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siniylev [52]
3 years ago
15

Which equation is used to calculate the magnetic force on a charge moving in a magnetic field? F = |q|vBcos F = |q|vBsin F = |q|

Bcos F = |q|Bsin
Physics
2 answers:
anygoal [31]3 years ago
8 0
The answer : F = |q| B sin
sergiy2304 [10]3 years ago
7 0

The answer is B) F = |q|vBsin.

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a woman is swimming across a cold lake. her body temperature is 98 degrees fahrenheit , and the lake water is at 60 degrees fahr
sergiy2304 [10]

Answer:

the woman has more thermal energy becuase of her body temp.

Explanation:

none

5 0
3 years ago
What is the momentum of a 3 kg bowling ball moving at 3 m/s?
Nataly [62]

Explanation:

<h3>p = mv</h3>

  • <em>p</em> denotes momentum
  • <em>m</em> denotes mass
  • <em>v</em> denotes velocity

→ p = 3 kg × 3 m/s

→ <u>p</u><u> </u><u>=</u><u> </u><u>9</u><u> </u><u>kg</u><u>.</u><u>m</u><u>/</u><u>s</u>

<u>Option</u><u> </u><u>D</u><u> </u><u>is</u><u> </u><u>corre</u><u>ct</u><u>.</u>

5 0
3 years ago
A brick is dropped from a high scaffold. a. What is its velocity after 4.0s ?
Ilia_Sergeevich [38]

Answer:

A: 1.962

B: 3.924

Explanation:

g = G *M /R^2

g = 9.807*M/R^2 the gravitational constant of ground level on earth is about 9.807

g = 9.807*5lbs/R^2 the average brick is about 5 pounds.

g = 9.807*5*10^2.   I'm assuming the height is around ten feet to help you out.

with these numbers plugged in you get an acceleration of 0.4905 a final velocity after 4 seconds 1.962. It's height fallen after 4 seconds is 3.924.

( M = whatever the brick weighs it's not specified in the question)

(R = the distance from the ground or how high the scaffold is)

(hopefully you can just plug your numbers in there hope this helps)

6 0
3 years ago
A motorcycle of mass 100 kilograms travels around a flat, circular track of radius 10 meters with a constant speed of 20 meters
JulijaS [17]

Answer:

100/10 = 10 , 10 × 10 = 100÷20 = 5

I'm pretty sure its wrong

8 0
3 years ago
Read 2 more answers
What is the specific heat of an object if it takes 1200 J to raise the temperature of a 20 kg object by 6 degrees C?
ch4aika [34]

Answer:

c=10\ J/kg^{\circ} C

Explanation:

Given that,

Heat required, Q = 1200 J

Mass of the object, m = 20 kg

The increase in temperature, \Delta T=6^{\circ} C

We need to find the specific heat of the object. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C

So, the specific heat of the object is 10\ J/kg^{\circ} C.

5 0
2 years ago
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