An alloy that contains mainly copper and tin is

You have a chart that illustrates a series of species with their common ancestor. This chart is a

emmasim [6.3K]

The answer is *C. Cladogram*.

8 0

3 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

In a rigid container, as the temperature of a gas decreases, the pressure of the gas will decrease. t f

Sergio039 [100]

The answer is true: the pressure of a gas will decrease as temperature decreases in a rigid container.

This is one of the central gas laws called the Gay-Lussac law that states for a given gas at a constant volume, the pressure of the gas is directly proportional to its temperature. We also know that as temperature reduces, so too does molecular interaction. Increased temperature results in increased pressure, and decreased temperature therefore results in decreased pressure.

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3 years ago

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If the magnitude of the resultant force is to be 500n, directed along the positive y axis, determine the magnitude of force f an

scoundrel [369]

The answer is 0=45.2 degrees

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Sound waves are mechanical waves in which the particles in the medium vibrate in a direction parallel to the direction of energy

ch4aika [34]

C longitudnal waves

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