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Brut [27]
3 years ago
8

Su una strada rettilinea ci sono due semafori che distano 98 m e che diventano verdi contemporaneamente. Dal primo semaforo part

e un’auto che sì muove alla velocità costante di 43 km/h; dall’altro semaforo parte, nel verso opposto, un camper che viaggia alla velocità costante di 36 km/h.
Determina quanto tempo dopo l'accensione del verde i due automezzi sì affiancano e quanta strada ha percorso l'automobile in quel momento.
Physics
1 answer:
pantera1 [17]3 years ago
3 0

Answer:

Le due rette rappresentano le leggi del moto di due punti

materiali (per esempio due macchine) che si muovono di moto

rettilineo uniforme. La retta meno pendente (velocità più bassa)

è riferita alla macchina che parte da una posizione iniziale più

avanti rispetto alla seconda macchina. Però, la seconda

macchina, avendo una velocità maggiore della prima (retta più

pendente), nonostante parta più indietro rispetto alla prima, dopo

un certo tempo la raggiungerà. Pertanto, il punto P, intersezione

delle due rette, rappresenta la situazione fisica in cui la seconda

macchina raggiunge la prima. In particolare, le coordinate del

punto P (tempo e spazio), rappresentano l’istante e la posizione

in cui la seconda macchina raggiunge la prima.

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Answer:

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Explanation:

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mixer [17]

Answer:

Power = 21[W]

Explanation:

Initial data:

F = 35[N]

d = 18[m]

In order to solve this problem we must remember the definition of work, which tells us that it is equal to the product of a force for a distance.

Therefore:

Work = W = F*d = 35*18 = 630 [J]

And power is defined as the amount of work performed in a time interval.

Power = Work / time

Time = t = 30[s]

Power = 630/30

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3 years ago
A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 2.50 × 1012 m/s2 in a machine. If the proton h
Veronika [31]

Answer:

(A) Speed will be 44.18\times 10^4m/sec

(b) Change in kinetic energy = 1560\times 10^{-19}      

Explanation:

We have given mass of proton m=1.67\times 10^{-27}kg

Acceleration of the proton a=2.50\times 10^{12}m/sec^2

Initial velocity u = 1.60\times 10^4 m/sec

Distance traveled by proton s = 3.90 cm = 0.039 m

(a) From third equation of motion we know that

v^2=u^2+2as

v^2=(1.60\times 10^4)^2+2\times 2.5\times 10^{12}\times 0.039

v=44.18\times 10^4m/sec

(b) Initial kinetic energy KE_I=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (1.6\times 10^4)^2

Final kinetic energy KE_F=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (44.18\times 10^4)^2

So change in kinetic energy \Delta KE=KE_F-KE_I=\frac{1}{2}\times 1.6\times 10^{-27}\times 10^8\times (44.18^2-1.6^2)=1560\times 10^{-19}J

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