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barxatty [35]
3 years ago
6

A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. What is the vertical compon

ent of the cannonball’s velocity? What is the horizontal component of the cannonball’s velocity?
473.8 m/s; 473.8 m/s
-525.2 m/s; 435.5 m/s
0 m/s; 670 m/s
-378 m/s; 378 m/s
Physics
1 answer:
elixir [45]3 years ago
6 0

Answer:

<h3>473.8 m/s; 473.8 m/s</h3>

Explanation:

Given the initial velocity U = 670m/s

Horizontal velocity Ux = Ucos theta

Vertical component of the cannon velocity Uy = Usin theta

Given

U = 670m/s

theta = 45°

horizontal component of the cannonball’s velocity = 670 cos 45

horizontal component of the cannonball’s velocity = 670(0.7071)

horizontal component of the cannonball’s velocity = 473.757m/s

Vertical component of the cannonball’s velocity = 670 sin 45

Vertical component of the cannonball’s velocity  = 670 (0.7071)

Vertical component of the cannonball’s velocity  = 473.757m/s

Hence pair of answer is 473.8 m/s; 473.8 m/s

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A vector is 253m long and points in a 55.8degree direction. Find the y-component of the vector.
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209 m

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What is the current when the resistance is 5 ohms and the voltage is 10 volts?
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The electron gun in a television tube accelerates electrons (mass = 9.11x10^-31 kg, charge = 1.6 x 10^-19C) from rest to 3 x10^7
stich3 [128]

Answer:

Electric field acting on the electron is  127500 N/C.

Explanation:

It is given that,

Mass of an electron, m=9.11\times 10^{-31}\ kg

Charge on electron, q=1.6\times 10^{-19}\ C

Initial speed of electron, u = 0

Final speed of electron, v=3\times 10^7\ m/s

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}

a=2.25\times 10^{16}\ m/s^2

According to Newton's law, force acting on the electron is given by :

F = ma

F=9.1\times 10^{-31}\times 2.25\times 10^{16}

F=2.04\times 10^{-14}\ N

Electric force is given by :

F = q E, E = electric field

E=\dfrac{F}{q}

E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.

8 0
3 years ago
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