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2 years ago

A pendulum with a period of 1 s on earth where the acceleration

Physics

1 answer:

Svetradugi [14.3K]2 years ago

6 0

A pendulum with a period of ones on earth where the acceleration

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An electron is released from rest in a weak electric field given by vector E = -1.30 10-10 N/C Äµ. After the electron has travel

Murljashka [212]

Answer:

v = 6.45 10⁻³ m / s

Explanation:

Electric force is

F = q E

Where q is the charge and E is the electric field

Let's use Newton's second law to find acceleration

F- W = m a

a = F / m - g

a = q / m E g

Let's calculate

a = -1.6 10⁻¹⁹ / 9.1 10⁻³¹ (-1.30 10⁻¹⁰) - 9.8

a = 0.228 10² -9.8

a= 13.0 m / s²

Now we can use kinematics, knowing that the resting parts electrons

v² = v₀² + 2 a y

v =√ (0 + 2 13.0 1.6 10⁻⁶)

v = 6.45 10⁻³ m / s

4 0

2 years ago

BEST ANSWER GETS MARKED AS BRAINLIEST!!!! Can someone PLZ help me come up with a study method, no matter what I try, I just can'

Leni [432]

For physics, I would recommend to just keep doing practice problems and reviewing notes. Repetition of the same concepts will help drill it into your brain. Hope that helps!

8 0

3 years ago

Read 2 more answers

What creates a magnetic field?

kolbaska11 [484]

Answer:

moving electric charges

4 0

2 years ago

Two long parallel wires carry currents of 20 a and 5.0 a in opposite directions. the wires are separated by 0.20 m. what is the

Alex777 [14]

The two wires carry current in opposite directions: this means that if we see them from above, the magnetic field generated by one wire is clock-wise, while the magnetic field generated by the other wire is anti-clockwise. Therefore, if we take a point midway between the two wires, the resultant magnetic field at this point is just the sum of the two magnetic fields, since they act in the same direction.

Therefore, we should calculate the magnetic field generated by each wire and then calculate their sum. We are located at a distance r=0.10 m from each wire.

The magnetic field generated by wire 1 is:
$B_1= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(20 A)}{2 \pi (0.10 m)}= 4 \cdot 10^{-5}T$

The magnetic field generated by wire 2 is:
$B_2= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(5.0 A)}{2 \pi (0.10 m)}= 1 \cdot 10^{-5}T$

And so, the resultant magnetic field at the point midway between the two wires is
$B=B_1 + B_2 = 4 \cdot 10^{-5} T + 1 \cdot 10^{-5}T=5 \cdot 10^{-5} T$

8 0

3 years ago

A baseball bat is 32 inches (81.3 cm) long and has a mass of 0.96 kg. Its center of mass is 22 inches (55.9 cm) from the handle

s344n2d4d5 [400]

Answer:

0.24 kgm²

Explanation:

$L$ = length of the bat = 81.3 cm = 0.813 m