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Vilka [71]
2 years ago
9

Karissa is conducting an experiment on the amount of salt that dissolves in water at different temperatures. She repeats her tes

ts several times using the same procedure.
What can Karissa do to further increase confidence in the results of her experiment?

have another scientist run the same exact procedure in her lab
include several other liquids for comparison
calculate the average amount of salt that dissolves to summarize findings
make sure the experiment follows a specific procedure that allows other scientists to produce the same findings
Physics
2 answers:
faltersainse [42]2 years ago
5 0

Answer:

C.) Calculate the average amount of salt that dissolves to summarize findings

Explanation:

Just telling you, nobody reads these things. They just look at the answer and say oh i got it right now but you should bc i got a 50 on this quiz and this was one of them that was right so believe me when i say, i got this question right. UwU <333333333

Oksanka [162]2 years ago
3 0

Answer:

idk

Explanation:

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Suppose that you lift four boxes individually, each at a constant velocity. The boxes have weights of 3.0 N, 4.0 N, 6.0 N, and 2
Alona [7]

Answer:

The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

Explanation:

Worked : work can be defined as the product of force and distance.

The S.I unit of work is Joules (J).

Mathematically it can be represented as,

W = F×d.................. Equation 1

d = W/F.............................. Equation 2

where W = work, F = force, d = distance.

<em>Given: W = 12 J</em>

(i) for the 3.0 N weight,

using equation 2

d = 12/3

d= 4 m.

(ii) for the 4.0 N weight,

d = 12/4

d = 3 m.

(iii) for the 6.0 N weight,

d = 12/6

d = 2 m.

(iv) for the 2.0 N weight,

d = 12/2

d = 6 m

Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

8 0
2 years ago
The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm
Nesterboy [21]

Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

Answer:

225 N

Explanation:

From Pascal's principle,

F/A = f/a ...................... Equation 1

Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.

Making f the subject of the equation,

f = F(a)/A ..................... Equation 2

Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².

Substituting into equation 2

f = 15000(3/200)

f = 225 N.

Hence the downward force that must be applied to small piston = 225 N

8 0
2 years ago
Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als
lesantik [10]

Answer:

b)

Explanation:

By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.

This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.

(As the positive charge would move away from positive charges and would  be attracted by negative ones).

So, the combination of answers that is true is b) (positive, negative, positive).

3 0
2 years ago
in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at
Tema [17]
<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:

</span>

3 0
2 years ago
In an adiabatic process oxygen gas in a container is compressed along a path that can be described by the following pressure p,
viktelen [127]

Answer:

Explanation:

In case of gas , work done

W = ∫ p dV , p is pressure and dV is small change in volume

the limit of integration is from Vi to Vf .

= ∫ p dV

=  ∫ p₀V^{-\frac{6}{5}  dV

= p₀ V^{-\frac{6}{5} +1} / ( \frac{-6}{5} +1 )

=  - 5p₀ V^{-\frac{1}{5}

Taking limit from Vi  to Vf

W = - 5 p₀ ( V_f^\frac{-1}{5} - V_i^{\frac{-1}{5}  ) ltr- atm.

7 0
2 years ago
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