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1 year ago

Does hot air rise? Why?

Physics

1 answer:

pashok25 [27]1 year ago

8 0

Answer: Yes

Explanation:

does hot air rise

er substances, expands when heated and contracts when cooled. Because there is more space between the molecules, the air is less dense than the surrounding matter and the hot air floats upward. This is the concept used in the hot air balloons.

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How do you make an electromagnet

Ivan

Putting a piece of iron or steel inside the coil makes the magnet strong enough to attract objects. The strength of an electromagnet can be increased by increasing the number of loops of wire around the iron core and by increasing the current or voltage.

7 0

2 years ago

Read 2 more answers

9) What will happen to the period of a pendulum if we change the rope of a pendulum with another which is four times with the in

Arturiano [62]

Double

Explanation:

Since the period T of a pendulum is given by

$T = 2\pi \sqrt{\dfrac{l}{g}}$

By increasing the length of the pendulum by 4, the period becomes

$T' = 2\pi \sqrt{\dfrac{4l}{g}} = 2\left(2\pi \sqrt{\dfrac{l}{g}}\right) = 2T$

You can see that the period doubles when we increase the length by a factor of 4.

5 0

2 years ago

A horizontal spring with spring constant 210 Nm is compressed by 20 cm and then used to launch a 250 g box across the floor. The

Damm [24]

Answer:

ugmd = 1/2 kx²

d = (1/2 kx²) / (ugm)

= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)

= 7.4 m

ugmd = 1/2 mv²

v = √2ugd

= √(2(0.23)(9.81 m/s²)(7.4 m)

= 5.8 m/s

Explanation:

3 0

2 years ago

The radioactive 60co isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequenc

Ivanshal [37]

1. Frequency: $3.23\cdot 10^{20} Hz$

The energy given is the energy per mole of particles:

$E=1.29\cdot 10^{11} J/mol$

1 mole contains a number of Avogadro of particles, $N_A$ , equal to

$N_A=6.022\cdot 10^{23}$ particles

So, by setting the following proportion, we can calculate the energy of a single photon:

$1.29 \cdot 10^{11} J/mol : 6.022 \cdot 10^{23} ph/mol = E_1 : 1 ph\\E_1 = \frac{(1.29\cdot 10^{11} J/mol)(1 ph)}{6.022\cdot 10^{23} ph/mol}=2.14\cdot 10^{-13} J$