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Vladimir [108]
3 years ago
11

SOMONE PLEASE HELP ASAP!!!!

Physics
2 answers:
Kazeer [188]3 years ago
8 0

Answer:

the shiny mirror

Solnce55 [7]3 years ago
5 0

Answer:

i think its shiny mirror

Explanation:

'='

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How will creat thunderstrom​
emmasim [6.3K]

Answer:

the air has to be unstable as well as it needs to be moved upwards.

Explanation:

it needs to be moved upwards and also needs to have unstable air.

3 0
2 years ago
A motorcycle accelerates from 10. m/s to 25 m/s in 5.0 seconds. What is the average acceleration of the bike?
valentinak56 [21]

If a motorcycle accelerates from 10 m/s to 15 m/s in 5 seconds, the average acceleration of the bike is 3 m/s/s. This only means that the motorcycle’<span>s velocity will increase 3 m/s every second.  You need to divide 15 m/s which is the difference of 10 and 25 to 5 seconds.</span>

3 0
4 years ago
Read 2 more answers
Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum
Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

\frac{1}{2}mv^2 = mgh

where

m is the mass of the man

v is the speed after jumping

g=9.8 m/s^2 is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s

In the acceleration phase, we know that the initial velocity is

u=0

And the force exerted on the floor is 2.3 times the gravitational force, so

F=2.3 mg

This means the net force on you is

F_{net} = F-mg=2.3mg-mg=1.3 mg

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g

Now we can use the  following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

v^2-u^2=2as

where s is the quantity we want to find. Solving for s,

s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m

2)

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

\Delta h = -0.77 m

This means that the lowest point reached by the center of mass above the floor during the crouch is

h'=h+\Delta h = 1.0 - 0.77 = 0.23 m

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

3 0
3 years ago
PLEASE HELP ME ON THIS ONE ANYONE
andreev551 [17]

Answer:

A & B

Explanation:

A & B Would be the right answer since Morse code cannot be represented through the height of the fire.

6 0
2 years ago
The average weight of a particular box of crackers is 26.0 ounces with a standard deviation of 0.5 ounce. The weights of the
Olenka [21]

Answer:

(a) 99.865%

(b) 0.135%

Explanation:

Given that the weight of the boxes are normally distributed.

The average weight of the particular box,

\mu=26.0 ounce

The standard deviation of weight,

\sigma=0.5 ounce.

Let z be the standard normal variable,

z=\frac{x-\mu}{\sigma}

And, the probability of the boxes having weight x ounces is

P(z)=\frac{1}{2\pi}e^{-\frac{z^2}{2}}

For x=24.5,

z=\frac{x-\mu}{\sigma}=\frac{24.5-26}{0.5}=-3

(a) For the boxes having weight more than 24.5 ounces:z>-3

So, the probability of boxes for z>-3 is

P(z>-3)=\int_{-3}^{\infty}\left(\frac{1}{2\pi}e^{-\frac{z^2}{2}}\right)dx

=0.99865

So, the percent of the boxes weigh more than 24.5 ounces is 99.865%.

(b) For the boxes having weight less than 24.5 ounces: z<-3

So, the probability of boxes for z<-3 is

P(z-3}=1-0.99865

=0.00135

So, the percent of the boxes weigh more than 24.5 ounces is 0.135%.

8 0
4 years ago
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