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Topic: **How to solve problem solving in physics****Question:**
This is it:
A flower pot falls from a window 25.0 meters above the ground
How fast is the flowerpot moving when it strikes the ground?
How much time does someone have to get out of the way?
How in the world do I go about solving this??

June 17, 2019 / By Courtney

D = 0.5at^2 + Vo *t And Vo = 0 since the pot starts from rest on the window sill 25 m = 0.5 * 9.8 m/s * t^2 Solve for t. ( answer to 2 part) V = At V = 9.8 m/s^2* t from above. For answer to 1st part

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take note that when you write these solutions all preceding text after V, I , and R are all subscripted (e.g. Vr1 should be written as V with a subscript of "R1", and R1 should be written as "R" with a subscript of "1") 1) let R1,R2,R3,R4 be 4.9, 9.1, 3.6, and 8.6 ohms respectively Given Ir2 = 0.25 A Solution: V=IR Vr2=(Ir2)(R2)=(0.25)(9.1)=2.275 V since R2 // R1 (read as R2 parallel to R1) Vr1=Vr2=2.275 V let R5 be the equivalent resistance of R1 and R2 R5=(R1)(R2) / (R1+R2)= (4.9)(9.1) / (4.9+9.1) = 3.185 ohms let R6 be the equivalent resistance of R3 and R4 R6=(R3)(R4) / (R3+R4)= (3.6)(8.6) / (3.6+8.6) = 2.538 ohms let It be the total current across the circuit since R5 and R6 are now in series Ir5=Ir6=It It=Vr5 / R5 = 2.275 / 3.185 = 0.714 A let Req be the equivalent resistance of the circuit Req=R5 + R6 = 3.185 + 2.538 = 5.723 ohms let Vt be the total applied voltage Vt=(It)(Req)=0.714 * 5.723 = 4.086 V 2) as previously solved Vr1 = Vr2 = 2.275 V Vr3 = Vr4 = Vt - Vr1 = Vt - Vr2 = 4.086 - 2.275 = 1.811 V 3) Ir1 = Vr1 / R1 = 2.275 / 4.9 = 0.464 A Ir2 = Vr2 / R2 = It - Ir1 = 2.275 / 4.9 = 0.714 - 0.464 = 0.25 A (same as given value) Ir3 = Vr3 / R3 = 1.811 / 3.6 = 0.503 A Ir4 = Vr4 / R4 = It - Ir3 = 1.811 / 8.6 = 0.714 - 0.503 = 0.211 A 4) i'll leave the sketch to you, that's too basic... *voltage across parallel resistances are the same *voltage is additive on series *current is additive on parallel *resistances in series is additive what grade are you in?

till now fixing any situation convert each and every unit into S.I. unit(m.ok.s. equipment). or use the comparable equipment for each unit. be conscious: one million knot = one million.852 km/hr one million: GIVEN velocity of deliver w.r.t earth =Va=7Km/hr NW velocity of flow =3 kn =(3*one million.852) = 4.fifty six km/hr SW be certain the two vectors alongside X and Y axes. The components alongside x axis could upload up jointly being interior the comparable course while the climate alongside the y axis are in opposite course so locate there distinction. as a result the cost of the deliver w.r.t earth may be the ensuing of the two values alongside the x & y axes. V= (Vx^2+Vy^2)^one million/2. 2: draw each and each vector as stated interior the question such that the tail of the 2nd touches the pinnacle of the 1st. the ensuing ofthe vectors i.e. the vector drawn from th tail of the 1st vector to the pinnacle of the final vector could be your answer. this way you get the two significance and th course at as quickly as

You solve it by knowing the physics and deriving the equations to work. The physics starts with definitions. distance = average speed X time traveled, in shorthand that's S = Vavg T. S = 25 m. You need to find T and V, the end speed upon impact. When acceleration A is assumed to be constant, we can write Vavg = (V + U)/2 where U = 0 is the initial speed; so Vavg = V/2. And now we have S = (V/2) T and we have both V and T where we want them, in the derived equation, so we can solve for them. One more bit of physics to learn...V = AT = gT; where A is acceleration and A = g is the acceleration due to gravity, which is about 9.8 m/s^2 near Earth's surface. So now S = (V/2) T = gT/2 * T = 1/2 gT^2 and, ta da, you have S and g; so you can solve for T = sqrt(2S/g) = ? seconds. You can do the math. ANS. Now that you have T, you can find speed at impact V = gT = ? mps. Plug and chug. ANS. There you are...how you go about solving this; derived from distance = average speed X time traveled, which is the physics.

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Use the following formulae. 1. s=ut +1/2 a*t*t s=distance covered (here it is height - 25M) u= initial velocity (here it is 0 ) t = time taken for the fall a = acceleration ( here it is 'g' - 9.81 m/sec.Square) From this equation, you can calculate the time of fall. 2. v=u+at v = final velocity a = acceleration ( here it is 'g' - 9.81 m/sec.Square) u= Initial velocity = 0 t = time of fall (calculated above)

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Basic equal and opposite reaction 30 kg x 500 m/s = 2000 kg x V m/s (15000 kg m/s) / 2000 kg = V m/s 7.5 m/s = V m/s The cannon has a velocity of 7.5 m/s

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