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grigory
2 months ago
8

A car with an initial velocity of 16.0 meters per second east slows uniformly to 6.0 meters per second east in 4.0 seconds. What

is the acceleration of the car during this 4.0-second interval?
Physics
1 answer:
serg [3.5K]2 months ago
5 0
(6-16)/4.0=-2.5 m/s²
The car's acceleration is -2.5 m/s²
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A negatively charged object is located in a region of space where the electric field is uniform and points due north. the object
serg [3582]
- The greatest potential energy increase occurs when the charge travels north. This happens because the charge is negative, which means it gains potential energy when moving in the same direction as the field (in contrast, a positive charge moving along the field loses potential energy, converting it to kinetic energy). The potential energy gained is calculated as the charge multiplied by the distance moved:
\Delta U = e d

- The next largest increase occurs as the charge moves east. Here, the change in potential energy is actually zero since the charge moves perpendicular to the field, traversing points with constant potential. Therefore, there is no variation in potential energy in this case:
\Delta U = 0

- Finally, when the charge moves south, it experiences a reduction in potential energy. This is due to moving against the electric field, and since it is a negative charge, it loses potential energy in this direction, which transforms into kinetic energy. Thus, in this scenario:
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5 0
1 month ago
To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Softa [3030]

Answer:

H = 10.05 m

Explanation:

The stone reaches the top of the flagpole at both t = 0.5 s and t = 4.1 s

therefore, the total duration of the upwards motion above the peak of the pole is provided as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

this indicates the speed at the flagpole's top

at this point we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

the height of the flagpole is stated as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
2 months ago
A proton moves along the x-axis with vx=1.0×107m/s. As it passes the origin, what are the strength and direction of the magnetic
inna [3103]

Answer:

At this position, the magnetic field equals ZERO

Explanation:

The magnetic field produced by a moving charge is described as

B = \frac{\mu_0 qv sin\theta}{4\pi r^2}

Here, we determine the direction of the magnetic field using

\hat B = \hat v \times \hat r

Thus, we find

\hat B = \hat i \times (\hat i + 0\hat j + 0\hat k)

Leading to a magnetic field of ZERO

Consequently, when the charge moves in the same line as the given position vector, the magnetic field will be nonexistent

3 0
1 month ago
In the reaction at Blood Falls, iron and oxygen combine to form iron oxide, which is called rust (water is also present). The re
Maru [3345]
A reactant is any substance that takes part in a chemical reaction. Conversely, a product is what emerges from the chemical transformation. A familiar example of a chemical change is rust formation. In this reaction, oxygen and iron, which serve as the reactants, react to produce a substance known as iron oxide, or rust.
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Which statements describe properties of stars check all that apply
Keith_Richards [3271]

Answer:

Stars generate energy by the process of nuclear fusion.

They are large entities composed of gaseous elements.

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Conversely, it is true that stars predominantly consist of hydrogen and helium (two hydrogen nuclei can fuse to become helium), which implies that a star is essentially an enormous ball of gas without a solid surface suitable for standing on.

As for the presence of water on a star, it is simply impossible. The extreme temperatures found in stars are far too high for water to exist in any liquid state on their surfaces.

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1 month ago
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