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Pepsi [2]
3 years ago
11

One of the purposes of an experiment is to determine whether the dependent variable affects the independent variable. Please sel

ect the best answer from the choices provided T F
Physics
2 answers:
vredina [299]3 years ago
8 0

Answer: False

An independent variable in a scientific experimentation process is the one which can be changed or manipulated in a scientific experiment the effect of such a change can be observed on the dependent variable. The dependent variable is a variable which cannot be changed manually but the changes occur in the dependent variable are because of the independent variable. The changes that occur in the dependent variable are the outcomes of the experimentation process.

For example, the effect of the water and sunlight on the plant height can be used to explain the dependent and independent variables of the experiment. The dependent variable is the height of the plant which is dependent upon the independent variables such as water and sunlight.  

On the basis of the above explanation, One of the purposes of an experiment is to determine whether the dependent variable affects the independent variable is false.

sweet-ann [11.9K]3 years ago
7 0
The answer is False. False is the answer
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A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its
gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

d = 2\pi r ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

T = \frac{2\pi r}{v}

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
v = \sqrt{\frac{Gm}{r}}

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s

Now, we can solve for the period:

T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}

7 0
2 years ago
Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f
elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

W_T+W_F=K_f-K_i (1)

Where:

W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to \frac{1}{2}mv^{2}, so if the initial velocity v_i is equal to zero, the initial kinetic energy K_i is equal to zero.

Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0
2 years ago
The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
vovikov84 [41]

Gravitational force is given by, F= G\frac{mM}{R^{2}}

Where, m and M are the masses of the objects, R is the distance between them and G gravitational constant.

Gravitational force of the star on planet 1, F_{1}= G\frac{m_{1}M}{R^{2}}

Gravitational force of the star on planet 2, F_{2}= G\frac{3m_{1}M}{(3R)^{2}}

Ratio, \frac{F_{1}}{F_{2}}= \frac{\frac{Gm_{1}M}{R^{2}}}{\frac{G3m_{1}M}{(3R)^{2}}}

\frac{F_{1}}{F_{2}}=  \frac{3}{1}

Therefore, the gravitational force of the star on the planet 1 is three times that on planet 2.

6 0
2 years ago
Read 2 more answers
Which temperature is the hottest? 98 F or 39 C or 303K?<br> F= 1.8C + 32<br> C= (F-32)/1.8
sergejj [24]

Answer:

The hottest temperature is  T_2 = 39^o C

Explanation:

From the question we are given

    T_1 =  98 F

  T_2 =  39^oC

  T_3 =  303 \  K

Generally converting T_3 to  Fahrenheit

    T_3' =  (T_3 -273 ) * \frac{9}{5}  + 32

=> T_3' =  (303 -273 ) * \frac{9}{5}  + 32

=> T_3' = 86 F

Converting  T_2 to  Fahrenheit

      T_2' =  T_2 * \frac{9}{5}  + 32

=> T_2' =  39 * \frac{9}{5}  + 32

=> T_2' =102.2 F  

Now comparing  the temperature  in Fahrenheit we see that T_2  is the hottest

3 0
2 years ago
Which action describes an experimental investigation?
EastWind [94]
I think the answer is c
3 0
2 years ago
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