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3 years ago

Which is an example of a multicellular organism? A. gorilla B. amoeba C. bacterium D. paramecium

Physics

1 answer:

Natali5045456 [20]3 years ago

8 0

A gorrila is a multicellular organism

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A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug

natali 33 [55]

Answer:

Approximately $1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}$ , assuming friction between the vehicle and the ground is negligible.

Explanation:

Let $m$ denote the mass of the vehicle. Let $v$ denote the initial velocity of the vehicle. Let $k$ denote the spring constant (needs to be found.) Let $x$ denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

$\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}$ .

Initial kinetic energy ( ${\rm KE}$ ) of the vehicle:

$\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}$ .

When the vehicle is brought to a rest, the elastic potential energy ( $\text{EPE}$ ) stored in the spring would be:

$\displaystyle \frac{1}{2}\, k\, x^{2}$ .

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial $\text{KE}$ of the vehicle should be equal to the ${\rm EPE}$ of the vehicle. In other words:

$\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}$ .

Rearrange this equation to find an expression for $k$ , the spring constant:

$\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}$ .

Substitute in the given values $m = 1508\; {\rm kg}$ , $v \approx 1.908\; {\rm m\cdot s^{-1}}$ , and $x = 6.87\; {\rm m}$ :

$\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}$

8 0

2 years ago

Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr

Sedbober [7]

Answer:

(A) ratio of electric force to weight will be $23.469\times 10^{-10}$

(b) Electric field will be $E=4.26\times 10^{10}N/C$

Explanation:

We have given mass of bee = 100 mg = $m=100\times 10^{-3}=0.1kg$

Charge on bee $q=23pC=23\times 10^{-12}C$

Electric field E = 100 N/C

Weight of the bee $W=mg=0.1\times 9.8=0.98N$

Electric force on the bee $F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N$

So the ratio of electric force on the bee and weight is $=\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}$

(B) To hold the bee in air electric force must be equal to weight of bee

So $mg=qE$

$0.1\times 9.8=23\times 10^{-12}E$

$E=4.26\times 10^{10}N/C$

4 0

3 years ago

In tae-kwon-do, a hand is slammed down onto a target at a speed of 10 m/s and comes to a stop during the 7.0 ms collision. Assum

wlad13 [49]

Answer:

(A) Impulse = 9Ns

(B) F = 1286N

Explanation:

Impulse = change in momentum = m(v-u)

v = 0 (the hand comes to a stop)

u = -10m/s

Mass = 0.9kg

Impulse = 0.9 ×(0- (-10))

= 9Ns

(B) F×t = Impulse

F = Impulse/ t

t = 7ms = 7×10-³

F = 9/ (7×10-³)

F = 1286N.

5 0

3 years ago

The lantern rises up because the __________ of the _________inside the lantern is _________ compared to the air around it

Vika [28.1K]

Density, fire, less dense

7 0

3 years ago

Read 2 more answers

un bus sale de florencia hacia el municipio de san vicente del caguan con una velocidad de 30km/h. 2 horas mas tarde sale otro b

goldenfox [79]

Recuerda que distancia=velocidad x tiempo ( $d=vt$ ). vamos a llamar $d _{1}$ a la distancia que reccore el primer bus, y $d _{2}$ la distancia que recoore el segundo. Los se econtraran cuando la ditancia recorrida por cada bus sea la misma, es decir $d _{1} =d _{2}$ .
Ahora sabemos que la velocidad del primer bus es 30 km/h y su tiempo es $t$ ; en cuanto al segundo bus, su velocidad es 45 km/h y como sale 2 horas mas tarde su tiempo será $t-2$ . Ahora podemos reemplazar los valores en nuestra ecuación de de distanica para hallar el tiempo:
$30t=45(t-2)$
$30t=45t-90$
$30t-45t=-90$
$-15t=-90$
$t= \frac{-90}{-15}$
$t=6$
Los buses se encontraran en 6 horas, 4 horas después de haber salido el segundo bus.

Ahora para hallar la distancia a la que los buses estaran de <span>san vicente del caguan solo necesitaremos reemplazar la el tiempo en nuestra equación de distancia para el primer bus:
</span> $d _{1} =(30)(6)$
$d _{1} =180$
<span>En el punto de encuentro los buses estarán a 180 km de San Vicente del Caguan.
</span>

7 0

3 years ago