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3 years ago

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

Physics

2 answers:

lesya [120]3 years ago

8 0

Answer:

169.4 m/s

Explanation:

Given that the angle of projectile is θ_1 =450°

The speed of body at maximum height is U cosθ_1 = 150 m/s

The angle in second trail is θ_2 =37°

From the given data U cosθ_1 = 150 m/s

U = 150 m/s / cosθ_1

= 150m/s / cos45°

=212.13 m/s

The velocity of the projectile at maximum height in second trail= Ucos(θ_2)

=212.13 m/s×cos37°

=169.4 m/s

juin [17]3 years ago

7 0

Answer:

Explanation:

Given

For first case

launch angle $\theta =45^{\circ}C$

at highest point $h=150 m/s$

$150=u\cos 45$

$u=\frac{150}{\cos 45}=212.132 m/s$

For second case

$\theta _2=37^{\circ}C$

at highest Point velocity is $u\cos \theta _2$

$=212.132\times \cos 37$

$=169.41 m/s$

as there is no acceleration in x direction therefore horizontal velocity is same

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Answer:

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$K = -22.41 +273.15 =250.739 K$

Explanation:

For this case we want to create a function like this:

$Z = a C + b$

Where Z represent the degrees for the Z scale C the Celsius grades and tha valus a and b parameters for the model.

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The melting point of iron is 1538 °C

We know the following equivalences:

-195.8 °C = 0 °Z

1538 °C = 1000 °Z

Let's say that one point its (1538C, 1000 Z) and other one is (-195.8 C, 0Z)

So then we can calculate the slope for the linear model like this:

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And now for the slope we can use one point let's use for example (-195.8C, 0Z), and we have this:

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$b = 0.577*195.8 =112.931 Z$

So then our lineal model would be:

$Z = 0.577 C +112.931$

Part A

The boiling point of water is 100C so we just need to replace in the model and see what we got:

$Z = 0.577*(100) +112.931=170.631 Z$

Part B

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$Z-112.931 = 0.577 C$

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$K = -22.41 +273.15 =250.739 K$

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Answer

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Answer:

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