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3 years ago

A projectile is launched at an angle above the

1 answer:

7 0

The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

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Which method will provide the most emotional support in a new exercise program?

enyata [817]

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3 years ago

A spring loaded toy shoots straight upward with a velocity of 4.5 m/s. Determine the maximum height it reaches. Determine the ti

Angelina_Jolie [31]

Recall that

${v_y}^2-{v_{0y}}^2=2a_y(y-y_0)$

At its maximum height $y_{\mathrm{max}}$ , the toy will have 0 vertical velocity, so that

$-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)y_{\mathrm{max}}$

$\implies y_{\mathrm{max}}=1.0\,\mathrm m$

For the toy to reach this maximum height, it takes time $t$ such that

$\dfrac{0+v_0}2t=y_{\mathrm{max}}\implies t=0.46\,\mathrm s$

which means it takes twice this time, i.e. $t=0.92\,\mathrm s$ , for the toy to reach its original position.

The velocity of the toy when it falls 1.0 m below its starting point is

${v_y}^2-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(0-1.0\,\mathrm m)$

$\implies{v_y}^2=39.85\,\dfrac{\mathrm m^2}{\mathrm s^2}$

$\implies v_y=-6.4\,\dfrac{\mathrm m}{\mathrm s}$

where we took the negative square root because we expect the toy to be moving in the downward direction.

6 0

3 years ago

two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco

Reika [66]

Answer:

$t=750s$

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

$\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt$

$x_f$ is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

$x_0_1=6km\\x_0_2=0$

We have $x_f_1=x_f_2$ . Thus, solving for t:

$x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s$

8 0

3 years ago

A crate is pulled with a force of 165 N at an angle 30 ° northwest. What is the resultant horizontal force on the crate?

Ad libitum [116K]

Answer:

Resultant horizontal force = 143 N

Explanation:

Since the a gle is 30° northwest, then it means the resultant force will be horizontal and as such;

Resultant horizontal force = 165 * cos 30

Resultant horizontal force = 142.89

Approximating to a whole number gives;

Resultant horizontal force = 143 N

5 0

2 years ago

A housefly walking across a clean surface can accumulate a significant positive or negative charge. In one experiment, the large

Lisa [10]

Answer:

$4.56*10^8 \text{ electrons.}$

Explanation:

Since the fly accumulated a positive charge of +73pC, it must have lost an equal number of negative charge of -73pC to the surface (because the housefly was neutral to begin with).

Therefore, to answer our question we have to ask ourselves <em>how many electrons combine to make -73pC of charge? </em>

The answer is since one electron carries a charge of $-1.6*10^{-19}C$ , the number $n$ of electrons that make up -73pC $(-73*10^{-12}C)$ are

$n= \dfrac{-73*10^{-12}C}{-1.6*10^{-19}C/e}$

$\boxed{n= 4.56*10^8e.}$

Thus, the housefly lost about 456 million electrons to the surface!

4 0

4 years ago