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4 years ago

HELP I DONT HAVE THAT MUCH POINTS LEFT BUT 15 POINTSSS

Physics

2 answers:

Aliun [14]4 years ago

4 0

Answer:

Explanation:

the answers are all not good, but I would say A is the best

andrey2020 [161]4 years ago

4 0

The answer is is c your welcome

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Water flowing through a garden hose of diameter 2.76 cm fills a 20.0-L bucket in 1.45 min. (a) What is the speed of the water le

mafiozo [28]

Answer:

v = 31.84 cm/s or 0.318 m/s

the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s

Explanation:

Given;

Diameter of hose d = 2.76 cm

Volume filled V = 20.0 L = 20,000 cm^3

Time t = 1.45 min = 105 seconds

The volumetric flow rate of water is;

F = V/t = 20,000cm^3 ÷ 105 seconds

F = 190.48 cm^3/s

The volumetric flow rate is equal the cross sectional area of pipe multiply by the speed of flow.

F = Av

v = F/A

Area A = πd^2/4

Speed v = F/(πd^2/4)

v = 4F/πd^2 ......1

Substituting the given values;

v = (4×190.48)/(π×2.76^2)

v = 31.83767439628 cm/s

v = 31.84 cm/s or 0.318 m/s

the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s

3 0

3 years ago

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

8 0

3 years ago

What is the gravitational field strength at a distance of 60.0 km above the surface of the earth

zloy xaker [14]

Answer:

g=9.64m/s^2.

Explanation:

Gravitational field strength (in other words, gravitational acceleration) is given as follows:g=GMR2g=R2GMwhere G=6.674×10−11m3kg⋅s2G=6.674×10−11kg⋅s2m3 is the gravitational constant, M=5.972×1024kgM=5.972×1024kg is the mass of the Earth, and R=6.371×106m+0.06×106m=6.431×106mR=6.371×106m+0.06×106m=6.431×106m is the distance from the center of the Earth to the required point above the surface (radius plus 60 km).

6 0

2 years ago

A 45.0 g hard-boiled egg moves on the end of a spring with force constant 25.0 N/m. Its initial displacement 0.500 m. A damping

leva [86]

Answer:

0.02896 kg/s

Explanation:

$A_1$ = Initial displacement = 0.5 m

$A21$ = Final displacement = 0.1 m

t = Time taken = 0.5 s

m = Mass of object = 45 g

Displacement is given by

$x=Ae^{-\dfrac{b}{2m}t}cos(\omega t+\phi)$

At maximum displacement

$cos(\omega t+\phi)=1$

$\\\Rightarrow A_2=A_1e^{-\dfrac{b}{2m}t}\\\Rightarrow \dfrac{A_1}{A_2}=e^{\dfrac{b}{2m}t}\\\Rightarrow ln\dfrac{A_1}{A_2}=\dfrac{b}{2m}t\\\Rightarrow b=\dfrac{2m}{t}\times ln\dfrac{A_1}{A_2}\\\Rightarrow b=\dfrac{2\times 0.045}{5}\times ln\dfrac{0.5}{0.1}\\\Rightarrow b=0.02896\ kg/s$