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3 years ago

6. Velocity is a ____________. a) vector b) value c) arrow d) none of these

Physics

2 answers:

erastova [34]3 years ago

6 0

A) vector because a vector is giving the magnitude or you could kind of view as a size with a direction.

Olenka [21]3 years ago

4 0

<span>the speed of something in a given direction. so i think none of these</span>

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What would be the speed of an object just before hitting the ground if dropped 91.5 meters

Aleks04 [339]

Answer:

about 42.35 m/s

Explanation:

Use the equation for accelerated motion (g), and with zero initial velocity that doesn't include time:

$v_f^2=v_i^2+2\,a\,\Delta x$

which for our case would reduce to:

$v_f^2=v_i^2+2\,a\,\Delta x\\v_f^2=0+2\,9.8\,(91.5)\\v_f^2= 1793.4\\v_f=\sqrt{1793.4} \\v_f \approx 42.35$

then the velocity just before hitting would be about 42.35 m/s

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3 years ago

In flight, a rocket is subjected to four forces; weight, thrust, lift, and drag. Forces are vector quantities that have both a m

Flura [38]

Answer:

thrust

Explanation:

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2 years ago

Read 2 more answers

A 0.37-kg object connected to a light spring with a force constant of 23.2 N/m oscillates on a frictionless horizontal surface.

mars1129 [50]

Answer:

a) v = 31.67 cm / s , b) v = -29.36 cm / s , c) v= 29.36 cm/s, d) x = 3.46 cm

Explanation:

The angular velocity in a simple harmonic movement is

w = √ K / m

w = √ 23.2 / 0.37

w = 7,918 rad / s

a) the expression against the movement is

x = A cos (wt + Ф)

Speed is

v = dx / dt = - A w sin (wt + Ф)

The maximum speed occurs for cos = ± 1

v = A w

v = 4.0 7,918

v = 31.67 cm / s

b) as the object is released from rest

0 = -A w sin (0+ Фi)

sin Ф = 0

Ф = 0

The equation is

x = 4.0 cos (7,918 t)

v = -4.0 7,918 sin (7,918 t)

v = - 31.67 sin (7.918t)

Let's look for the time for a displacement of x = 1.5 cm, remember that the angles must be in radians

7,918 t = cos⁻¹ 1.5 / 4.0

t = 1,186 / 7,918

t = 0.1498 s

We look for speed

v = -31.67 sin (7,918 0.1498)

v = -29.36 cm / s

c) if the object passes the equilibrium equilibrium position again at this point the velocity has the same module, but the opposite sign

v = 29.36 cm / s

d) let's look for the time for the condition v = v_max / 2

31.67 / 2 = 31.67 sin ( 7,918 t)

7.918t = sin⁻¹ 0.5

t = 0.5236 / 7.918

t = 0.06613

With this time let's look for displacement

x = 4.0 cos (7,918 0.06613)

x = 3.46 cm

6 0

3 years ago

On the Moon, the acceleration due to the effect of gravity is only about 1/6 of that on Earth. An astronaut whose weight on Eart

vazorg [7]

And on the moon, the with is 595/6 N

3 0

3 years ago

At what height h above the ground does the projectile have a speed of 0.5v?

maw [93]

Answer:

$h=\dfrac{3v^2}{8g}$

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

$v=u+at$

$v=u-gt$

Initial speed of the projectile is v and final speed is 0.5 v.

$0.5v=v-gt$

$t=\dfrac{v}{2g}$

g is the acceleration due to gravity

Let h is the height above the ground. Using the second equation of motion as :

$h=vt-\dfrac{1}{2}gt^2$

$h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2$

$h=\dfrac{3v^2}{8g}$

So, the height of the projectile above the ground is $\dfrac{3v^2}{8g}$ . Hence, this is the required solution.

6 0

3 years ago