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Anna35 [415]
3 years ago
7

What formula allows you to calculate the x component of a projectile?

Physics
2 answers:
Kay [80]3 years ago
8 0

X-component of a projectile in flight =

(initial x-component)

plus

(initial horizontal component of velocity) x (flight time so far)

RideAnS [48]3 years ago
4 0

Answer:

Explanation:

X- component = Horizontal component

Initial velocity = U

X- component =Ucos x angle

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Points A and B lie within a region of space where there is a uniform electric field that has no x- or z-component; only the y-co
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Answer:

(a) Ey is negative

(b) The magnitude of the electric field is E = 171.429 V/m

(c) The potential difference between points B and C is 17.1429 V

Explanation:

(a) Here, we have the potentials given by;

V_A - V_B = +12.0V with point A at y = 8.00 cm and point B at point y = 15.0 cm

where point B is at a higher potential than point A, that is the electric potential is from;

B with y = 15.0 cm to A with y = 8.0 cm which means

E_y decreases as y increases or E_y  is negative.

(b) The magnitude of the electric field is given by

The work done to move a charge from B to A is

W_{BA} = - \Delta U where

\Delta U = U_a -U_b = q_0E(y_b-y_a)

V_{BA} = \frac{\Delta U}{q_0} = \frac{q_0E(y_b-y_a)}{q_0}  = E(y_b-y_a)

∴ E = \frac{V_{BA}}{(y_b-y_a)}

E = \frac{12 \hspace{0.09cm}V}{(0.015\hspace{0.09cm} m -0.008\hspace{0.09cm} m)}

E = 171.429 V/m

(c) Here we have point C x = 5.00 cm and y = 5.00 cm

Therefore we have the distance from B to C given by

y_b-y_c = 15.00 \hspace{0.09cm}cm - 5.00  \hspace{0.09cm}cm = 10.00 \hspace{0.09cm} cm

Where 10.00 cm = 0.01 m

E = V/Δy

Therefore, V = Δy·E

For V_{BC}, Δy = y_b-y_c  = 0.01 \hspace{0.09cm} m and we have,

V_{BC} = E\times (y_b-y_c)

V_{BC} = 171.429\times (0.015-0.005) = 17.1429\hspace{0.09cm}V

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3 years ago
Two waves collide and the temporary combined waves that results is smaller that the original wave.
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A.

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Visible Light Spectrum (VLS).
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Now the same particle is removed from the thread and placed over the center of a charged plate. Are there any conditions under w
labwork [276]

Answer:

changing the direction of the electric potential, we can get the particle to be in balance between the electric force, the weight and the thrust.

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In these cases we can use Newton's second law, where the acceleration is zero

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We replace

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Therefore, by changing the direction of the electric potential, we can get the particle to be in balance between the electric force, the weight and the thrust.

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The correct answer should be c.The kinetic energy of the water molecules decreases.

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