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3 years ago

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It's not D

1 answer:

DanielleElmas [232]3 years ago

5 0

A. it provides support

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Can someone please help me solve. like now? or in a few minutes tops

mafiozo [28]

Hmmmm...I’m not sure but you will will get the right answer don’t give up!

6 0

3 years ago

Read 2 more answers

Determine the magnitude of the force that must be applied to a wooden block whose weight has a magnitude of 8N, so that it acqui

Lostsunrise [7]

Answer:

0.4 N

Explanation:

The weight mg is 8 N, so the mass is:

mg = 8 N

m = 8 N / g

m = 8 N / (10 m/s²)

m = 0.8 kg

The force needed to accelerate it to 0.5 m/s².

F = ma

F = (0.8 kg) (0.5 m/s²)

F = 0.4 N

5 0

3 years ago

An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that

marishachu [46]

Answer:

$\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}$ .

The unit of both sides of this equation are $\rm s$ .

Explanation:

The unit of the left-hand side is $\rm s$ , same as the unit of $T$ .

The following makes use of the fact that for any non-zero value $x$ , the power $x^{-1}$ is equivalent to $\displaystyle \frac{1}{x}$ .

On the right-hand side of this equation:

$\pi$ has no unit.
The unit of $r$ is $\rm m$ .
The unit of $G$ is $\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}$ , which is equivalent to $\rm m^{3} \cdot kg^{-1} \cdot s^{-2}$ .
The unit of $m$ is $\rm kg$ .

$\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}$ .

Hence, the unit on the right-hand side of this equation is also $\rm s$ .

8 0

3 years ago

An object is moving in a straight line along the y axis. As a function of time, its position is given by the equation y=3.0+4.8x

romanna [79]

Answer:

Explanation:

<u>Instant Velocity and Acceleration </u>

Give the position of an object as a function of time y(x), the instant velocity can be obtained by

$v(x)=y'(x)$

Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by

$a(x)=v'(x)=y''(x)$

We are given the function for y

$y(x)=3.0+4.8x +6.4x^2$

Note we have changed the last term to be quadratic, so the question has more sense.

The velocity is

$v(x)=y'(x)=4.8+12.8x$

And the acceleration is $a(x)=v'(x)=12.8$

5 0

4 years ago

A motorcyclist traveling due north at 50km/h. the wind appears to come from north west at 60km/h . what is the true velocity of

pav-90 [236]

Answer:

6o km/h wind - cyclist wind 50 km/h = 10 km/h Head-Wind.

At NW (45 deg) True wind vel = 10 / (cos 45) = 14.1 km/h.

Explanation:

6 0

3 years ago