Ask question

Ask question

All categories

3 years ago

7

Write a letter to your friend telling him how you are keeping active with your studies at home due to the Coronavirus Pandemic o

utbreak

1 answer:

tangare [24]3 years ago

7 0

Answer:

HERPES CURE OR ANY DISEASES CONTACT_________________________ ROBINSONBUCLER ((G MAIL.)) COM ……………

Explanation:

You might be interested in

Which sentence in the passage can be used to conclude that Eris is a dwarf planet and not a planet?

Lelu [443]

Eris is slightly more massive than Pluto. However, both of them are smaller than Earth's Moon.
This should conclude that Eris is a dwarf planet.

3 0

2 years ago

Read 2 more answers

Help on finding kinetic energy??

Jobisdone [24]

Trick question? In order to have kinetic energy, an object must be moving. Therefore, in this case, kinetic energy would be 0. If it were asking about potential energy, it would be a different story.

8 0

2 years ago

Read 2 more answers

A student connects four AA batteries (1.5 V each) in series to light up a light bulb. The circuit has a resistance of 35 2. How

allsm [11]

Answer:

D

Explanation:

6/35=0.17

8 0

3 years ago

A treatment in which an electrical current is applied to the brain is

Elza [17]

A treatment in which an electrical current is applied to the brain is;

D. Electroconvulsive therapy

<u>An electric current would be transmitted through the brain without anesthesia through this method. However, there may be side effects of possible seizures or broken bones. </u>

3 0

2 years ago

Read 2 more answers

A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

6 0

3 years ago