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4 years ago

5

Your physics textbook is sliding to the right across the table draw the vectors starting at the black dot. the location and orie

ntation of the vectors will be graded. the exact length of your vectors will not be graded but the relative length of one to the other will be graded.

1 answer:

liberstina [14]4 years ago

6 0

You would take a black dot at the top right of the y axis and dray it to the the far right of your x axis. hope this helps have a nice day and God bless

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Give an example of a balanced force and explain what makes it a balanced force

Stella [2.4K]

When two forces are the same strength but act in opposite direction, they are called balanced forces. Again, tug-of-war is a perfect example. If the people on each side of the rope are pulling with the same strength, but in the opposite direction, the forces are balanced. The result is no motion.

5 0

4 years ago

Gravel is being dumped from a conveyor belt at a rate of 10 ft3/min, and its coarseness is such that it forms a pile in the shap

AURORKA [14]

Answer: The height of the pile increasing is increasing at $0.16\frac{ft}{min}$

Explanation:

Given

Rate at which gravel is being dumped , $\frac{\mathrm{d} V}{\mathrm{d} t}=10\frac{ft^{3}}{min}$

=>Rate of increase of volume of cone= $\frac{\mathrm{d} V}{\mathrm{d} t}=10\frac{ft^{3}}{min}$

If height of the cone at any instant is h then the diameter of cone is also h

Volume of cone , $V=\frac{\pi r^{2}h}{3}=\frac{\pi h^{3}}{4\times 3}=\frac{\pi h^{3}}{12}$

Now differentiate both sides w.r.t time(t)

$\frac{\mathrm{d} V}{\mathrm{d} t}=\frac{\pi h^{2}}{4}\frac{\mathrm{d} h}{\mathrm{d} t}$

Therefore at h = 9 ft

$10=\frac{\pi \times 9^{2}}{4}\times \frac{\mathrm{d} h}{\mathrm{d} t}$

=> $\frac{\mathrm{d} h}{\mathrm{d} t}=0.16\frac{ft}{min}$

Thus the height of the pile increasing is increasing at $0.16\frac{ft}{min}$

6 0

4 years ago

I need to know which atom is left after the decay

serious [3.7K]

Answer:

the answer is b

Explanation:

3 0

3 years ago

Two clear but non-mixing liquids each of depth 15 cm are placed together in a glass container. The liquids have refractive indic

vladimir1956 [14]

Answer:

The apparent depth d = 19.8495 cm

Explanation:

The equation for apparent depth can be expressed as:

$d = \dfrac{d_1} {\mu_1}+\dfrac {d_2}{\mu_2}$

here;

$d_1 = d_2 = 15 \ cm$

$\mu_1$ = refractive index in the first liquid = 1.75

$\mu_2$ = refractive index in the second liqquid= 1.33

∴

$d = \dfrac{15}{1.75}+\dfrac{15}{1.33}$

$d = 15( \dfrac{1}{1.75}+\dfrac{1}{1.33})$

$d = 15( 0.5714 +0.7519)$

d = 15(1.3233 ) cm

d = 19.8495 cm

6 0

3 years ago

(1 point) A gun has a muzzle speed of 90 meters per second. What angle of elevation θ, where 0≤θ≤π4, should be used to hit an ob

Likurg_2 [28]

Answer:

0.122 radians

Explanation:

Given that the muzzle speed is, $u=90m/s$

And the distance of the object is, $s=200 m$

And acceleration due to gravity is $g=9.8 m/s^{2}$

Therefore negative acceleration due to gravity for upward motion.

Now,

The x component of velocity will be, $u_{x}=ucos\theta$

The y component of velocity will be, $u_{y}=usin\theta$

And the horizontal displacement is given. Now horizontal displacement is,

$s=u_{x}\times t$

Here, $u_{x}$ is horizontal velocity and t is time.

Substitute all the values in above equation, we get

$s=ucos\theta\times t\\200=90cos\theta\times t\\t=\dfrac{200}{90cos\theta}\\t=\dfrac{20}{9cos\theta}$

Now, in y direction we can calculate displacement and in y direction displacement is zero.

$s_{y}=u_{y}t+\frac{1}{2}at^{2}$

Substitute all the variables.

$0=usin\theta t+\frac{1}{2}at^{2}\\0=90sin\theta (\frac{20}{9cos\theta})+\frac{1}{2}(-9.8 m/s^{2})\times (\frac{20}{9cos\theta})^{2} \\200tan\theta=4.9\times \frac{400}{81}\times \frac{1}{cos^{2}\theta}\\200sin\theta=\frac{24.1975}{cos\theta}\\ 2sin\theta cos\theta=0.241975\\sin2\theta=0.241975\\2\theta=sin^{-1}(0.241975)\\ \theta=0.122 radians$

Therefore the required angle of elevation is 0.122 radians.

8 0

4 years ago