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2 years ago

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Question 5: A heavy rope is flicked upwards, creating a single pulse in the rope. Make a drawing of the rope and indicate the fo

llowing in your drawing
a. the direction of motion of the pulse
b. Amplitude
c. Pulse length
d. Position of the rest

1 answer:

Anastaziya [24]2 years ago

8 0

Answer:

the direction of motion of the pulse

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A spherical balloon has a radius of 7.40 m and is filled with helium. Part A How large a cargo can it lift, assuming that the sk

malfutka [58]

Answer:

The mass of the cargo is $M = 188.43 \ kg$

Explanation:

From the question we are told that

The radius of the spherical balloon is $r = 7.40 \ m$

The mass of the balloon is $m = 990\ kg$

The volume of the spherical balloon is mathematically represented as

$V = \frac{4}{3} * \pi r^3$

substituting values

$V = \frac{4}{3} * 3.142 *(7.40)^3$

$V = 1697.6 \ m^3$

The total mass the balloon can lift is mathematically represented as

$m = V (\rho_h - \rho_a)$

where $\rho_h$ is the density of helium with a value of

$\rho_h = 0.179 \ kg /m^3$

and $\rho_a$ is the density of air with a value of

$\rho_ a = 1.29 \ kg / m^3$

substituting values

$m = 1697.6 ( 1.29 - 0.179)$

$m = 1886.0 \ kg$

Now the mass of the cargo is mathematically evaluated as

$M = 1886.0 - 1697.6$

$M = 188.43 \ kg$

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3 years ago

Two blocks of masses m and M are connected by a string and pass over a frictionless pulley. Mass m hangs vertically, and mass M

horsena [70]

Answer:

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

$sin\theta - \mu_k cos\theta = 1$

Explanation:

Force of friction on M mass so that it will move down the inclined plane is given as

$F_f = \mu Mgcos\theta$

now if it is moving down the inclined plane at constant speed

so we will have

$Mgsin\theta - T - \mu mgcos\theta = 0$

on other side the mass "m" will go up at constant speed

so we have

$T - mg = 0$

so we have

$Mgsin\theta = \mu Mgcos\theta + mg$

so we have

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

for special case when m = M

then we have

$sin\theta - \mu_k cos\theta = 1$

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3 years ago

Animals need oxygen to complete the many chemical reactions taking place in the body. An animal inhales oxygen and exhales _____

mezya [45]

Carbon dioxide and nitrogen

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3 years ago

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The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

(c) Escape velocity on earth is $3.563\times 10^4m/sec$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

(c) Escape velocity is given by

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

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What impact can a very large volcanic eruption have on earths climate

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Answer:

It could instantly kill all life on Earth
It could block out the sun and kill off all plants
It could block out the sun and freeze everything
It could throw sulfur millions of miles and kill everything that 'inhales' it

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3 years ago