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SSSSS [86.1K]
3 years ago
9

Explain why it is important to keep all variables that might affect the dependent variable, other than the manipulated variable

constant?
Physics
2 answers:
zlopas [31]3 years ago
7 0
So that when you see a result of an experiment or equation, you know WHICH variable caused it.
Inga [223]3 years ago
5 0
<span>You have multiple confounding variables, you cannot accurately conclude the relationship between the manipulated and dependent variable because the other variables that are not controlled for could be the reason for seeing a certain change</span>
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what is the name of the area around a charged object where the object can exert a force on other charged objects?
shusha [124]

Answer:

An electric field is a region around a charged object where the object's electric force is exerted on other charged objects. Electric fields get weaker the farther away they are from the charge. An electric field is invisible. You can use the field line to represent it.

Explanation:

4 0
2 years ago
Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.18 m away from a waterfall 0.294 m in heigh
Karolina [17]

Answer:

v = 7.65 m/s

t = 0.5882 s

Explanation:

We are told that the salmon started downstream, 3.18 m away from a waterfall.

Thus, range = 3.18 m

Since the horizontal velocity component is constant, then;

Range = vcosθ × t

Thus,

vcosθ × t = 3.18 - - - (eq 1)

We are told the salmon reached a height of 0.294 m

Thus, using distance equation;

s = v_y•t + ½gt²

g will be negative since motion is against gravity.

s = v_y•t - ½gt²

Thus;

0.294 = v_y•t - ½gt²

v_y = vsinθ

Thus;

0.294 = vtsinθ - ½gt² - - - (eq 2)

From eq(1), making v the subject, we have;

v = 3.18/tcosθ

Plugging into eq 2,we have;

0.294 = (3.18/tcosθ)tsinθ - ½gt²

0.295 = 3.18tanθ - ½gt²

We are given g = 9.81 m/s² and θ = 45°

0.295 = (3.18 × tan 45) - ½(9.81) × t²

0.295 = 3.18 - 4.905t²

3.18 - 0.295 = 4.905t²

4.905t² = 2.885

t = √2.885/4.905

t = 0.5882 s

Thus;

v = 3.18/(0.5882 × cos45)

v = 7.65 m/s

8 0
3 years ago
If a system has 475 kcal of work done to it, and releases 5.00 × 102 kJ of heat into its surroundings, what is the change in int
Andrews [41]
First, we convert kcal to joules:
1 kcal = 4.184 kJ
475 kcal = 1987.4 kJ

Now, calculating the change in internal energy:

ΔU = Q + W; where Q is the heat supplied to the system and W is the work done on the system.

ΔU = -500 + 1987.4
ΔU = 1487.4 kJ
4 0
3 years ago
Read 2 more answers
Help !!
Vaselesa [24]
The answer is ; 6cm

Hope this helps!
Please give Brainliest!

This is because of the diagram below:

6 0
2 years ago
A 10-cm long solenoid has 100 turns and a radius of 5 cm. If it carries a current of 2 A, What is the magnetic field B inside th
OLga [1]

Hi there!

We can use Ampère's Law:

\oint B \cdot dl = \mu_0 i_{encl}

B = Magnetic field strength (B)
dl = differential length element (m)
μ₀ = Permeability of free space (T/Am)

Since this is a closed-loop integral, we must integrate over a closed loop. We can integrate over a rectangular-enclosed area of the rim of the solenoid - ABCD - where AD and BC are perpendicular to the solenoid.

Thus, the magnetic field is equivalent to:
\oint B \cdot dl = \int\limits^A_B {B} \, dl  + \int\limits^B_C {B} \, dl   + \int\limits^C_D {B} \, dl   + \int\limits^D_A {B} \, dl

Since AD and BC are perpendicular, and since:
\oint B \cdot dl = B \cdot L = BLcos\phi

BLcos(90) = 0

If perpendicular to the field, the equation equals 0.

Additionally, since AB is outside of the solenoid, there is no magnetic field present, so B = 0. The only integral we integrate now is:
\oint B \cdot dl = \int\limits^C_D {B} \, dl

Which is horizontal and inside the solenoid. Let the distance between C and D be 'L', and the enclosed current is equivalent to the number of loops multiplied by the current:

B L = \mu_0 Ni

N = # of loops per length multiplied by the length, so:
BL = \mu_0 nL i \\\\B = \mu_0ni

Plug in the given values and solve. Remember to convert # of loops to # of loops per unit length.

B = \mu_0 (100/0.1)(2) = (4\pi *10^{-7})(1000)(2) = \boxed{0.00251 T}

6 0
2 years ago
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