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4 years ago

Air resistance is a type of friction true or false

Physics

2 answers:

Illusion [34]4 years ago

7 0

Air resistance is a type of friction.

TEA [102]4 years ago

7 0

Answer:

True

Explanation:

In fluid dynamics, drag (sometimes called air resistance, a type of friction, or fluid resistance, another type of friction or fluid friction) is a force acting opposite to the relative motion of any object moving with respect to a surrounding fluid.

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I don't get this question could someone help out

ratelena [41]

Answer:

that would probably be Rock A is harder than Rock B

Explanation:

Because if Rock A can scratch Rock B then it obviously means that Rock A is harder.

Right?

Hope This Helps You Out♡

8 0

3 years ago

One terminal of a car battery is said to be connected to “ground.” since it is not really connected to the ground, what is meant

GalinKa [24]

This expression means that the negative terminal (-) is connected to the metal chassis or engine, which means that all voltages used for the electrical devices in the car are measured with respect to the car's chassis or engine.
Today's vehicles have a negative ground system, which means that the vehicle's steel frame or chassis is directly connected to the negative side of the battery via the negative battery cable.

3 0

3 years ago

Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g

ollegr [7]

Answer:

The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

velocity of wave on the string with greater tension;

$v_1 = \sqrt{\frac{T_1}{\mu }$

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

$v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}$

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

$v_2^2 = \frac{T_2v_1^2}{T_1} \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s$

Fundamental frequency of wave on the string with greater tension;

$f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1} =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz$

Beat frequency = F₂ - F₁

= 270.6 - 258

= 12.6 Hz

Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0

4 years ago

To live deep ocean what condition do a fish need to be adapted for

tekilochka [14]

High pressure
Low oxygen
Darkness

7 0

3 years ago

A thermos contains m1 = 0.89 kg of tea at T1 = 31° C. Ice (m2 = 0.075 kg, T2 = 0° C) is added to it. The heat capacity of both w

lorasvet [3.4K]

Answer:

a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

b) T = 295.37 K

Explanation:

Given;

Initial temperature of tea T1 = 31 C

Initial temperature of ice T2 = 0 C

Mass of tea m1 = 0.89 kg

Mass of ice m2 = 0.075kg

The heat capacity of both water and tea c = 4186 J/(kg⋅K)

the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg

And T = the final temperature of the mixture

Heat loss by tea = heat gained by ice

m1c∆T1 = m2c∆T2 + m2Lf

m1c(T1-T) = m2c(T-T2) + m2Lf

m1cT1 - m1cT = m2cT - m2cT2 + m2Lf

m1cT + m2cT = m1cT1 + m2cT2 - m2Lf

T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

Substituting the values;

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)

T = 22.37 °C

T = 273 + 22.37 K

T = 295.37 K

4 0

3 years ago

Read 2 more answers