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Y_Kistochka [10]
2 years ago
6

Tell me the max amount you should owe on this card.

Physics
1 answer:
olga nikolaevna [1]2 years ago
8 0

Answer:

the max is 2,500 or less

Explanation:

because you cant owe anymore

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Only 5 question plz answer
Aleks04 [339]
It is subduction;) good luck on the others my man
4 0
3 years ago
Read 2 more answers
Your car is initially at rest when you hit that gas and the car begins to accelerate. The forward force of the car is 5630 N whi
Neporo4naja [7]
330...................
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2 years ago
The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on a
vampirchik [111]

Answer:

F = 1.5 \times 10^{-16} N

this force is 1.68 \times 10^{13} times more than the gravitational force

Explanation:

Kinetic Energy of the electron is given as

KE = 1 keV

KE = 1 \times 10^3 (1.6 \times 10^{-19}) J

KE = 1.6 \times 10^{-16} J

now the speed of electron is given as

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

now we have

v = \sqrt{\frac{2 KE}{m}}

v = 1.87 \times 10^7 m/s

now the maximum force due to magnetic field is given as

F = qvB

F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})

F = 1.5 \times 10^{-16} N

Now if this force is compared by the gravitational force on the electron then it is

\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}

\frac{F}{F_g} = 1.68 \times 10^{13}

so this force is 1.68 \times 10^{13} times more than the gravitational force

4 0
3 years ago
In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it
bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

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

$\Rightarrow v = \sqrt{2gh}$

      $=\sqrt{2 \times 9.8 \times 0.35}$      (h = 35 cm = 0.35 m)

      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}

      $=\frac{0.175}{0.525}+0$

     = 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

 

3 0
3 years ago
What is the net force on a car if the force of friction is 15 N and the forward force due to the engine is 20 N?
anygoal [31]
D. 35n forwards....................
8 0
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