A car accelerates from 0 m/s to 20 m/s in 5 seconds,

A table is pushed 15m across a room with a force of 50n.

kari74 [83]

Answer:

The work is 750 J

The power is 37.5 watt

Explanation:

Mechanical Work and Power

Mechanical work is the amount of energy transferred by a force. It's a scalar quantity, with SI units of joules.

Being $\vec F$ the force vector and $\vec s$ the displacement vector, the work is calculated as:

$W=\vec F\cdot \vec s$

If both the force and displacement are parallel, then we can use the equivalent scalar formula:

W=F.s

Power is the amount of energy transferred or converted per unit of time. In the SI, the unit of power is the watt, equal to one joule per second.

The power can be calculated as:

$\displaystyle P=\frac {W}{t}$

Where W is the work and t is the time.

a)

The table is pushed d=15 m across a room with a force F=50 N, thus the work is:

W=50 * 15 = 750 J

The work is 750 J

b)

The time needed to push the table is t=20 seconds, thus:

$\displaystyle P=\frac {750}{20}=37.5$

The power is 37.5 watt

7 0

2 years ago

Where are you most likely to find a trench?

telo118 [61]

You are most likely to find a trench at a divergent boundary between two plates.

*additional note

You don't have to but if you want to I recommend downloading this:

https://phet.colorado.edu/en/simulation/legacy/plate-tectonics

I'm currently in seventh grade and the assignments we are given in science is focused on earth and its layers. (because of schools closing ya know, online assignments and stuff, our teacher gave this helpful simulation sort of game)

Hope that helps! (it's basically a simulation of how plates form the way they do)

5 0

2 years ago

If you push a book a against a wall hard enough , it will not slide down even through gravity is pulling it

sattari [20]

Yes it will not hope this helps you can i get brainlist(-;

7 0

3 years ago

Two identical air-filled parallel-plate capacitors C1 and C2, each with capacitance C, are connected in series to a battery that

solmaris [256]

Answer:

a) The ratio is $\frac{U}{U_{0}} =\frac{2}{(1+\frac{1}{\kappa})}$ .

b) The total stored energy increases after the dielectric is inserted.

Explanation:

Before the dielectric is inserted:

We have two parallel plate capacitors ( $C_{1}$ and $C_{2}$ ), each with a capacitance C, connected in series to a battery that has voltage V.

The two capacitors can be replaced by an equivalent one. The equivalent capacitance for two capacitors in series is

$\frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{2}{C}$

The potential differences across the two capacitors are:

$\Delta V_{1}=\frac{Q}{C_{1}}=\frac{Q}{C}$ and $\Delta V_{2}=\frac{Q}{C_{2}}=\frac{Q}{C}$

The total potential difference $\Delta V$ is the sum of the potential differences across each capacitor:

$\Delta V=\Delta V_{1}+ \Delta V_{2}=2\frac{Q}{C}$

The initial stored energy $U_{0}$ is:

$U_{0}=\frac{1}{2}C_{eq} \Delta V^{2}=\frac{1}{2}\frac{C}{2}(2\frac{Q}{C})^{2} \ \Longrightarrow\ U_{0}=\frac{Q^{2}}{C}$

Using $\Delta V=V$ instead we have that $U_{0}=\frac{1}{4}CV^{2}$ .

After the dielectric is inserted:

A dielectric with $\kappa > 1$ is inserted between the plates of one of the capacitors while the two capacitors are connected to the battery. The total potential difference reamins the same. If Q is the charge in the plates without the dielectric, in the capacitor with the dielectric the charge increases to $\kappa Q$ and the capacitance to $\kappa C$ .

Now we have that the equivalent capacitance is:

$\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{\kappa C}=\frac{1}{C} (1+\frac{1}{\kappa})$

So the stored energy after the dielectric is inserted $U$ becomes:

$U=\frac{1}{2}C_{eq}\Delta V^{2}=\frac{1}{2} \frac{C}{(1+\frac{1}{\kappa})}V^{2} \Longrightarrow\ U=\frac{1}{2(1+\frac{1}{\kappa})}CV^{2}$

So, the ratio $\frac{U}{U_{0}}$ is:

$\frac{U}{U_{0}}=\frac{\frac{1}{2(1+\frac{1}{\kappa})}CV^{2}}{\frac{1}{4}CV^{2}} \Longrightarrow\ \frac{U}{U_{0}}=\frac{2}{(1+\frac{1}{\kappa})}$

Because $\frac{U}{U_{0}} >1$ if $\kappa >1$ this means that the total stored energy increases after the dielectric is inserted.

8 0

3 years ago

Match the words givne below to the appropriate blanks in the sentences on the right. Make certain each sentence is complete befo

Pepsi [2]

Answer: forward

The child will continue to move forward at the before the crash speed. This is because the child at the instant of head on collision of the car will tend to continue the state of uniform motion the car exhibited as newton's first law states. Newtons first law: A body will continue to be at rest or in uniform motion with constant velocity/speed unless acted upon by an external force.

Explanation:

The child will continue moving forward until he/she encounters some obstruction or resistance which could be dangerous. In a safety design car, the child should collide with the SRS airbag right in front which gets activated for situations where the acceleration of the car is above 250m/s².

Thank you for reading and I hope this was helpful.

6 0

3 years ago