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3 years ago

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bit.♠ly/3♠vhMu♠vJ remove symbols before searching or it wont work, there was a bug stoping me from attaching the image so there

it is

2 answers:

dmitriy555 [2]3 years ago

3 0

Answer:

k and...

Explanation:

Sophie [7]3 years ago

3 0

Answer:

no thank you.

explanation: Do not want to

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If a 5.5 kg object experiences 15 N of force for .15 seconds what is the speed change

Nady [450]

The speed change : Δv = 0.41 m/s

<h3>Further explanation</h3>

Given

mass = 5.5 kg

Force = 15 N

time = 0.15 s

Required

the speed change

Solution

Newton 2nd's law

Impulse and momentum

F = m.a

F = m . Δv/t

F.t = m.Δv

Input the value :

15 N x 0.15 s = 5.5 kg x Δv

Δv = 0.41 m/s

8 0

3 years ago

HELP ASAP <br> Describe when contact metamorphism occurs?

diamong [38]

Contact metamorphism occurs adjacent to igneous intrusions and results from high temperatures associated with the igneous intrusion. Since only a small area surrounding the intrusion is heated by the magma, metamorphism is restricted to the zone surrounding the intrusion, called a metamorphic or contact aureole

5 0

3 years ago

Read 2 more answers

A 0.15 g honeybee acquires a charge of 22 pC while flying. The electric field near the surface of the earth is typically 100 N/C

Rus_ich [418]

Answer:

$1.50\ *10^{-6} }$

Explanation:

Given

e=100 N/C

M=0.15 g

$q=\ 22\ pC\\=\ 22\ *10^{-2}$

The ratio of the electric force on the bee to the bee's weight can be determined by the following formula

$\frac{fe}{M*9.81}$

$\frac{22*10^{-12\ *\ 100} }{0.15*\ 10^{-3} *\ 9.81}$

$=\ 1.50\ *10^{-6}$

4 0

4 years ago

Hannah just finished building a house of cards that stands four stories high. She is worried that it will fall down. Which of th

Tju [1.3M]

A.) <span>If no unbalanced force acts upon the house of cards, then it will remain standing forever.

[ Other Statements doesn't make any sense in Physics, they can be true in some situation & can't be false, option A is the only answer ]
Hope this helps!</span>

6 0

4 years ago

Read 2 more answers

Can someone help me?

Leviafan [203]

Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.

In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

$V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t$

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

$V_y.t = \frac{2d}{t} .t = 2d$

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.

4 0

3 years ago