234 oz to tons using the table method (US)

A student walks 4 block east, 7 blocks west, 1 blocks east then 2 blocks west in an hour. Her average speed is____ block/hour

Gemiola [76]

Average speed = (total distance) / (total time)

Average speed = (4+7+1+2 blox) / (1 hour)

Average speed = 14 blocks/hour

I'm gonna go out on a limb here and take a wild guess:

I'm guessing that there's another question glued onto the end of this one, and it asks you to find either her displacement or her average velocity. I'm so sure of this that I'm gonna give you the solution for that too. If there's no more question, then you won't need this, and you can just discard it. I won't mind.

Average velocity = (displacement) / (time for the displacement)

"Displacement" = distance and direction from the start point to the end point, regardless of how she got there.

Displacement = (4E + 7W + 1E + 2W)

Displacement = (5E + 9W)

Displacement = 4 blocks west

Average velocity = (4 blocks west) / (1 hour)

Average velocity = 4 blocks/hour West

4 0

2 years ago

he tune-up specifications of a car call for the spark plugs to be tightened to a torque of 47 N⋅m . You plan to tighten the plug

S_A_V [24]

Answer:

207.4 N

Explanation:

The torque $\tau$ on a body is

$\tau = r* F$ where r is the radius vector from the point of rotation to the point at which force F is applied.

The product of r and F is equal to the product of magnitude of r and F multiplied by the sine of angle between both vectors.

Therefore, torque is also given by

$\tau = rF\sin \theta$

Where $\theta$ is the angle between r and F.

Use the expression of torque.

Substitute L for r in the equation $\tau = rF\sin \theta$

$\tau = LF\sin \theta$

Where L is the length of the wrench.

Making F the subject

$F = \frac{\tau }{{L\sin \theta }}$

Force required to pull the wrench is given as,

$F = \frac{\tau }{{L\sin \theta }}$

Substitute $47{\rm{ N}} \cdot {\rm{m}}$ for $\tau$ , 25 cm for L, and 115o for $\theta$

$\begin{array}{c}\\F = \frac{{47{\rm{ N}} \cdot {\rm{m}}}}{{\left( {25{\rm{ cm}}} \right)\sin {{115}^{\rm{o}}}}}\left( {\frac{{1{\rm{ cm}}}}{{{{10}^{ - 2}}{\rm{ m}}}}} \right)\\\\ = 207.435{\rm{ N}}\\\\ \approx 207.4{\rm{ N}}\\\end{array}$

6 0

2 years ago

at the extact same time, youj drop a piece of notebook paper and your science out the window. the science book hit first? why?

jeyben [28]

The piece of paper has less mass and will glide down the window, whereas the textbook will go straight to the ground. Since the textbook has more mass and less ways of it being able to 'glide' the textbook will hit the ground first.

8 0

3 years ago

A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h

hram777 [196]

Answer:

$v_{2}=3.5 m/s$