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bija089 [108]
3 years ago
5

The breaks on a 15,680 N car exert a stopping force of 640 N.

Physics
1 answer:
BartSMP [9]3 years ago
8 0
The brakes on a 15,680 N car exert a stopping force of 640 N. The car's velocity changes from 20.0 m/s to 0 m/s.
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How many hydrogens are in 3 molecules of H20?
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Answer:

1 atom of oxygen

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Each water molecule contains 3 atoms making the H2O formula

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A student is trying to determine the acceleration of a feather as she drops it to the ground. if the student is looking to achie
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The coordinate system should have the origin at the point where the feather is dropped and the downward direction is to be taken as positive.

All falling bodies experience acceleration towards the center of the Earth due to the force of gravitational attraction exerted on the object by the Earth. A feather, when dropped experiences an acceleration in the downward direction. Since the acceleration of the feather is in the downward direction, a feather, when dropped with zero initial velocity, has its velocity vector directed in the direction of its acceleration.

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Which factors allow for life on Earth​
Elan Coil [88]

Are there supposed to be multiple choices for this question?

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3 years ago
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A uniform electric field of magnitude 7.0 ✕ 104 N/C passes through the plane of a square sheet with sides 5.0 m long. Calculate
Vadim26 [7]

Answer:

1.52*10^6 Nm^2/C

Explanation:

Given that:

Electrical field E = 7.0 * 10^{-4}N/C

square side l = 5.0 m

Area A = 5.0 * 5.0

= 25.0 m²

Angle ( θ ) between area vector and E = (90° - 60°)

= 30°

The flux \phi_E can now be determined by using the expression

\phi_E = E*A*Cos \theta

\phi_E = 7.0 * 10^{-4}N/C *25.0m*Cos 30^0

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\phi_E = 1.52*10^6 Nm^2/C

5 0
3 years ago
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A ship leaves a harbor and sails at 23.5 ∘ to the north of due west. After traveling 575 km, how far west is the ship from the h
solniwko [45]

Answer:

D = 527.31 Km

Explanation:

given,

angle of ship, θ = 23.5° N of W

distance travel in the direction = 575 Km

Distance of ship in west from harbor = ?

now,

Distance of the ship in the west direction

D = d cos θ

d = 575 Km

θ = 23.5°

inserting all the values

D = 575 x cos 23.5°

D = 575 x 0.91706

D = 527.31 Km

Hence, the distance travel by the ship in west from harbor is equal to D = 527.31 Km  

6 0
3 years ago
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