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2 years ago

How long does it take for a train to increase its velocity from 10m/s to 40m/s if it accelerates at 3 m/s

Physics

1 answer:

DIA [1.3K]2 years ago

6 0

Answer:

Explanation:

Givens

Vi = 10 m/s

Vf = 40 m/s

a = 3 m/s^2

Formula

a = (vf - vi) /t Substitute the givens into this formuls

Solution

3 = (40 - 10) / t Multiply both sides by t

3*t = t(40 - 10)/t Combine. Cancel t's on the right

3*t = 30 Divide by 3

3t/3 = 30 / 3

Answer: t = 10 seconds.

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3 years ago

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An airplane flies at a speed of 250 mi/hr, 40 degrees south of east. A wind blows at a speed of 35 mi/hr 30 degrees west of sout

alina1380 [7]

Compute the components of the given vectors. Let $P$ denote the plane's velocity vector, and $W$ the wind. Then

$P=\langle P_x,P_y\rangle=\langle250\cos(-40^\circ),250\sin(-40^\circ)\rangle$

$\implies P=\langle191.5,-160.7\rangle$

$W=\langle35\cos(-120^\circ),35\sin(-120^\circ)\rangle=\langle-17.5,-30.3\rangle$

The resultant velocity (rounded) is

$P+W=\langle174,-191\rangle$

with magnitude $\sqrt{174^2+(-191)^2}=258$ and direction $\tan^{-1}\frac{-191}{174}=-47.7^\circ$ , or about 258 mi/hr at 47.7 degrees south of east.

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3 years ago

Help and show work thanks

Zepler [3.9K]

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6 0

3 years ago

a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w

nataly862011 [7]

Answer:

40 cm

Explanation:

We are given that

Load=800 N

Effort=200 N

Load distance=10 cm

We have to find the effort distance.

We know that

$load\times load\;distance=Effort\times effort\;distance$

Using the formula

$800\times 10=200\times effort\;distance$

Effort distance= $\frac{800\times 10}{200}$

Effort distance= $\frac{8000}{200}$

Effort distance=40 cm

Hence, the effort distance will be 40 cm.

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3 years ago

Find the sum of the vectors:11 km N ,11km E

vladimir1956 [14]

The resultant vector is 11√2 km due north east.

<h3><u>Explanation:</u></h3>

The vector is a type of quantity which has both magnitude and direction. This quantities when expressed needs to specify both magnitude and direction.

We need to calculate the magnitude and direction separately.

Here firstly for the magnitude,

The magnitudes are both 11 km and they are at right angles to each other.

So, the resultant magnitude = √(11² +11²) km

=11√2 km

Now for the direction, one vector is due north and the other is due east.

So the resultant vector is due north east.

So the final vector is 11√2 km due North-East.

6 0

3 years ago