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Lelechka [254]
3 years ago
9

Which describes an object in projectile motion? Check all that apply.

Physics
1 answer:
Elodia [21]3 years ago
5 0

Answer:

Gravity acts to pull the object down.

The object’s inertia carries it forward.

The path of the object is curved.

Explanation:

The motion of a projectile consists of two separate motions:

- A uniform motion along the horizontal direction, where the velocity is constant; since there are no forces along this direction, the velocity does not change, and so the object continues its motion for inertia --> so, the statement  "The object’s inertia carries it forward" is true.

- A uniformly accelerated motion along the vertical direction, with a constant downward acceleration (g=9.8 m/s^2, acceleration due to gravity). So, the vertical velocity changes, due to the presence of the gravity that acts to pull the object down.

- As a result of the combination of these two motions, the object follows a curved path (in particular, it is a parabolic path).

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Read 2 more answers
At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r
Oksana_A [137]

Answer:

<u>For proton:</u>

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

<u>For electron:</u>

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s    

Explanation:

<u>For a proton we have</u>:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.                

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

W = \Delta Vq    (1)                                        

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C    

V: is the potential    

Also, the work is equal to:

W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}     (2)      

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

v_{a}: is the velocity in the point a

v_{b}: is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

\Delta Vq = \frac{1}{2}mv_{a}^{2}

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}

v_{a} = 2.9 \cdot 10^{4} m/s

<u>For an electron we have</u>:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}

Since v_{a} = 0 (starts from rest) and m_{e} = 9.1x10⁻³¹ kg (electron's mass), we have:

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}    

v_{b} = 1.2 \cdot 10^{6} m/s

I hope it helps you!

6 0
4 years ago
How much kinetic energy does a care 1000 kg car moving at 7 m/s have?
aalyn [17]
Hey there!

<span>How much kinetic energy does a care 1000 kg car moving at 7 m/s have?

Answer:
</span><span>A. 24500 J

Hope this helps
Have a great day (:
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Answer:

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Explanation:

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Answer:

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Explanation:

hope this helps....

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