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2 years ago

If a person suffers from mysophobia what do they fear?

2 answers:

4 0

Mysophobia, or the fear of germs, refers to an unhealthy fear of contamination. ... However, if you suffer from mysophobia, these normal concerns become overblown. The phobia is common, affecting even celebrities such as Howie Mandel,

hope this helps!

Liula [17]2 years ago

4 0

Mysophobia is a mental disorder that causes an affected person to have a massive fear of dirt or otherwise contamination. This is also, informally, called "germophobia". Mysophobia and OCD disorder are commonly mistaken for each other.

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A ball is dropped and bounces off the floor. Its speed is the same immediately before and immediately after the collision. Which

Korolek [52]

Answer:

option B.

Explanation:

The correct answer is option B.

when the ball drops, the velocity of the ball before the collision is v

After the collision, the velocity of the ball is the same but in the opposite direction.

Impulse delivered to the ball and the floor, in this case, is not zero.

The magnitude of the momentum remains the same but the direction of the ball changes.

7 0

3 years ago

Hellp.....its physics

blondinia [14]

Answer:

i know it

Explanation:

18.493 its the answer

brainliest pls

4 0

3 years ago

Read 2 more answers

A car moving in the positive direction with an initial velocity of 26 m/s slows down at a constant rate of -3 m/s2. What is its

Greeley [361]

The answer is 5 m/s
$- 3 = \frac{x - 26}{ 7} \\ x = \frac{7 \times( - 3)}{1} + 26 = - 21 + 26 \\ = 5$

good luck

8 0

3 years ago

Fission fusion worksheet answers

fredd [130]

Http://tomschoderbekchem.blogspot.com/2014/09/nuclear-fission-and-fusion-worksheet.html

5 0

3 years ago

Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

<u>For the course of upward acceleration:</u>

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0

3 years ago