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3 years ago

A projectile is thrown at an angle 30° from horizontal. Which statement about its vertical component of velocity is true?

Physics

2 answers:

Veseljchak [2.6K]3 years ago

8 0

idk it's wrong chose ...remains constant...

Lady_Fox [76]3 years ago

7 0

Vertical component of velocity at the highest point becomes zero but horizontal component of velocity remains constant throughout the projectile motion.
Hope this helps!

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the external circuit is directed away from the positive terminal and toward the negative terminal of the battery

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3 years ago

Modern oil tankers weigh over a half-million tons and have lengths of up to one-fourth mile. Such massive ships require a distan

sergiy2304 [10]

(a) $-1.46\cdot 10^{-4} m/s^2$

The average acceleration of the ship is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time elapsed

Here we have:

$u=34 km/h =9.44 m/s$ is the initial velocity

v = 0 is the final velocity

$t=18 min =64800 s$ is the time elapsed

Substituting, we find

$a=\frac{0-9.44 m/s}{64800 s}=-1.46\cdot 10^{-4} m/s^2$

(b) 4.72 m/s

Assuming the acceleration is uniform, the average velocity of the ship is given by:

$v_{avg} = \frac{v+u}{2}$

where

v is the final velocity

u is the initial velocity

Here we have:

v = 0

u = 9.44 m/s

So the average velocity of the ship is

$v_{avg} = \frac{0+9.44 m/s}{2}=4.72 m/s$

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4 years ago

While running a 100m race, a runner runs from the 20m to 30m mark in 77 frames of a video record. If the video camera recorded d

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Answer:

Speed of the runner during video interval is 6.49 m/s

Explanation:

According to the problem,

Number of frames recorded by camera in 1 second = 50

Time takes by camera to record 1 frame = (1/50) s

Time taken by camera to record 77 frames, t = $\frac{1}{50}\times 77$ s

Distance covered by the runner during the video recording, d = 10 m

Speed, v = $\frac{Distance}{time}=\frac{d}{t}$

Substitute the values of d and t in the above equation.

v = $\frac{10}{\frac{77}{50} }$

v = 6.49 m/s

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4 years ago

Find the work done by the force F = xyi +(y-x)j over the straight line from (-1,1)to (3,-3). The amount of work done is ___?

dalvyx [7]

Answer:

amount of work done is $W = \frac{-4}{3}$

Explanation:

Formula for work done by force field

$W = \int F. dr = \int_{t_o}^{t_1} f(r(t)) r'(t) dt$

where

r(t) is parametrization of line

as it is straight line so

$r(t) = 1- t) r_o + tr_1 0 \leq t \leq 1$

thus,

r(t) = (1-t)(-1,1) + t(3,-3)

= (-1+t,1-t) + (3t - 3t)

= (-1+t +3t, 1-t-3t)

r(t) = (4t -1, 1- 4t)

r'(t) = (4,-4)

putting value in above integral

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$= \int_{0}^{1} (-16 t^2 + 8t -1,2-8t) .(4,-4) dt$

$=\int_{0}^{1} (4(-16t^2 +8t -1) -4(2-8t)) dt$

$= 4[ -16 \frac{t^3}{3} + 16\frac{t^2}{2} - 3t]_{0}^{1}$

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4 years ago