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3 years ago

Pressing two objects together with more force __________ friction

Physics

1 answer:

AysviL [449]3 years ago

6 0

Pressing two objects together with more force Increase friction

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A playground merry-go-round of radius 2.00 m has a moment of inertia I = 275 kg · m2 and is rotating about a frictionless vertic

ryzh [129]

Answer:

0.2932 rad/s

Explanation:

r = Radius = 2 m

$I_i$ = Initial angular momentum = $275\ kgm^2$

$\omega_i$ = Initial angular velocity = 14 rev/min

$I_f$ = Final angular momentum

$\omega_f$ = Final angular velocity

Here the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow 275\times 14\times \dfrac{2\pi}{60}=(275+275(2)^2)\omega_f\\\Rightarrow \omega_f=\dfrac{275\times 14\times \dfrac{2\pi}{60}}{275+275(2)^2}\\\Rightarrow \omega_f=0.2932\ rad/s$

The final angular velocity is 0.2932 rad/s

7 0

3 years ago

Where is heat transferred by conduction in a lava lamp?

Semmy [17]

It is transferred by direct contact, say you touched it, you would feel the heat

8 0

3 years ago

Find the wavelength λ of the 80.0-khz wave emitted by the bat. express your answer in millimeters.

vfiekz [6]

Answer:

4.29 millimeters

Explanation:

Bats emit ultrasound waves: in air, ultrasound waves travel at a speed of

$v=343 m/s$

The frequency of the waves emitted by this bat is:

$f=80.0 kHz = 80,000 Hz$

Therefore we can find the wavelength of the wave emitted by the bat by using the relationship between speed, frequency and wavelength:

$\lambda=\frac{v}{f}=\frac{343 m/s}{80,000 Hz}=4.29\cdot 10^{-3} m=4.29 mm$

4 0

3 years ago

Analyzing a blood sample usually involves which type of separation method?

schepotkina [342]

Answer:(D.)

Explanation:

4 0

3 years ago

An insulating cup contains 200 grams of water at 25 ∘C. Some ice cubes at 0 ∘C is placed in the water. The system comes to equil

Nataly [62]

Answer:

The amount of ice added in gram is 32.77g

Explanation:

This problem bothers on the heat capacity of materials

Given data

Mass of water Mw= 200g

Temperature of water θw= 25°c

Temperature of ice θice= 0°c

Equilibrium Temperature θe= 12°c

Mass of ice Mi=???

The specific heat of ice Ci= 2090 J/(kg ∘C)

specific heat of water Cw = 4186 J/(kg ∘C)

latent heat of the ice to water transition Li= 3.33 x10^5 J/kg

heat heat loss by water = heat gained by ice

N/B let us understand something, heat gained by ice is in two phases

Heat require to melt ice at 0°C to water at 0°C

And the heat required to take water from 0°C to equilibrium temperature

Hence

MwCwΔθ=MiLi +MiCiΔθ

Substituting our data we have

200*4186*(25-12)=Mi*3.3x10^5+

Mi*2090(12-0)

837200*13=Mi*3.3x10^5+Mi*2090

10883600=332090Mi

Mi=10883600/332090

Mi= 32.77g

4 0

3 years ago