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vodomira [7]
4 years ago
11

Suppose we repeat the experiment from the video, but this time we use a rocket three times as massive as the one in the video, a

nd in place of water we use a fluid that is twice as massive (dense) as water. If the new fluid leaves the rocket at the same speed as the water in the video, what will be the ratio of the horizontal speed of our rocket to the horizontal speed of the rocket in the video after all the fluid has left the rocket?
Physics
1 answer:
aliina [53]4 years ago
8 0

Answer:

2/3

Explanation:

It is important to bear in mind that momentum is generated by the fluid. Momentum is proportional to the mass and velocity, therefore increasing the mass(density) of the liquid by a factor of 2, while the velocity is constant, increases the momentum of fluid (and rocket) by 2. For any closed system, the momentum is conserved. With that in mind, the momentum of the fluid is equal to but opposite in direction to the momentum of the rocket. Velocity of the rocket given by v=p/m (Where p=momentum of the rocket initially and m is its initial mass). The mass of the rocket increases 3 times. So....

Initially:

vi=p/m

Now:

vf=2p/3m

<h3>The ratio is given by:</h3>

vf:vi = 2p/3m:p/m

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Answer:It's important to listen to your body and what it's really saying. Don't listen, and your exercise could wind up working against you, actually triggering muscle loss and fat gain. Plus, pushing too hard comes with a huge risk of injury-and all of your training won't do you much good if you're sidelined on race day.

Explanation:

7 0
3 years ago
Un resorte se monta horizontalmente con su extremo izquierdo fijo. Conectando una balanza de
zavuch27 [327]

Translation of important question part (Google translation used)

Force of 6N causes a displacement of 0.03m. remove the balance and connect a body 0.5 kg to the end, pull it to move it 0.02 m, release it and see how it oscillates.

(a)Determine the spring constant.

(b)Calculate angular velocity, frequency, and oscillation period

Answer:

(a)K=200 N/m

(b) w= 20 rad/s f=3.2 Hz T=0.3125 s

Explanation:

(a)

From Hooke's law, we deduce that F=kx where F is applied force, k is spring constant and x is extension of the spring. Making k the subject of the formula then k=F/x and substituting F with 6 N and x with 0.03 m then k=6/0.03=200 N/m

(b)

Angular velocity, w is given by w= \sqrt{\frac {k}{m}} where m is the mass and k is spring constant calculated in part a above. Substituting mass with 0.5 kg and k with 200 N/m then

w= \sqrt{\frac {200}{0.5}}=20 rad/s

We know that frequency, f is given by f=\frac {w}{2\pi} and substituting 20 rad/s for w then

f=\frac {20}{2\pi}=3.1830988618379067153776752674502872406891\approx 3.2 Hz

Finally, oscillation period, T is usually the reciprocal of frequency hence T=1/f and substituting f with 3.2 Hz then T=1/3.2=0.3125 s

7 0
3 years ago
A tiger leaps with a horizontal speed of 4.5 m/s from a boulder and lands 15 meters away.What is the vertical velocity with whic
Gelneren [198K]

Answer:

v_oy = 16.33 m/s

Explanation:

To find the vertical velocity of the tiger, you use the information about the horizontal velocity and maximum horizontal distance traveled.

You use the following formula for the range of the trajectory:

x_{max}=\frac{2v_{ox}v_{oy}}{g}     ( 1 )

v_ox: horizontal initial velocity = 4.5m/s

v_oy: vertical initial velocity = ?

g: gravitational acceleration = 9.8m/s^2

x_max: range of the trajectory = 15 m

You do v_oy the subject of the formula ( 1 ) and you replace the values of the other parameters in order to calculate v_oy:

v_{oy}=\frac{gx_{max}}{2v_{ox}}=\frac{(9.8m/s^2)(15m)}{2(4.5m/s)}\\\\v_{oy}=16.33\frac{m}{s}

hence, the initial vertical velocity of the tiger is 16.33m/s

8 0
3 years ago
How does light reflect from the surface of carbon, copper wire, and the silicon lump? Does it appear metallic or nonmetallic (is
HACTEHA [7]

Answer:

When a light wave hits a metallic surface, the electrons on the surface are pushed and ... in a process called "total internal reflection", making that surface look shiny too.

Explanation:

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4 0
3 years ago
During normal beating, the heart creates a maximum 4.00-mV potential across 0.300 m of a person’s chest, creating a 1.00-Hz elec
erik [133]

Answer:

(a). The maximum electric field strength is 0.0133 V/m.

(b). The maximum magnetic field strength in the electromagnetic wave is 4.433\times10^{-11}\ T

(c). The wavelength of the electromagnetic wave is 3\times10^{8}\ m

Explanation:

Given that,

Maximum potential = 4.00 mV

Distance = 0.300\ m

Frequency = 1.00 Hz

(a). We need to calculate the maximum electric field strength

Using formula of the potential difference

\Delta V=Ed

E=\dfrac{\Delta V}{d}

E=\dfrac{4.00\times10^{-3}}{0.300}

E=0.0133\ V/m

(b). We need to calculate the maximum magnetic field strength in the electromagnetic wave

Using formula of the maximum magnetic field strength in the electromagnetic wave

B=\dfrac{E}{c}

Put the value into the formula

B=\dfrac{0.0133}{3\times10^{8}}

B=4.433\times10^{-11}\ T

(c). We need to calculate the wavelength of the electromagnetic wave

Using formula of wavelength

c=f\lambda

\lambda=\dfrac{c}{f}

Put the value into the formula

\lambda=\dfrac{3\times10^{8}}{1.00}

\lambda=3\times10^{8}\ m

Hence, (a). The maximum electric field strength is 0.0133 V/m.

(b). The maximum magnetic field strength in the electromagnetic wave is 4.433\times10^{-11}\ T

(c). The wavelength of the electromagnetic wave is 3\times10^{8}\ m

4 0
3 years ago
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