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4 years ago

Why collaboration in science is critical to success in the scientific community

1 answer:

5 0

Collaboration in science is important because if only one scientist does an experiment, and gets a result, he/she could have messed. So this is where collaboration comes in. A few other scientists could try the experiment, and if they get the same answers, the result may be proven correct.

Hope this helped!

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A turtle and a rabbit are in a 150 meter race. The rabbit decides to give the turtle a 1 minute head start. The turtle moves at

yan [13]

Answer:

a) $s_{T} = 30\,m$ , b) $t = 5\,min$ , c) $\Delta t = 6.667\,s$ , d) $\Delta s_{R} = 33.333\,m$ , e) $t' = 11.667\,s$ , f) The rabbit won the race.

Explanation:

a) As turtle moves at constant speed, its position is determined by the following formula:

$s_{T} = v_{T}\cdot t$

Where:

$t$ - Time, measured in seconds.

$v_{T}$ - Velocity of the turtle, measured in meters per second.

$s_{T}$ - Position of the turtle, measured in meters.

Then, the position of the turtle when the rabbit starts to run is:

$s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)$

$s_{T} = 30\,m$

The position of the turtle when the rabbit starts to run is 30 meters.

b) The time needed for the turtle to finish the race is:

$t = \frac{s_{T}}{v_{T}}$

$t = \frac{150\,m}{0.5\,\frac{m}{s} }$

$t = 300\,s$

$t = 5\,min$

The time needed for the turtle to finish the race is 5 minutes.

c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:

$v_{R} = v_{o,R} + a_{R}\cdot \Delta t$

Where:

$v_{R}$ - Final velocity of the rabbit, measured in meters per second.

$v_{o,R}$ - Initial velocity of the rabbit, measured in meters per second.

$a_{R}$ - Acceleration of the rabbit, measured in $\frac{m}{s^{2}}$ .

$\Delta t$ - Running time, measured in second.

$\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}$

$\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }$

$\Delta t = 6.667\,s$

The time taken by the rabbit to reach maximum speed is 6.667 s.

d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:

$v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}$

$\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}$

Where $\Delta s_{R}$ is the travelled distance of the rabbit from rest to maximum speed.

$\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}$

$\Delta s_{R} = 33.333\,m$

The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.

e) The time required for the rabbit to finish the race can be determined by the following expression:

$t' = \frac{\Delta s_{R}}{v_{R}}$

$t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }$

$t' = 11.667\,s$

The time required for the rabbit from rest to maximum speed is 11.667 seconds.

f) The animal with the lowest time wins the race. Now, each running time is determined:

Turtle:

$t_{T} = 300\,s$

Rabbit:

$t_{R} = 60\,s + 6.667\,s + 11.667\,s$

$t_{R} = 78.334\,s$

The rabbit won the race as $t_{R} < t_{T}$ .

7 0

4 years ago

A small ferryboat is 4.70 m wide and 6.10 m long. When a loaded truck pulls onto it, the boat sinks an additional 5.00 cm into t

geniusboy [140]

Answer:

M = 1433.5 kg

Explanation:

This exercise is solved using the Archimedean principle, which states that the hydrostatic thrust is equal to the weight of the desalinated liquid,

B = ρ g V

with the weight of the truck it is in equilibrium with the push, we use Newton's equilibrium condition

Σ F = 0

B-W = 0

B = W

body weight

W = M g

the volume is

V = l to h

rho_liquid g (l to h) = M g

M = rho_liquid l a h

we calculate

M = 1000 4.7 6.10 0.05

M = 1433.5 kg

6 0

4 years ago

What would Hubble's constant be if we found one galaxy moving away at 30,000 km/s at a distance of 600 Mpc?

lora16 [44]

Answer:

H₀ = 1.6 x 10⁻¹⁸ s⁻¹

Explanation:

The Hubble's Constant can be found by the following formula:

$v = H_o D\\\\H_o = \frac{v}{D}$

where,

H₀ = Hubble's Constant = ?

v = speed of galaxy = 30000 km/s = 3 x 10⁷ m/s

D = Distacance = 600 Mpc = (6 x 10⁸ pc)(3.086 x 10¹⁶ m/1 pc)

D = 18.52 x 10²⁴ m

Therefore,

$H_o = \frac{3\ x\ 10^7\ m/s}{18.52\ x\ 10^{24}\ m}$

H₀ = 1.6 x 10⁻¹⁸ s⁻¹

3 0

3 years ago

In being served, a tennis ball is accelerated from rest to a speed of 31.6 m/s. The average power generated during the serve is

joja [24]

Answer:

The force acting on the ball is 92.4 N.

Explanation:

Given that,

Initial speed of the ball, u = 0

Final speed of the ball, v = 31.6 m/s

The average power generated during the serve is 2920 W. Power generated by an object is given by :

$P=\dfrac{W}{t}$

W is the work done, W = Fd

$P=\dfrac{Fd}{t}$

Since, $v=\dfrac{d}{t}$

So,

$P=F\times v$

F is the force acting on the ball

$F=\dfrac{P}{v}\\\\F=\dfrac{2920\ W}{31.6\ m/s}\\\\F = 92.4\ N$

So, the force acting on the ball is 92.4 N. Hence, this is the required solution.

7 0

3 years ago

At takeoff, an aircraft travels at 62 m/s, so that the air speed relative to the bottom of the wing is 62 m/s. Given the sea lev

olganol [36]

Answer:

the aircraft must travel at a speed of 73.4 m/s in order to create the ideal lift.

Explanation:

We will use Bernoulli's theorem in order to determine the pressure lift:

ΔP = 1/2 (ρ)(v₂² - v₁²)

the generated pressure lift is ΔP = 1000 N/m²

Therefore,

1000 = 1/2(ρ)(v₂² - v₁²)

v₂² - v₁² = 2000 / ρ

v₂² = (2000 N/m² / 1.29 kg/m³) + (62 m/s)²

v₂ = √[ (2000 N/m² / 1.29 kg/m³) + (62 m/s)² ]

v₂ = 73.4 m/s

Therefore, the aircraft must travel at a speed of 73.4 m/s in order to create the ideal lift.

5 0

3 years ago