Ask question

Ask question

All categories

3 years ago

7

An object is thrown vertically upward such that it has a speed of 25 m/s when it reaches two thirds of its maximum height above

the launch point. Determine this maximum height.
A) 96m
B) 32m
C) 48 m
D) 75m
E) 64 m

1 answer:

AlekseyPX3 years ago

4 0

Answer:

v0 = 25 m/s

vf = 0 m/s

a = -9.80 m/s^2

change in x = 31.89m

but that's only 1/3 of the hight, so i time it by 3 to get 96m

You might be interested in

How do you compare the energies of the different parts of the electromagnetic spectrum?

faltersainse [42]

The different types of radiation in electromagnetic spectrum are compared by the amount of energy found in the photons.

Radio waves have photons with low energies, microwave photons have a little more energy than radio-waves. Infrared photons still have more energy, then comes visible, ultraviolet, x-rays and the most energetic of all, gamma rays.

The energy associated with electromagnetic radiation is proportional to frequency and inversely proportional to wavelength. So, electromagnetic waves with shorter wavelengths have more energy.

On one end of the electromagnetic spectrum are radio waves, which have wavelengths billions of times longer than those of visible light. On the other end of the spectrum are gamma rays with wavelengths billions of times smaller than those of visible light.

To know more about electromagnetic spectrum:

brainly.com/question/27839167

#SPJ4

5 0

1 year ago

An ore sample weighs 17.50 N in air. When the sample

zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T= $W-F_b$

$F_b$ =buoyancy force

T=Tension force

W=Weight

By using the formula

$11.2=17.5-F_b$

$F_b=17.5-11.2=6.3$ N

$F_b=V_{object}\times \rho_{water}\cdot g$

Where

$V_{object}$ =Volume of object

$\rho_{water}=1000 kgm^{-3}$ =Density of water

$g=9.8 ms^{-2}$ =Acceleration due to gravity

Substitute the values then we get

$6.3=9.8\times 1000\times V_{object}$

$V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3$

Volume of sample= $6.43\times 10^{-4} m^3$

Density of sample, $\rho_{object}=\frac{Mass}{volume_{object}}$

Where mass of ore sample=1.79 kg

Substitute the values then, we get

$\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3$

Density of the sample= $2.78\times 10^{3} kgm^{-3}$

7 0

3 years ago

A ball is rolling across the floor at a constant speed. What will happen to the ball if it is exposed

ELEN [110]

I think it’s speed will increase, if I understood it correctly

7 0

3 years ago

Read 2 more answers

Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 19% of the

Zepler [3.9K]

Answer:

51.94°

Explanation:

$I_0$ = Unpolarized light

$I_2$ = Light after passing though second filter = $0.19I_0$

Polarized light passing through first filter

$I_1=\frac{I_0}{2}$

Polarized light passing through second filter

$I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.19I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.19I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.19\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.38}\\\Rightarrow \theta=51.94^{\circ}$

The angle between the two filters is 51.94°

5 0

3 years ago

Suppose that a comet that was seen in 563 A.D. by Chinese astronomers was spotted again in year 1951. Assume the time between ob

Mars2501 [29]

Answer:

$a=2.77*10^{13}m$

$R_a=5.49*10^{13}m$

Explanation:

The period of the comet is the time it takes to do a complete orbit:

T=1951-(-563)=2514 years

writen in seconds:

$2514years*\frac{3,154*10^7s}{1year}=7.93 *10^{10}s$

Since the eccentricity is greater than 0 but lower than 1 you can know that the trajectory is an ellipse.

Therefore, if the mass of the sun is aprox. 1.99e30 kg, and you assume it to be much larger than the mass of the comet, you can use Kepler's law of periods to calculate the semimajor axis:

$T^2=\frac{4\pi^2}{Gm_{sun}}a^3\\ a=\sqrt[3]{\frac{Gm_{sun}T^2}{4\pi^2} } \\a=1.50*10^{6}m$

Then, using the law of orbits, you can calculate the greatest distance from the sun, which is called aphelion:

$R_a=a(1+e)\\R_a=2.77*10^{13}(1.986)\\R_a=5.49*10^{13}m$

8 0

3 years ago