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3 years ago

14

What is the difference between the control group and the experimental group in an experimental study?

1 answer:

Igoryamba3 years ago

6 0

Answer:a control group has no effects or changes whilst the experimental group has changes or something different from a control group

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1. There is a famous intersection in Kuala Lumpur, Malaysia, where thousands of vehicles pass each hour. A 750 kg Tesla Model S

tigry1 [53]

Solution :

Let the positive $x-axis$ is along the East and the positive $y$ direction is along the north.

Given :

Mass of the Tesla car, $m_1$ = $750 \ kg$

Mass of the Ford car, $m_2 = 1250 \ kg$

Now let the initial velocity of Tesla car in the south direction be = $-v_1j$

The initial momentum of Tesla car, $p_1 = -750 \ v_1$

Let the initial velocity of Ford car in the east direction be = $v_2 \ i$

So the initial momentum of the Ford car is $p_2=1250\ v_2 \ i$

Therefore, the initial velocity of both the cars is $p_i = p_1+p_2$

$=1250 \ v_2 \ i - 750\ v_1 \ j$

Now the final velocity of both the cars is $v = 18 \ m/s$

So the vector form is :

$v = 18\cos 32\ i-18 \sin 32 \ j$

$= 15.26 \ i - 9.54 \ j$

Therefore the momentum after the accident is

$p_f=(m_1+m_2) \times v$

$=(750+1250) \times (15.26 \ i - 9.54 \ j)$

$= 30520\ i -19080\ j$

According to the law of conservation of momentum, we know

$p_i = p_f$

$1250 \ v_2 \ i - 750\ v_1 \ j$ $= 30520\ i -19080\ j$

$1250 \ v_2 = 30520$

$v_2=24.4 \ m/s$

From, $750\ v_1 = 19080$

We get, $v_1=25.4 \ m/s$

Therefore the speed of Tesla car before collision = 25.4 m/s

The speed of ford car before collision = 24.4 m/s

6 0

3 years ago

On planet X, the absolute pressure at a depth of 2.00 m below the surface of a liquid nitrogen lake is 5.00 × 105 N/m2 . At a de

eimsori [14]

Answer:

300000.01008 Pa

123.76237 m/s²

Explanation:

$\rho$ = Density of liquid nitrogen = 808 kg/m³

h = Depth

g = Acceleration due to gravity

P = Atmospheric pressure

Absolute Pressure is given by

Below 2 m from surface

$P_a=P+\rho gh\\\Rightarrow 5\times 10^5=P+808g\times 2$

Below 5 m from surface

$P_a=P+\rho gh\\\Rightarrow 8\times 10^5=P+808g\times 5$

Subtracting the above equations we get

$-3\times 10^5=-808g3\\\Rightarrow g=\frac{3\times 10^5}{808\times 3}\\\Rightarrow g=123.76237\ m/s^2$

The acceleration due to gravity on the planet is 123.76237 m/s²

Equating the value of g in the first equation

$5\times 10^5=P+808\times 123.76237\times 2\\\Rightarrow P=5\times 10^5-808\times 123.76237\times 2\\\Rightarrow P=300000.01008\ Pa$

The atmospheric pressure on the planet is 300000.01008 Pa

7 0

4 years ago

An arrow is shot vertically upward at a rate of 250ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in

SIZIF [17.4K]

Answer:

The arrow is at a height of 500 feet at time t = 2.35 seconds.

Explanation:

It is given that,

An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s

The projectile formula is given by :

$h=-16t^2+v_ot$

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

$-16t^2+250t=500$

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds

So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.

6 0

4 years ago

Read 2 more answers

I would like to know the answers to both 10 & 11

oksian1 [2.3K]

Answer:

picture is not showing up sorry

Explanation:

7 0

4 years ago

28.2 km equal how many meters

gogolik [260]

Answer:exactly 28200 meters

Explanation:

6 0

3 years ago