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3 years ago

Help a homie plz

Physics

1 answer:

Sever21 [200]3 years ago

4 0

Answer:

the 4 resistor wired in parallel is equal to 23.946ohm

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Do you think distance and time are relevant terms in describing motion?

olya-2409 [2.1K]

Answer:

yes

Explanation:

because motion is relevant

6 0

3 years ago

Read 2 more answers

Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict

GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = $m$

Mass of the heavier cart = $2m$

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ $KE = \frac{mv^2}{2}$

⇒ $KE = \frac{mv^2}{2}\times \frac{m}{m}$

⇒ $KE = \frac{m^2v^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(mv)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}$

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ $K=\frac{(F\times t)^2}{2m}$ ...equation (i)

KE (K1) of mass 2m.

⇒ $K_1=\frac{(F\times t)^2}{2\times 2m}$

⇒ $K_1=\frac{(F\times t)^2}{4m}$ ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ $\frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}$

⇒ $\frac{K_1}{K} =\frac{2m}{4m}$

⇒ $\frac{K_1}{K} =\frac{2}{4}$

⇒ $\frac{K_1}{K} =\frac{1}{2}$

⇒ $K_1=\frac{K}{2}$

⇒ $K_1=\frac{1}{2}K$

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0

3 years ago

Please help on this one?

vlada-n [284]

Answer:

idk

Explanation:

7 0

3 years ago

Plz help asap

kolezko [41]

Answer:

Given the body part: They would administer a vaccine
Plate tectonics: It describes earthquakes

5 0

3 years ago

Saturated water vapor at 200 kPa is condensed into a saturated liquid via a constant-pressure process inside of a piston-cylinde

evablogger [386]

Answer:

The process is not possible

Explanation:

if we want to determine if the process is possible , we can check with the second law of thermodynamics

ΔS≥ ∫dQ/T

for a constant temperature process ( condensation)

ΔS≥ 1/T ∫dQ

and from the first law of thermodynamics

ΔH = Q - ∫VdP , but P=constant → dP=0 → ∫VdP=0

Q=ΔH

then

ΔS≥ΔH/T

from steam tables

at P= constant = 200 Kpa → T= 120°C = 393 K

at P= constant → H vapor = 2201.5 kJ/kg , H liquid = 1.5302 kJ/kg

, S vapor= 7.1269 kJ/kg , S liquid 1.7022 kJ/kg

therefore

ΔH = H vapor - H liquid = 2201.5 kJ/kg - 1.5302 kJ/kg = 2199.9698 kJ/kg

ΔS = S vapor - S liquid = 7.1269 kJ/kg - 1.7022 kJ/kg = 5.4247 kJ/kg

therefore since

ΔS required = ΔH/T = 2199.9698 kJ/kg/(393 K)= 5.597 kJ/kg K

and

ΔS= 5.4247 kJ/kg ≤ ΔS required=5.597 kJ/kg K

the process is not possible

5 0

3 years ago