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2 years ago

Do you thin it is possible to have the benefits of the Agricultural and Industrial revolutions without the environmental costs?

Physics

1 answer:

Julli [10]2 years ago

4 0

Answer:No

Explanation:There is no current technology which supports zero waste exhaust

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Jenny pushes a 40 N crate down the hall 2m. How much work did she do?

Shalnov [3]

Work done = force x distance = 40 x 2 = 80 Joules.

8 0

2 years ago

Which statement is true about Gram negative organisms?

aliina [53]

Answer:

B, C and D are true.

<h3>Explanation:</h3>

A is false because they appear a pale reddish colour not purple.

3 0

2 years ago

When the spring, with the attached 275.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the

topjm [15]

Answer:

The period of oscillation is 1.33 sec.

Explanation:

Given that,

Mass = 275.0 g

Suppose value of spring constant is 6.2 N/m.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=\sqrt{\dfrac{k}{m}}$

Where, m = mass

k = spring constant

Put the value into the formula

$\omega=\sqrt{\dfrac{6.2}{275.0\times10^{-3}}}$

$\omega=4.74\ rad/s$

We need to calculate the period of oscillation,

Using formula of time period

$T=\dfrac{2\pi}{\omega}$

Put the value into the formula

$T=\dfrac{2\pi}{4.74}$

$T=1.33\ sec$

Hence, The period of oscillation is 1.33 sec.

4 0

2 years ago

A 241 kg mass is lifted 1.8 m. What is the potential energy of the mass (in J)?

Korolek [52]

Answer:

mass is lifted 1.8 m. What is the potential energy of the mass 4. A 100 kg

6 0

2 years ago

Read 2 more answers

What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa

dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

$E=3.315\times10^{-11}\ J$

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

6 0

2 years ago