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3 years ago

What is the average speed of a car that travels 15 km in 4000. seconds?

Physics

1 answer:

Katyanochek1 [597]3 years ago

8 0

Answer:

266.66

Explanation:

and you divide 4000 by 15 to get your answer

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A 6.00-μf parallel-plate capacitor has charges of 40.0 μc on its plates. how much potential energy is stored in this capacitor?

dedylja [7]

Thew energy stored in a capacitor of capacitance $C$ and voltage between the plates $V$ is

$E=\frac{1}{2} CV^2=\frac{1}{2C} Q^2$ .

Substituting numerical value

$E=\frac{1}{2*6*10^{-6}} (40*10^{-6})^2\\ E=133.33\; \mu J$

7 0

3 years ago

An ideal spring hangs from the ceiling. A 1.95 kg mass is hung from the spring, stretching the spring a distance d=0.0865 m from

uranmaximum [27]

Answer:

kinetic energy = 0.1168 J

Explanation:

From Hooke's law, we know that ;

F = kx

k = F/x

We are given ;

Mass; m = 1.95 kg

Spring stretch; d = x = 0.0865

So, Force = mg = 1.95 × 9.81

k = 1.95 × 9.81/0.0865 = 221.15 N/m

Now, initial energy is;

E1 = mgL + ½k(x - L)²

Also, final energy; E2 = ½kx² + ½mv²

From conservation of energy, E1 = E2

Thus;

mgL + ½k(x - L)² = ½kx² + ½mv²

Making the kinetic energy ½mv² the subject, we have;

½mv² = mgL + ½k(x - L)² - ½kx²

We are given L=0.0325 m

Plugging other relevant values, we have ;

½mv² = (1.95 × 9.81 × 0.0325) + (½ × 221.15(0.0865 - 0.0325)² - ½(221.15 × 0.0865²)

½mv² = 0.62170875 + 0.3224367 - 0.82734979375

½mv² = 0.1168 J

7 0

3 years ago

The table below shows data of sprints of animals that traveled 75 meters. At each distance marker, the animals' times were recor

mr_godi [17]

Answer:

Explanation:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

5 0

3 years ago

Read 2 more answers

A long. 1.0 kg rope hangs from a support that breaks, causing the rope to fall, if the pull exceeds 43 N. A student team has bui

raketka [301]

Answer:

6.8 m/s2

Explanation:

Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is

W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N

For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of

a = F/m = 13.6 / 2 = 6.8 m/s2

So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail

8 0

3 years ago

Using a force of 28.0 Newton, a student pulls a 70.0 Newton weight along the tabletop for a distance of 15.0 meters in 3.0 secon

Musya8 [376]

Answer:

140 watt

Explanation:

We are given that

Force applied by student ,F=28 N

Weight pulled by students=70 N

Displacement,s=15 m

Time=3 s

We have to find the power developed by the student.

Work done=w= $F\times s$

Work done by the student= $28\times 15=420 J$

Power= $\frac{work\;done}{time}$

Using the formula

Power= $\frac{420}{3}=140watt$

Hence, the power developed by the students=140 watt

8 0

3 years ago