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tia_tia [17]
3 years ago
9

Under what conditions will an object experience a gravitational force?

Physics
1 answer:
Radda [10]3 years ago
6 0
A) an object with mass > 0 in a gravitational field
b) an object with an electric charge not 0 in an electric field
c) a moving object with an electric charge not 0 in a magnetic field
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What feelings do learned helplessness and depression have in common?
8_murik_8 [283]

Answer:

Learned helplessness can have a profound impact on mental health and well-being. People who experience learned helplessness are also likely to experience symptoms of depression, elevated stress levels, and less motivation to take care of their physical health. Not everyone responds to experiences the same way.

Explanation:

Brainlist will be very accepted ;D *wink*

6 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
What acceleration will a force of 20 newtons cause if applied to a go-kart with a mass of 20 kilograms?
iren2701 [21]

Answer:

1m/s is the acceleration used. C

Explanation:

please mark brainliest

3 0
3 years ago
How are light waves used to bring far away objects into view and how does the eye translate them?
adelina 88 [10]

Answer:

When focused light is projected onto the retina, it stimulates the rods and cones. The retina then sends nerve signals are sent through the back of the eye to the optic nerve. The optic nerve carries these signals to the brain, which interprets them as visual images.

Explanation:

Hope it will help u

5 0
3 years ago
A spring with a spring constant of 120 J/m2 is fixed to a wall, free to oscillate. On the other end, a ball with a mass of 1500
Neporo4naja [7]

Answer:

A. 4.47 m/s

Explanation:

As the ball oscillates, it mechanical energy, aka the total kinetic and elastics energy stays the same. For the ball to be at maximum speed, its elastic energy i 0 and vice versa. When the ball is at rest, its kinetic energy is 0 and its elastic energy is at maximum at 50 cm, or 0.5 m

1500 g = 1.5 kg

E_e = E_k

kx^2/2 = mv^2/2

120*0.5^2/2 = 1.5*v^2/2

15 = 0.75v^2

v^2 = 15 / 0.75 = 20

v = \sqrt{20} = 4.47 m/s

5 0
3 years ago
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