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4 years ago

Which of the following is a pair of vector quantities?

Physics

2 answers:

hammer [34]4 years ago

6 0

Answer:

velocity and displacement answer

Explanation:

thanks me

ale4655 [162]4 years ago

5 0

Velocity and displacement

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Which statements are true about the speed of a wave?

Pie

I guess b tell me if I’m wrong

4 0

3 years ago

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An airplane moving horizontally at 150 m/s drops a package from an altitude is 490 m. How long does it

Debora [2.8K]

Answer:

640

Explanation:

cause you should just add the 2

5 0

3 years ago

A particle moves with acceleration function a(t) = 5 + 4t - 2t^2. Its initial velocity v(0) = 3 m/s and its initial displacement

eimsori [14]

Answer:

Its position after 4 seconds is 62 meters.

Explanation:

It is given that,

The acceleration of the particle is given by equation :

$a(t)=5+4t-2t^2$

Also, $a=\dfrac{dv}{dt}$

$v=\int\limits {a.dt}$

$v=\int\limits {(5+4t-2t^2).dt}$

$v=5t+2t^2-\dfrac{2}{3}t^3+c$

At t = 0, $v(0)=3\ m/s$ . So, c = 3

$v=5t+2t^2-\dfrac{2}{3}t^3+3$

Also, $v=\dfrac{ds}{dt}$ , s is the position

$s=\int\limits {v.dt}$

$s=\int\limits {(5t+2t^2-\dfrac{2}{3}t^3+3).dt}$

$s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+c'$

At t = 0, $s(0)=10\ m$ . So, c' = 10

$s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+10$

At t = 4 s

$s=\dfrac{5}{2}(4)^2+\dfrac{2}{3}(4)^3-\dfrac{(4)^4}{6}+3(4)+10$

s = 62 m

So, at t = 4 seconds the position of the particle is 62 meters. Hence, this is the required solution.

5 0

3 years ago

10 POINTS BE FAST

lapo4ka [179]

Answer:

I think it's c or A hopes this help

4 0

3 years ago

42,43 and 44 please

Mariana [72]

42) The sailboat travels east with velocity $v_e=30 mph$ , while the current moves south with speed $v_s=30 mph$ . Since the two velocities are perpendicular to each other, he resultant velocity will be given by the Pytagorean theorem:
$v= \sqrt{v_e^2+v_s^2}= \sqrt{(30)^2+(30)^2}=42 mph$
and the direction is in between the two original directions, therefore south-east. So, the correct answer is
D) 42 mph southeast

43) Since the light moves by uniform motion, we can calculate the distance corresponding to one light year by using the basic relationship between velocity, space and time. In fact, we know the velocity:
$v=300,000,000 m/s=3 \cdot 10^8 m/s$
and the time is one year, corresponding to:
$t=32,000,000 s=3.2 \cdot 10^7 s$
therefore, the distance corresponding to one light year is:
$S=vt=(3\cdot 10^8 m/s)(3.2 \cdot 10^7 s)=9.6 \cdot 10^{15}m$
Therefore, the correct answer is D.

44) For the purpose of the problem, we can assume that the light travels instantaneously from the flash to us (because the distances involved are very small), so the time between the flash and the thunder corresponds to the time it took for the sound to travel to us.
The speed of sound is
$v=1100 ft/s=750 mph=335 m/s$
And since the time between the flash and the thunder is t=3 s, the distance is
$S=vt=(335 m/s)(3 s)=1005 m=0.62 miles$
Therefore, the correct answer is A) 3/5 mile.

6 0

3 years ago