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maria [59]
3 years ago
9

I need help ! if anyone answers this its worth 40 points please help !!

Physics
1 answer:
Sliva [168]3 years ago
3 0
Double displacement...I think
You might be interested in
Q1.
Vesnalui [34]

Answer:

P = 450 J

Explanation:

Given that,

Mass of a child, m = 18 kg

The vertical distance from the top to the bottom of the slide is 2.5 metres.

The Gravitational field strength = 10 N/kg

We need to find the decrease in gravitational potential energy of the child sliding from the  top to the bottom of the slide.

The formula for the gravitational potential energy is given by :

P = mgh

Substituting all the values,

P = 18 kg × 10 m/s² × 2.5 m

P = 450 J

Hence, the decrease in gravitational potential energy is 450 J.

4 0
2 years ago
10. What is the acceleration of a 1000 kg car subject to a 500 N net force?
Simora [160]

Answer:

0.5m/s^2

Explanation:

We can use the formula [ F = ma ] but solve for "a" since that is what we are looking for.

F = ma

F/m = a

We know the net force and mass so substitute those values and simplify.

500/1000 = 0.5m/s^2

Best of Luck!

3 0
2 years ago
To understand the interplay of observations and models you must first be able to distinguish between things that we observe and
Svetllana [295]

Answer: a) Observation

b) observation

c) observation

d) observation

e) inference

f) inference

g) inference

h) inference

Explanation:

observation: The photosphere is made mostly of hydrogen and helium,The photosphere emits mostly visible light,The corona is hotter than the photosphere, The Sun emits neurtrinos

inferences: The Sun generates energy by fusing hydrogen into helium,The core temperature is 10 million k, The convection zone is cooler than the radiation zone,The composition of the photosphere is the same as that the gas cloud that have birth to our solar system.

3 0
3 years ago
working alongside the pharmacist , one of the duties a medication reconciliation technician would perform is to a ) deal with va
Vanyuwa [196]

Answer:

okkiikkkkkkkkoiiiiiiii8iiii

5 0
2 years ago
A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th
Stells [14]

Answer:

The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     V=\frac{V_{o}}{\kappa}

The capacitance is modified as:

                                              C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}

The stored energy without the dielectric is

                                               PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}

The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

5 0
3 years ago
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