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statuscvo [17]
3 years ago
13

A test charge is introduced into the electric field of a charge. It feels a force of 2 F where the electric feild is 4 E. What w

ould the force on the test charge be where the electric feild is 1 E
Physics
1 answer:
Elza [17]3 years ago
3 0

Answer:

0.5 F

Explanation:

since electric field is defined as force per charge!

so, force on test charge is directly proportional to electric field

here new electric field becomes 1/4 of old one

i.e, 4E-->1E

so force is also will become 1/4 of old one

i.e, 2F--> 1/4 × 2F =0.5 F

✌️:)

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A baseball is moving at a speed of 2.2\,\dfrac{\text{m}}{\text{s}}2.2 s m ​ 2, point, 2, space, start fraction, m, divided by, s
Reika [66]

Answer:

 a = 1,008 10⁻³ m / s²

Explanation:

For this exercise, let's use the kinematic relations of accelerated motion

           v² = v₀² - 2 a x

The negative sign is because the acceleration is opposite to the speed, the final speed is zero

          0 = v₀² - 2 a x

          a = v₀² / 2x

   Let's reduce the magnitudes to the SI system

          x = 2.4mm (1m / 10³mm) = 2.4 10⁻³m

Let's calculate

         a = 2.2²/2 2.4 10⁻³

        a = 1,008 10⁻³ m / s²

5 0
3 years ago
Explain why atmospheric pressure changes as atmospheric depth changes
Pavel [41]

The force of the collisions creates pressure on the container. ... Explain why atmospheric pressure changes as depth changes. Atmospheric pressure increases as depth increases because at lower levels of the atmosphere, there is more air above that is being pulled down by gravitational force.

7 0
3 years ago
What pressure, in millimeters of mercury (mm Hg), is equivalent to 2.13 atmospheres
Sati [7]
The "Standard Atmosphere" is 760 mm Hg . 2.13 times that pressure is (2.13 x 760) = 1,618.8 mm Hg.
6 0
3 years ago
A toroid having a square cross section, 5.49 cm on a side, and an inner radius of 18.1 cm has 450 turns and carries a current of
joja [24]

Answer:

(a) 4.27 x 10^-4 Telsa

(b) 3.28 x 10^-4 Telsa

Explanation:

side of square, a = 5.49 cm

inner radius, r = 18.1 cm = 0.181 m

number of turns,N = 450

current, i = 0.859 A

(a)

The magnetic field due to a solenoid due to inner radius is

B = \frac{\mu_{0}Ni}{2\pi r_{inner}}

B = \frac{4\pi\times 10^{-7}\times 450\times 0.859}{2\pi \times 0.181}

B = 4.27 x 10^-4 Telsa

(b)

The outer radius is R = 18.1 + 5.49 = 23.59 cm = 0.236 m

The magnetic field due to the outer radius is

B = \frac{\mu_{0}Ni}{2\pi r_{outer}}

B = \frac{4\pi\times 10^{-7}\times 450\times 0.859}{2\pi \times 0.236}

B = 3.28 x 10^-4 Tesla

7 0
3 years ago
A particle moves along a straight path through a displacement d = 2.5i + cj while a force F = 8.5i + -8.5j acts on it. The displ
Zanzabum

Answer:

Explanation:

Work is defined as the scalar product of force and distance

W=F•d

Given that

F = 8.5i + -8.5j. +×-=-

F=8.5i-8.5j

d = 2.5i + cj

If the work in the practice is zero, then W=0

therefore,

W=F•ds

0=F•ds

0=(8.5i -8.5j)•(2.5i + cj)

Note that

i.i=j.j=k.k=1

i.j=j.i=k.i=i.k=j.k=k.j=0

So applying this

0=(8.5i -8.5j)•(2.5i + cj)

0= (8.5×2.5i.i + 8.5×ci.j -8.5×2.5j.i-8.5×cj.j)

0=21.25-8.5c

Therefore,

8.5c=21.25

c=21.25/8.5

c=2.5

7 0
3 years ago
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