A net force of 540 N is applied to a frictionless sled with a mass of 250 kg. What acceleration will the sled have?A. 5.5 m/s^2
B. 2.2 m/s^2
C. 4.5 m/s2
D. 3.3 m/s^2

Answer:

The thickness of sponge is the dependent variable in this investigation

Explanation:

This is because in the investigation, the amount of mass is the property to be varied while the thickness of the sponge is the property to be checked.

The thickness of the sponge is dependent on the amount of mass placed on it, so the thickness of the sponge depends on the mass.

Thus, the thickness of the sponge is the dependent variable in this instance while the amount of mass is the independent variable in this investigation.

Final answer:

In this scientific investigation about the relationship between the thickness of a sponge and the mass placed upon it, the 'dependent variable' is the sponge's thickness, and the 'independent variable' is the mass applied. The hypothesis may be that there's an exponential relationship between the mass and the sponge's thickness.

Explanation:

In the given scientific investigation, the dependent variable is the thickness of the sponge. This is because the thickness is the aspect of the experiment that changes (or is measured) based on the manipulation of another variable, which in this case is the amount of mass placed on top of the sponge, known as the independent variable. The investigator has control over the independent variable and observes whether changes to it cause a reaction or change in the dependent variable. In other words, the thickness of the sponge 'depends' upon the mass placed on it during the experiment. A possible hypothesis for this experiment may be that the thickness of the sponge decreases (get thinner) as more mass is placed upon it, suggesting an exponential relationship between the mass and the sponge's thickness.

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The rate at which the voltage of the given circuit is changing is gotten to be;

dV/dt = 0.452 V/s

We are given;

Current; I = 3 A

Resistance 1; R1 = 4Ω

Resistance 2; R2 = 3Ω

dR1/dt = 0.4 Ω/s

dR2/dt = 0.2 Ω/s

dI/dt = 0.02 A/s

Now, formula for voltage with resistors in parallel is;

1/V = (1/I)(1/R1 + 1/R2)

Plugging in the relevant values, we can find V;

1/V = (1/3)(1/4 + 1/3)

Simplifying this gives;

1/V = 0.194

Now, we want to find the rate at which the voltage is charging, we need to find dV/dt.

Thus, let us differentiate 1/V = (1/I)(1/R1 + 1/R2) with respect to t to get;

(1/V)²(dV/dt) = [(1/i²)(di/dt)(1/R1 + 1/R2)] + (1/I)[(1/R1²)(dR1/dt) + (1/R2²)(dR2/dt)]

Plugging in the relevant vies gives us;

0.194²(dV/dt) = [(1/3²)(0.02)(¼ + ⅓)] + (⅓)[(1/3²)(0.4) + (1/4²)(0.3)]

>> 0.037636(dV/dt) = 0.001296 + 0.0157

>> dV/dt = 0.016996/0.037636

>> dV/dt = 0.452 V/s

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Answer:

Explanation:

As we know that two resistors are in parallel

so we have

where we know that

so we have

now to find the rate of change we have

now from above equation we have

24.7.....................................

Answer:

See the anwers below

Explanation:

1) Series Circuit

2) Insulator Ohms

3) Moving Electrons

4) Chemical

5) Mechanical

6) Volts

7) Amps

8) Ohms

9) Voltage

10) Short Circuit

11) Open Circuit

12) Circuit Breaker, Fusible Link, or a Fuse