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zmey [24]
3 years ago
10

A package is dropped from a helicopter moving upward at 15m/s. If it takes 10 s before the package strikes the ground, how high

above the ground was the ground was the package when it was released if air resistance is negligible?
(A) 408 m
(B) 272 m
(C) 204 m
(D) 340 m
Physics
2 answers:
Vlad1618 [11]3 years ago
6 0

Answer:

A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released? (Disregard air resistance.)

Show step by step please.

Note: The answer is given it's should be 1000 m ??

This what i can up with so see what it is kid

Explanation:

Anni [7]3 years ago
4 0

Answer:

The right answer is (D) 340m

Explanation:

If the package is dropped from a helicopter moving upward and the air resistance is negligible, the only acting force in the package is the gravity force. Therefore you can consider this as a Vertical Projectile Motion problem.

The initial speed of the package V₀ is 15m/s.

The acceleration of the package G is 9.8m/s².

The initial hight is Y₀. Therefore:

Y(t)=Y₀+V₀·t-0.5·G·t²

But we know than T(10s)=0m

0m=Y₀+150m-490m

Y₀=340m

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7 0
3 years ago
A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
4 years ago
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