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4 years ago

How much work is done when a 214 newton force pushes a sleeping cow 37m across a field.

1 answer:

8 0

Hello!

Answer:
7918 J

Explanation:

We are assuming that the floor (field) is completely horizontal since there's no information about that in the statement.

We are going to use the following formula:

$W= F . Cos \alpha . D$

Where:

$F=214 N$
$\alpha =0$ º
$D= 37m$

Then, by substituting we have:

$W=214N . Cos (0).37m= 7918 N.m=7918 J$

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A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

Answer:

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

Explanation:

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0

4 years ago

What is the distance to a star whose parallex is 0.1 sec?

Arte-miy333 [17]

Answer:

$30.86\times 10^{13} km$

Explanation:

Given the parallex of the star is 0.1 sec.

The distance is inversely related with the parallex of the star. Mathematically,

$d=\frac{1}{P}$

Here, d is the distance to a star which is measured in parsecs, and P is the parallex which is measured in arc seconds.

Now,

$d=\frac{1}{0.1}\\d=10 parsec$

And also know that,

$1 parsec=3.086\times 10^{13} km$

Therefore the distance of the star is $30.86\times 10^{13} km$ away.

6 0

3 years ago

True or false Carbon in the form of carbon dioxide is needed for both processes of photosynthesis and cellular respiration True

Anna007 [38]

Yes carbon dioxide is needed for photosynthesis while cellular respiration needs oxygen and dispurses carbon dioxide

5 0

3 years ago

What information do you need to describe an object's location

stira [4]

longitude and latitude

4 0

3 years ago

If there are 20 waves in a 5-second time frame, what is the frequency?

creativ13 [48]

Your answer would be 0.20 Hz

3 0

3 years ago

Read 2 more answers