All categories

3 years ago

Energy may be measured in?

Physics

1 answer:

hram777 [196]3 years ago

6 0

Answer:

Energy May be measured in joule

You might be interested in

A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc

marin [14]

Answer:

Time period of the osculation will be 0.0671 sec

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So $kx=mg$

$k\times 0.04=0.012\times 9.8$

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

$T=2\pi \sqrt{\frac{m}{k}}$

$=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec$

4 0

3 years ago

As the wavelength increases, the frequency (2 points) Select one:

LiRa [457]

A.
According to the equation: v=λ*f
λ stands for wavelength. wavelength increases, the frequency will decrease. lower frequency deserves low energy.

7 0

3 years ago

When have you experienced an increase in kinetic<br> energy within a system?

Mars2501 [29]

Answer:

If a man starts running on a boat with an acceleration a with respect to the boat, there is no external force that acts on the Boat+Man system

8 0

3 years ago

How many air molecules are in a 13.0×12.0×10.0 ft room (28.2 L=1 ft3)? Assume atmospheric pressure of 1.00 atm, a room temperatu

aksik [14]

Answer:

1963.93 Moles

Explanation:

-We know the standard conversion ratio for the volume of a mole is $1 Mole=22.4L$

Given volume of rooms as $13.0ft\times12.0ft\times10ft=1560 ft^3$

Convert the volume into liters: $1560\times28.2l=43992l$

#From our conversion ratio above, we get the volume of air molecules in moles as:

$V_m=\frac{43992L}{22.4L}\times 1\ mole\\\\=1963.93\ Moles$

Hence, the volume of air molecules is 1963.93 Moles

4 0

2 years ago

Consider an e. coli to be a cylinder with a diameter of 1 micrometer (um) and with a length of 2 micrometer (um).

julia-pushkina [17]

Answer:

$A=6.28\ \mu m^2$

Part 1

$V=1.57 \ \mu m^3$

Part 2

$A=6.28\ \mu m^2$

Explanation:

Given that

Diameter,d=1 μm

Length ,l=2 μm

As we know that volume of cylinder given as

$V=\pi r^2l$

$V=\pi \times 0.5^2\times 2 \ \mu m^3$

$V=1.57 \ \mu m^3$

Surface area,A

A=π d l

$A=\pi \times 1 \times 2\ \mu m^2$

$A=6.28\ \mu m^2$

Part 1

$V=1.57 \ \mu m^3$

Part 2

$A=6.28\ \mu m^2$

4 0

3 years ago

Energy may be measured in?​

Energy may be measured in?