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3 years ago

13

A projectile is launched with an initial velocity of 25m/s at 35 degrees above the horizontal. assuming that the projectile retu

rns to the same height, what is the final velocity of this projectile

1 answer:

Dovator [93]3 years ago

3 0

Speed is the same as the initial: 25m/s.

*if* you need vectors though:

final velocity = (25*cos(35), -25*sin(35) ) m/s

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Which property do metalloids share with nonmetals

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Metalloids are all solid at room temperature. Some metalloids, such as silicon and germanium, can act as electrical conductors under the right conditions, thus they are called semi-conductors. Silicon for example appears lustrous, but is not malleable or ductile (it is brittle - a characteristic of some nonmetals).

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4 years ago

Read 2 more answers

On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole

ohaa [14]

The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
$F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\$
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. $\alpha$ is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
$g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\
g_0=g'\\
g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\
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w^2r=g_0(a-1)
$
Our final equation is:
$g=g_0-g_0(a-1)cos^2(\alpha)$
Colatitude is:
$\alpha_c=90^\circ-\alpha$
The answer is:
$g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)$

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4 years ago

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4 years ago

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3 years ago

We want to find how much charge is on the electrons in a nickel coin. Follow this method. A nickel coin has a mass of about 4.2

monitta

Answer:

The number of atoms is $N = 4.37*10^{22} \ atoms$

Explanation:

From the question we are told that

The mass of coin is $m_n = 4.2g$

Number of atom in one mole = $n =6.02*10^{23} \ atoms$

Molar mass of nickel $M = 57.8g$

Now the relation to obtain the number of atom in the nickel coin is

$N = \frac{Mass \ of Nickel\ coin}{Molar \ mass\ of nickel } * No\ of\atoms \ in \ \ one\ mole\ of\ nickel$

$= \frac{4.2}{57.8}* 6.02*10^{23}$

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3 years ago