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3 years ago

Which variable is the result of the experiment?

Physics

2 answers:

Cloud [144]3 years ago

5 0

Answer:

The independent variable is the core of the experiment and is isolated and manipulated by the researcher. The dependent variable is the measurable outcome of this manipulation, the results of the experimental design.

lana66690 [7]3 years ago

4 0

Answer:

independent variable is the answer! hope that helps :D

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The diagram below shows the oscillation of a

galben [10]

Answer:

I think option B 0.125

Explanation:

but i am not sure so don't mind

8 0

3 years ago

A 2.0 kg block, initially moving at 10.0 m/s, slides 50.0 m across a sheet of ice beforecoming to rest. What is the magnitude of

inna [77]

Answer:

The magnitude of the average frictional force on the block is 2 N.

Explanation:

Given that.

Mass of the block, m = 2 kg

Initial velocity of the block, u = 10 m/s

Distance, d = 50 m

Finally, it stops, v = 0

Let a is the acceleration of the block. It can be calculated using third equation of motion. It can be given by :

$v^2-u^2=2ad$

$-u^2=2ad$

$a=\dfrac{-u^2}{2d}\\\\a=\dfrac{-(10\ m/s)^2}{2\times 50\ m}\\\\a=-1\ m/s^2$

The frictional force on the block is given by the formula as :

F = ma

$F=2\ kg\times (-1)\ m/s^2\\\\F=-2\ N$

|F| = 2 N

So, the magnitude of the average frictional force on the block is 2 N. Hence, this is the required solution.

8 0

4 years ago

2500m =_____km

Mekhanik [1.2K]

Answer:

1. 2.5km

2. 4.8m

3. 0.075 litre

4. 65000mg

5. 5600g

6. 0.5m

7. 63mm

Explanation:

1. 1000m in 1 km so divide by 1000

2. 100cm in m so divide by 100

3. 1000ml in 1l so divide by 1000

4. multiply the mass value by 1000

5.multiply the mass value by 1000

6. divide the length value by 100

7. multiply the length value by 10

hope this helps

5 0

3 years ago

A ball is dropped and falls with an acceleration of 9.8m/s2 downward. it hit the ground with a velocity of 49 m/s downward. how

Lelechka [254]

V=u+at
49=0+9.8t
t=49/9.8
t = 5 (seconds)

4 0

3 years ago

Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l

alex41 [277]

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:

The magnitude of the gravitational force between two objects is given by the formula

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A - Star A, we have:

$m_1=M_p\\m_2 = M_s\\r=R$

So the force is

$F_A=G\frac{M_p M_s}{R^2}=F_0$

For the system planet B - Star B, we have:

$m_1 = 4 M_p\\m_2 = M_s\\r=R$

So the force is

$F=G\frac{4M_p M_s}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

$m_1 = M_p\\m_2 = 4M_s\\r=R$

So the force is

$F=G\frac{M_p (4M_s)}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

$v=\sqrt{\frac{GM}{r}}$

For System A,

$M=M_s\\r=R$

So the tangential speed is

$v_A=\sqrt{\frac{GM_s}{R}}$

For system B,

$M=M_s\\r=R$

So the tangential speed is

$v_B=\sqrt{\frac{GM_s}{R}}=v_A$

So, the speed of planet B is the same as planet A.

For system C,

$M=4M_s\\r=R$

So the tangential speed is

$v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A$

So, the speed of planet C is twice the speed of planet A.

3 0

3 years ago