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3 years ago

Which is an example of qualitative data?

Physics

1 answer:

LiRa [457]3 years ago

6 0

The answer is C)the rating that the golfers give Callaway clubs

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Subduction occurs when:

Inessa05 [86]

Explanation:

a plate with oceanic crust sinks beneath a plate with

continental crust is the answer

5 0

3 years ago

A train moving at 5 m/sec passed a track gang and then accelerated uniformly a rate of 1.2 m/sec/sec for 5 min. How far did the

Amanda [17]

Answer:

S = V0 t + 1/2 a t^2

S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)

S = 1500 m + .6 * 90000 m = 55,500 m

Check: V0 = 5 m/s

V2 = V0 + a t = 5 + 1.2 * 300 = 365 m/s

Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s (note uniform motion)

S = 185 * 300 = 55,500 m

We calculated V2 above at 365 m/s the speed after 300 sec

3 0

3 years ago

When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is

padilas [110]

Answer:

$P_{C} = 3.2\, atm$

Explanation:

Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:

Bulb A (2 L, 2 atm) - Before opening:

$P_{A} \cdot V_{A} = n_{A} \cdot R_{u} \cdot T$

Bulb B (3 L, 4 atm) - Before opening:

$P_{B} \cdot V_{B} = n_{B} \cdot R_{u} \cdot T$

Bulbs A & B (5 L) - After opening:

$P_{C} \cdot (V_{A} + V_{B}) = (n_{A} + n_{B})\cdot R_{u} \cdot T$

After some algebraic manipulation, a formula for final pressure is derived:

$P_{C} = \frac{P_{A}\cdot V_{A} + P_{B}\cdot V_{B}}{V_{A}+V_{B}}$

And final pressure is obtained:

$P_{C} = \frac{(2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L)}{5\,L}$

$P_{C} = 3.2\, atm$

5 0

3 years ago

A string is waved up and down to create a wave pattern with a wavelength of 0.5 m. If the waves are generated with a frequency o

GrogVix [38]

Wave speed = (wavelength) x (frequency)

Wave speed = (0.5 m) x (2 /sec)

<em>Wave speed = 1 m/s</em>

4 0

3 years ago

A series-parallel circuit consists of two parallel circuits connected in series across a 45-V source. One parallel branch consis

musickatia [10]

Answer:

The answer to the question is

The current through R4 = 0.5865 mA

Explanation:

To solve this we list out the known thus

Voltage source = 45-V

Firsrt parallel circiuit has resistances of R1 = 17 kΩ and R2 = 23 kΩ

Second parallel circiuit has resistances of R3 = 45 kΩ and R4 = 55 kΩ

we first find the current flowing in the circuit by finding thr sum of the total resistance in eah parallel circuit

Firsrt parallel circiuit, for circuit in parallel, sum of resistance 1/RT1 = 1/ R1 + 1/R2 = 1/17+1/23 = 0.1023017 therefore RT1 = 1/0.1023017 = 9.775 kΩ

Similarly we have for the second parallel circuit 1/RT2 = 1/R3 + 1/R4

= 1/45 + 1/55 = 0.0404 Hence RT2 = 1/0.0404 = 24.75 kΩ

This means that the 9.775 kΩ and the 24.75 kΩ are in series hence total resistance of the circuit = sum of all resistances in series

= 9.775 kΩ + 24.75 kΩ = 34.525 kΩ

However current, I is given by V/R = 45-V/34.525 kΩ = 1.303 × 10⁻³ A or 1.303 mA

From the current divider rule, I4 = I × (R4/(R3+R4)

That is the currrent flowing throuhgh R4 = 1.303 × (45/(45+55)) = 0.5865 mA

3 0

4 years ago