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lorasvet [3.4K]

3 years ago

12

Metals can be described as

1 answer:

shutvik [7]3 years ago

7 0

They tend to be lustrous, ductile, malleable, and good conductors of electricity, while nonmetals are generally brittle (for solid nonmetals), lack lustre, and are insulators.

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I’ll give brainliest and 10 points

schepotkina [342]

I think it is D. I hope this helps

4 0

3 years ago

A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

5 0

3 years ago

If a rock were dropped from a building it would hit the ground with a certain kinetic energy and velocity. If it fell from a roo

STALIN [3.7K]

Answer:

no

Explanation:

because if you test it they hit the ground at the same time.

5 0

3 years ago

An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Dete

dezoksy [38]

Answer:43.34 m

Explanation:

Given

acceleration(a) $=2 m/s^2$

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

$v=0+2\times 6=12 m/s$

After this object will start moving under gravity

height reached in first 6 s

$s=ut+\frac{at^2}{2}$

$s=0+\frac{2\times 6^2}{2}$

s=36 m

After fuel run out distance traveled in upward direction is

$v^2-u^2=2as_0$

here v=0

u=12 m/s

$a=9.8 m/s^2$

$0-12^2=2(-9.8)(s)$

$s_0=\frac{144}{2\times 9.8}=7.34 m$

$s+s_0=36+7.34=43.34 m$

7 0

3 years ago

On the planet Xenophous a 1.00 m long pendulum on a clock has a period of 1.32 s. What is the free fall acceleration on Xenophou

Vilka [71]

The period of the pendulum is given by the following equation

T = 2π * sqrt (L/g)

Where g is the gravity (free fall acceleration)

L is the longitude of the pendulum

T is the period.

We find g.............> (T /2π)^2 = L/g

g = L/(T /2π)^2...........> g = 22.657 m/s^2

3 0

4 years ago