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3 years ago

8

Which tricks do you use to reduce pressure in everyday activities? Which tricks do you use to reduce pressure in everyday activi

ties?

1 answer:

Molodets [167]3 years ago

8 0

Answer:

same problem is here also bro i cannot find this plz help anyone

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If 25.0 g of water at 21c is mixed with 45.0 g of water at 75c, what is the final temperature of the mixture?

mrs_skeptik [129]

56C

Hope that helps, Good luck! (:

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3 years ago

B. Jerome plays middle linebacker for South's varsity football team. In a game against

weqwewe [10]

Answer:

Option D

670 Kg.m/s

Explanation:

Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)

Final momentum is also mv but v being westward direction, we take it negative

Final momentum=82*-2.5= -205 Kg.m/s

Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s

Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s

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3 years ago

Determine the value of the information in scientific notation. The moon is approximately 3. 63 × 108 meters from Earth. What is

Usimov [2.4K]

Answer:

d

Explanation:

7 0

2 years ago

I Need answer ASAP

anzhelika [568]

Answer: Convection and conduction

Tell me that I got it right??

Explanation

Mark me as Brainliest PLEASE I HAVE 0 BRAINLIEST

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3 years ago

Read 2 more answers

An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of

GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, $a=6\times 10^{15}\ m/s^2$

Charge on electron, $q=1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ kg$

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, $F=qvB\ sin\theta$

Here, $\theta=90$

So, $ma=qvB$

Where

v is the velocity of the electron

B is the magnetic field

$v=\dfrac{ma}{qB}$

$v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}$

v = 341250 m/s

or

$v=3.41\times 10^5\ m/s$

So, the speed of the electron is $3.41\times 10^5\ m/s$

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

7 0

3 years ago