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2 years ago

What investigations allow for the control of variables

Physics

2 answers:

lina2011 [118]2 years ago

7 0

Most as long the hypothesis is a good answer and can be answered

PolarNik [594]2 years ago

3 0

I believe that the statement is true. Investigations allow for the control of variables. Changing variables will lead you to observations that may prove your hypothesis

You might be interested in

To model this process, assume two charged spherical conductors are connected by a long conducting wire and a 1.20-mC charge is p

il63 [147K]

Answer:

Part a: The electric potential of each sphere is 1.35x10⁸V

Part b: The electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

Explanation:

As the complete question is not given, the similar question is attached herewith. The values are used as indicated in the given question

Let r_1 = 6 cm=0.06 m

r2 = 2 cm = 0.02 m

Q = 1.2 mC

Let q1 and q2 are the charges on each sphere.

q1 + q2 = 1.2 mC -------(1)

In the equilibrium, V1 = V2

k*q1/r1 = k*q2/r2

q1/0.06 = q2/0.02

q1/q2 = 0.06/0.02

q1/q2 = 3 ---------(2)

On solving equation 1 and 2

we get

q1 = 0.9 mC

q2 = 0.3 mC

V1 = k*q1/r1 = (9*10^9*0.9*10^-3)/0.06 = 1.35*10^8 Volts

V2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02 = 1.35*10^8 Volts

So the electric potential of each sphere is 1.35x10⁸V

Part b

Now the electric potential is given as

E1 = k*q1/r1^2 = 9*10^9*0.9*10^-3/0.06^2 = 2.25*10^9 N/C

E2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02^2 = 6.75*10^9 N/C

So the electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

7 0

3 years ago

A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a

katen-ka-za [31]

Answer:b

Explanation:

Given

mass of first cart $m_1=6 kg$

mass of second cart $m_2=3 kg$

velocity of first cart $v_1=3 m/s$

conserving momentum

$m_1v_1+m_2v_2=(m_1+m_2)v$

$6\times 3+3\times 0=(9)\cdot v$

$v=2 m/s$

Initial kinetic Energy $K.E._1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$

$K.E._1=\frac{1}{2}\cdot 6\cdot 3^2+0$

$K.E._1=27 J$

Final Kinetic Energy

$K.E._2=\frac{1}{2}(m_1+m_2)v^2$

$K.E._2=\frac{1}{2}(6+3)\cdot 2^2=18 J$

Ratio of initial Kinetic Energy to the Final Kinetic Energy

$=\frac{27}{18}=1.5$

6 0

3 years ago

A.) If its booster rockets accelerate the space shuttle at 15m/s2, how high will it be one minute after launch?

poizon [28]

Answer:

27,000 m

450 m/s

Explanation:

Assuming the initial velocity is 0 m/s:

v₀ = 0 m/s

a = 15 m/s²

t = 60 s

A) Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²

Δy = 27,000 m

B) Find: v_avg

v_avg = Δy / t

v_avg = 27,000 m / 60 s

v_avg = 450 m/s

5 0

3 years ago

A special rocket can produce 7.66 ✕ 10^5 N of instantaneous thrust with an exhaust speed of 3.05 ✕ 103 m/s in vacuum. What mass

adelina 88 [10]

The mass of fuel the engine burn each second to produce a thrust of 7.66×10⁵ N is 2.5×10² kg/s.

Mass can be defined as the quantity of matter contained in a body. The S.I unit of mass is kilogram(kg)

To calculate the mass the engine burns each seconds, we use the formula below.

Formual:

M = T/v............. Equation

Where:

M = Mass per seconds of the rocket
T = Thrust
v = Velocity

From the question,

Given:

T = 7.66×10⁵ N
v = 3.05×10³ m/s

Substitute these values into equation 1

M = (7.66×10⁵)/(3.05×10³)
M = 2.5×10² kg/s

Hence, the mass of fuel burned in each second is 2.5×10² kg/s.

Learn more about mass here: brainly.com/question/25121535

#SPJ1

8 0

1 year ago

Two service technicians are discussing the best way to thread a rod. Technician A says they should use a tap and die set. Techni

Musya8 [376]

Answer:

Both technicians are right, to be able to make a threaded joint you need to use the external thread on one part of the rod using the tap and die set, and on the other side of the rod you need to have an internal thread using the thread repair insert kit

8 0

3 years ago