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barxatty [35]
2 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]2 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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ikadub [295]

Answer:

d = 2.26 km

Explanation:

Let the child is moving with speed same as the speed of water flow

So here the position of child with respect to flow must be zero

And if the boat start at an angle with the vertical

so its relative speed with flow of water is given as

v_x = 24.8 sin\theta

v_y = 24.8 cos\theta

now the time to reach the child is given as

\frac{0.6}{24.8 cos\theta} = \frac{2.5}{24.8 sin\theta }

so now we have

\theta = 76.5 degree

So the time to catch the child is given as

t = \frac{0.6}{24.8 cos78.2}

t = 0.104 h

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distance moved by the boat in upstream direction given as

x = (24.8 sin 74.8 - 3.1)(0.104)

x = 2.18 km

In y direction the displacement of boat is

y = 0.6 km

net displacement of the ball is given as

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d = \sqrt{2.18^2 + 0.6^2}

d = 2.26 km

4 0
3 years ago
A 13.4-mH inductor carries a current i = <img src="https://tex.z-dn.net/?f=I_%7Bmax%7D" id="TexFormula1" title="I_{max}" alt="I_
Digiron [165]

The voltage across an inductor ' L ' is

V = L · dI/dt .

I(t) = I(max) sin(ωt)

dI/dt = I(max) ω cos(ωt)

V = L · ω · I(max) cos(ωt)

L = 1.34 x 10⁻² H

ω = 2π · 60 = 377 /sec

I(max) = 4.80 A

V = L · ω · I(max) cos(ωt)

V = (1.34 x 10⁻² H) · (377 / sec) · (4.8 A) · cos(377 t)

<em>V = 24.25 cos(377 t)</em>

V is an AC voltage with peak value of 24.25 volts and frequency = 60 Hz.

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3 years ago
Which object had more potential energy when it was lifted to a distance of 1000 centimeters? Show your calculation.
RSB [31]

Explanation:

We Know That

POTENTIAL ENERGY= MASS*g*HEIGHT

When the objects are lifted to same height then the object with heavier mass would have the highest potential energy

.

5 0
2 years ago
Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration
LenaWriter [7]

Answer:

c.5.6m/s^2

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

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Using the formula

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Hence, the acceleration of cheetah=5.6m/s^2

5 0
3 years ago
Which statement is correct?
miv72 [106K]
'A' is correct. B, C, and D are false statements.
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