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Oduvanchick [21]

3 years ago

13

At lift-off, an astronaut on the space shuttle experiences an acceleration of approximately 35 m/s2 upward. What force does an 8

0 kg astronaut experience during this acceleration?
I need answers ASAP!!

1 answer:

Naily [24]3 years ago

4 0

Acceleration=a=35m/s^2
Mass=m=80kg

According to Newton's second law

$\\ \sf\longmapsto F=ma$

$\\ \sf\longmapsto F=80(35)$

$\\ \sf\longmapsto F=2800N$

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Step-by-step explanation:

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3 years ago

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beks73 [17]

Answer:

The Value of $a=\frac{15}{4}$ .

Step-by-step explanation:

We have Named the figure please find the attachment for your reference.

Given:

PR = y

QR = a

RS = b

PS = z

PQ = x

QS = 15

∠P = 90°

∠R = 90°

∠Q = 60°

∠S = 30°

We need to find the Value of 'a'.

Solution:

Now we know that:

In Δ PQS

∠P = 90°

∠S = 30°

Now we know that;

$sin\ \theta = \frac{opposite\ side}{Hypotenuse}$

$sin \ S= \frac{PQ}{QS}$

Substituting the given values we get;

$sin\ 30\°=\frac{x}{15}$

Now we know that;

$sin\ 30\° = \frac12$

So we can say that;

$\frac{1}{2}=\frac{x}{15}\\\\x=\frac{15}{2}$

Now In Triangle PQR.

∠R = 90°

∠Q = 60°

So we can say that;

$Cos \theta = \frac{adjacent \ Side}{Hypotenuse}\\$

$Cos\ Q = \frac{QR}{PQ}$

Substituting the given values we get;

$cos 60\°= \frac{a}{x}$

Now we know that;

$cos 60\°= \frac12$

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So substituting the values we get;

$\frac{1}{2}=\frac{a}{\frac{15}{2}}$

By Using Cross Multiplication we get;

$a= \frac{1}{2}\times\frac{15}{2}\\\\a=\frac{15}{4}$

Hence The Value of $a=\frac{15}{4}$ .

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