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3 years ago

10

Three people push a piano on wheels with forces of 130 N to the right, 150 N to the left, and 165 N to the right. What is the st

rength and direction of the net force on the piano?

1 answer:

Irina-Kira [14]3 years ago

8 0

Answer:

45 N to the right possibly?

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2 years ago

How do you calculate the net force when there are multiple forces in different directions?

Artyom0805 [142]

To find $F_{net}$ we need to use vector addition and use the x and y components. First we subtract vector 2 from vector 5 which results in a vector with a length of 3 pointing directly east, then we use the distance formula to find the length of the net force $F_{net} = \sqrt{(3)^2+(4)^2} \\$ which gives $F_{net} = 5$ . We now have a magnitude but we also need a direction, since vector 4 and vector 5 are perpendicular. Using $\theta = \tan^{-1} (\frac{4}{3})$ where tan^-1(y/x) we get an angle of 53 degrees. The resultant force vector is 5 distance with an angle of 53 degrees north east.

4 0

3 years ago

The formula is x = 1/2 at^2 and I have managed to fill in the variables as this. d = 1/2 9.81 m/s^2 1^2

Artyom0805 [142]

Right, as you mentioned in the comments, you find $d$ by plugging in the different values of $t$ .

For $t=1\,\mathrm s$ , we have

$d=\dfrac12\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)(1\,\mathrm s)^2$

$d=\left(4.905\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(1\,\mathrm s^2\right)$

$d=4.905\,\mathrm m$

Similarly, for $t=2\,\mathrm s$ , you get

$d=\dfrac12\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(2\,\mathrm s\right)$

$d=\left(4.905\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(4\,\mathrm s^2\right)$

$d=19.62\,\mathrm m$

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3 years ago

Lance Armstrong bikes at a constant speed up the Col d’Izoard, a famous mountain pass. Assume his teammates do such a good job r

svetoff [14.1K]

Work done against gravity to climb upwards is always stored in the form of gravitational potential energy

so we can say

$W = mgh$

here h = vertical height raised

so here we know that

$h = 14.1 sin7.3 km$

here we have

$h = 1.79 km$

now from above equation

$W = (83 kg)(9.81 m/s^2)(1.79 \times 10^3 m)$

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so work done will be given by above value

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3 years ago

"In a pure substance all the particles must be identical; therefore pure substances are composed only of elements." Do you agree

kenny6666 [7]

I do not agree with the statement.
The "substance" can be a compound. It's "pure"
as long as there's nothing else in it but its name.

'Pure' water is 100% H₂O with nothing else in it.
'Pure' table salt is 100% NaCl with nothing else in it.
'Pure' carbon dioxide is 100% CO₂ with nothing else in it.

These example substances are all compounds, not elements.

6 0

3 years ago