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3 years ago

Can anyone pls answer this plss plss I'm not sure which one it is

Physics

1 answer:

lyudmila [28]3 years ago

5 0

Answer:

a) wet marble , dry marble, newspaper, and towel

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What cell organelle is necessary for cellular respiration

podryga [215]

It is the cytoplasm.

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3 years ago

Read 2 more answers

A 0.0780 kg lemming runs off a

kotegsom [21]

Answer:

5.01 J

Explanation:

Info given:

mass (m) = 0.0780kg

height (h) = 5.36m

velocity (v) = 4.84 m/s

gravity (g) = 9.81m/s^2

1. First, solve for Kinetic energy (KE)

KE = 1/2mv^2

1/2(0.0780kg)(4.84m/s)^2 = 0.91 J

so KE = 0.91 J

2. Next, solve for Potential energy (PE)

PE = mgh

(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J

so PE = 4.10 J

3. Mechanical Energy , E = KE + PE

Plug in values for KE and PE

KE + PE = 0.91J + 4.10 J = 5.01 J

4 0

3 years ago

Help with this physics task pls

cupoosta [38]

Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )

$t=\frac{v}{a}$ ; $d=s*(t-t_{0} )$

$k=\frac{2*U}{x^{2} }$ ; $T_{2}=\frac{P_{2}*V_{2}*T_{1} }{P_{1}*V_{1} } \\$

$L=\frac{F}{\pi*r*P}$ ; $d=\frac{w}{F*cos(o)}$

$t^{2}=\frac{2*x}{g} ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} } \\$

$h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}$

$h=\frac{m}{(1/2)*\pi *r^{2} } ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2} }$

$b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}$

$a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2} }{M}$

$v_{o}=\frac{x-\frac{1}{2}*a*t^{2} }{t} ; F=\frac{W+uNd}{d*cos(o)}$

10)

$h=\frac{E-\frac{1}{2}*m*v^{2} }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2} }{\frac{1}{2}m }$

11)

$N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2} }{\frac{1}{2}*k }$

12)

$x=x_{o} +\frac{v^{2-v_{o}^{2} } }{2a} ; m=\frac{P*A-F_{1}-F_{2} }{g}$

13)

$x_{o} = x-\frac{F}{k} ; u=\frac{cos(o)-\frac{a}{g} }{sin(o)}$

14)

$t=\frac{d}{v} +t_{o}$ ; $t_{o} = t-(\frac{v-v_{o} }{a} )$

15)

$F_{2}=\frac{W-F_{1} *d}{d}+F_{3} ; v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}$

16)

$y_{1}=y-\frac{u}{mg} ; x^{2} = \frac{2W}{k}+x_{o} ^{2}$

7 0

3 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0

3 years ago

What effect does the sun have on Earth's tides? A. The pull of the sun cancels the pull of the moon when they are on opposite si

Kisachek [45]

Answer:

Explanation:

because two vectors which align in the same line adds one to another

6 0

3 years ago

Can anyone pls answer this plss plss I'm not sure which one it is​

Can anyone pls answer this plss plss I'm not sure which one it is