43]

Answer:

The charge carried by the droplet is $1.330\times10^{-19}\ C$

Explanation:

Given that,

Distance =8.4 cm

Time = 0.250 s

Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude $5.92\times10^4\ N/C$ points straight down and if the mass of the droplet is $2.93\times10^{-15} kg$

We need to calculate the acceleration

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$8.4\times10^{-2}=0+\dfrac{1}{2}\times a\times(0.250)^2$

$a=\dfrac{8.4\times10^{-2}\times2}{(0.250)^2}$

$a=2.688\ m/s^2$

We need to calculate the charge carried by the droplet

Using formula of electric filed

$E=\dfrac{F}{q}$

$q=\dfrac{ma}{E}$

Put the value into the formula

$q=\dfrac{2.93\times10^{-15}\times2.688}{5.92\times10^4}$

$q=1.330\times10^{-19}\ C$

Hence, The charge carried by the droplet is $1.330\times10^{-19}\ C$

8 0

3 years ago

A 200g air-track glider is attached to a spring. The glider is pushed in 10cm and released. A student with a stopwatch finds tha

cricket20 [7]

The spring constant will be k= 5.5N/m for a 200g air track glider attached to a spring.

<h3>What is spring constant?</h3>

The spring constant, k, is a measure of the stiffness of the spring. It is different for different springs and materials.

Calculation for What is the spring constant

First step is to calculate the time period

T = 12 second/10

T = 1.2 second

Now let calculate the spring constant using this formula

$k=\dfrac{4\pi ^2m}{T^2}$

Where,

m=0.2kg

T=1.2second

k represent spring constant=?

Let plug in the formula

$k=\dfrac{4 \pi \times 0.2}{(1.2)^2}$

$k=\dfrac{39.48\times 0.2}{1.44}$

$k=\dfrac{7.90}{1.44}$

k=5.48 N/m

k=5.5 N/m ( Approximately)

Therefore the spring constant will be 5.5 N/m

To know more about spring constant follow

brainly.com/question/1968517

#SPJ4

4 0

2 years ago

Where do you see triangulation used on this structure? Explain how triangulation

Olegator [25]

I like your profile picture:)

3 0

3 years ago

A bullet of mass 11 g strikes a ballistic pendulum of mass 1.9 kg. The center of mass of the pendulum rises a vertical distance

Ymorist [56]

Answer:

The bullet's initial speed is 243.21 m/s.

Explanation:

Given that,

Mass of the bullet, $m_b=11\ g=0.011\ kg$

Mass of the pendulum, $m_p=19\ kg$

The center of mass of the pendulum rises a vertical distance of 10 cm.

We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,

$\dfrac{1}{2}(m_b+m_p)V^2 =(m_b+m_p)gh\\\\V=\sqrt{2gh} \ .................(1)$

V is the speed of the bullet and pendulum at the time of collision

Now using conservation of momentum as :

$m_bv=(m_p+m_b)V$

Put the value of V from equation (1) in above equation as :

$v=\dfrac{(m_p+m_b)}{m_b}\sqrt{2gh} \\\\v=\dfrac{(1.9+0.011)}{0.011}\sqrt{2\times 9.8\times 0.1}\\\\v=243.21\ m/s$

So, the bullet's initial speed is 243.21 m/s.

7 0

3 years ago