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2 years ago

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Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly ejects the water backward to prop

el itself forward. A 1.5 kg squid (not including water mass) can accelerate at 20 m/s2 by ejecting 0.15 kg of water.?

1 answer:

Setler [38]2 years ago

5 0

Answer:

a. FTh = 30 N

b. Fw = 30 N

c. a = 200 m/s2

Explanation:

See full explanation in the picture. Please rate as brainliest

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If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g

EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408 × 10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M_1, M_2$ are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

$\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}$

$\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2$

Since $M_2 = 2m_2$ and $r = R/3$

$\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5$

So gravity would have been decreased by a factor of 4.5

8 0

2 years ago

6. Starting from 1.5 miles away, a car drives toward a speed checkpoint and then passes it. The car travels at a constant rate o

loris [4]

For the answer to the question above,

<span>To be 0.1 miles away from the check point ,
the car has to travel 1.4 miles OR 1.6 miles. </span>

53 miles = 60 minutes

1.4 miles = 1.4 / 53 X 60 = 1.5849056 minutes OR 95.1 seconds

<span>1.6 miles = 1.6 /53 X 60 = 1.8113207 minutes OR 108.7 seconds
</span>So the answer is <span>95.1s and 108.7s
I hope my answer helped you</span>

7 0

2 years ago

A 40-mH ideal inductor is connected in series with a 50 Ω resistor through an ideal 15-V DC power supply and an open switch. If

sergey [27]

Answer:i=300 mA

Explanation:

Given

inductance(L)=40 mH

Resistor(R)= $50 \Omega$

Voltage(V)=15 V

Time constant( $\tau$ )= $\frac{L}{R}$

$\tau =\frac{40\times 10^{-3}}{50}=8\times 10^{-4}$

current $i_0=\frac{V}{R}$

$i_0=\frac{15}{50}=0.3 A$

Current as a function of time is given by

$i=i_0\left ( 1-e^{-\frac{t}{\tau }}\right )$

$i=0.3\times 0.9998$

i= 299.95 mA

6 0

2 years ago

The graph below shows the relationship between speed and time for two objects, A and B. Compare with the acceleration of object

kolbaska11 [484]

Answer:

A) greater

Explanation:

acceleration is calculated by dividing velocity over time..so by calculating, you find acceleration of A is greater than that of B

5 0

2 years ago

Read 2 more answers

What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?

DIA [1.3K]

The elastic potential energy of the spring is 6.8 J

Explanation:

The elastic potential energy of a compressed/stretched spring is given by the equation:

$E=\frac{1}{2}kx^2$

where

k is the spring constant

x is the elongation of the spring

The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):

$F=kx$

By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,

$k=\frac{15.0-0}{10.0-0}=1.5 N/m$

Therefore when the spring has a elongation of $x=3.0 m$ , its potential energy is

$E=\frac{1}{2}(1.5)(3.0)^2=6.8 J$

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0

2 years ago