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lubasha [3.4K]
3 years ago
7

A projectile is fired directly upwards at 49.4 m/s. A second projectile is dropped from rest at some higher elevation at the ins

tant the first projectile is fired and passes the first projectile 3.00 s later. From the frame of reference of the first projectile, what is the velocity of the second projectile as it passes by
Physics
1 answer:
8090 [49]3 years ago
8 0

Answer:

Explanation:

velocity of first projectile after 3 s

v = u - gt

v = 49.4 - 9.8 x 3

= 20 m /s

Velocity of second projectile after 3 s after being dropped from rest

v = u + gt

= 0 + 9.8 x 3

= 29.4 m /s

They will be moving in opposite direction at the time of meeting , so their relative velocity

= 20 + 29.4 = 49.4 m /s

From the frame of reference of the first projectile, the velocity of the second projectile will be 49.4 m /s .

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Explanation:A pascal is a pressure of one newton per square metre, or, in SI base units, one kilogram per metre per second squared.

I hope this helps.... I'm sorry if it doesn't

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What is Latent heat and also give types.<br>​
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2 years ago
The membrane that surrounds a certain type of living cell has a surface area of 4.7 x 10-9 m2 and a thickness of 1.3 x 10-8 m. A
vagabundo [1.1K]

Answer:

Q = 1.2*10⁻¹² C

Explanation:

  • For any capacitor, by definition the capacitance C is equal to the relationship between the charge on one of the conductors and the potential difference between them, as follows:

       C = \frac{Q}{V}  (1)

  • For the special case of a parallel plate capacitor, just by application of Gauss' law to a rectangular surface half out of the outer surface, and half inside it, it can be showed that the value of the capacitance C is a parameter defined only by geometric constants, as follows:

       C = \frac{\epsilon_{0}*\epsilon _{r} * A}{d}  (2)

  • So, due to the left sides in (1) and (2) are equal each other, right sides must be equal too.
  • Replacing ε₀, εr (dielectric constant), A, d and V by their values, we can solve for Q, as follows:

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3 years ago
Two particles, one with charge −7.97×10−6 C and the other with charge 6.91×10−6 C, are 0.0359 m apart. What is the magnitude of
kolbaska11 [484]

Answer:

-384.22N

Explanation:

From Coulomb's law;

F= Kq1q2/r^2

Where;

K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2

q1 and q2 = magnitudes of the both charges

r= distance of separation

F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2

F= -495.65 × 10^-3/ 1.29 × 10^-3

F= -384.22N

7 0
3 years ago
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