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3 years ago

What is the difference between a primary and a secondary cell?

1 answer:

5 0

Primary cells are the ones which cannot be recharged and have to be discarded after the expiration of the lifetime whereas, secondary cells need to be recharged when the charge gets over. Both the types of battery are used extensively in various appliances and these cells differ in size and material used in them.

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A 3047.8 kg truck has lost its brakes coming down a mountain. Fortunately, there is a ramp of thick gravel inclined at 9.5 degre

Yuliya22 [10]

Answer:

The work done by the gravel to stop the truck is 520.44 kJ

Explanation:

Step 1: Data given

Mass of the truck = 3047.8 kg

The ramp has an angle of 9.5 °

Velocity of the truck = 20.68 m/s

distance = 26.6 meters

Step 2: Calculate initial kinetic energy

sin 9.5° = 0.165

h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m

Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ = initial kinetic energy

Step 3: Calculate potential energy

Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ

Step 4: What work is done by the truck on the gravel?

Frictional energy Ef = 651.7 kJ - 131.26 kJ = 520.44 kJ

5 0

4 years ago

A mass of 10 g of oxygen fill a weighted pistonâ€“cylinder device at 20 kPa and 110Â°C. The device is now cooled until the tempe

mezya [45]

Answer:

The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

Explanation:

Given that,

Mass of oxygen = 10 g

Pressure = 20 kPa

Initial temperature = 110°C

Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

$\Delta V=V_{2}-V_{1}$

$\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})$

Put the value into the formula

$\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)$

$\Delta V=14.297\ L$

$\Delta V=14.3\times10^{-3}\ m^3$

Hence, The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

6 0

3 years ago

At the nose of a missile in flight, the pressure and temperature are 5.6 atm and 850°R, respectively. Calculate the density and

Contact [7]

To solve this problem we will apply the definition of the ideal gas equation, where we will clear the density variable. In turn, the specific volume is the inverse of the density, so once the first term has been completed, we will simply proceed to divide it by 1. According to the definition of 1 atmosphere, this is equivalent in the English system to

$1atm = 2116lb/ft^2$

The ideal gas equation said us that,

PV = nRT

Here,

P = pressure

V = Volume

R = Gas ideal constant

T = Temperature

n = Amount of substance (at this case the mass)

Then

$\frac{n}{V} = \frac{P}{RT}$

The amount of substance per volume is the density, then

$\rho = \frac{P}{RT}$

Replacing with our values,

$\rho = \frac{5.6*2116}{1716*850}$

$\rho = 0.00812slug/ft^3$

Finally the specific volume would be

$v = \frac{1}{\rho}$

$v = 123ft^3/slug$

6 0

3 years ago

A trait of an organism that helps it survive in its environment is calles

kompoz [17]

Answer is adaptation. An organism develops a trait over time to help survive in its environment called an adaptation. You could take a giraffe for example. A long time ago giraffes actually had short necks, but now since their food is higher they soon developed a longer neck and this is what we now see in the present. This goes for any artic animal. Polar bears and seals have a white fur adaptation to help them blend in with their environment. A chameleon changes colors in order to hide from predators and sneak up on prey. These are all adaptations

7 0

3 years ago

Read 2 more answers

A spring that is compressed 14.5 cm from its equilibrium position stores 2.99 J of potential energy. Determine the spring consta

strojnjashka [21]

Answer:

284.4233 N/m

Explanation:

k = Spring constant

x = Compression of spring = 14.5 cm

U = Potential energy = 2.99 J

The potential energy of a spring is given by