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dybincka [34]
3 years ago
7

. A helicopter is 8.50 m above the ground and climbing at 2.5 m/s. It drops a package from rest (relative to the helicopter). At

the moment that the package is released, what is its velocity relative to the ground?
Physics
1 answer:
Rudik [331]3 years ago
4 0

Answer:

2.5 m/s

Explanation:

The velocity of the package relative to the ground = the velocity of the package relative to the helicopter + the velocity of the helicopter relative to the ground

v = 0 m/s + 2.5 m/s

v = 2.5 m/s

At the moment it is released, the package is rising at 2.5 m/s.

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What does an unconformity in a sequence of rock layers reveal about geologic history?
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An unconformity, or abnormality in the layers of rocks, signifies that the process of layering was somehow interrupted. The only feasible choice for this would be (B), where erosion would have interrupted the layering process.
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3 years ago
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What is the average velocity of a car if it travels from position 25m to a position of -7m in 34 seconds?
zubka84 [21]

Answer:

vp = 0.94 m/s

Explanation

Formula

Vp = position/ time

position: Initial position - Final position

Position = 25 m - (-7 m) = 25 m + 7 m = 32 m

Then

Vp = 32 m / 34 seconds

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What shape is the graph produced by a force vs acceleration graph?
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6 0
2 years ago
A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar
m_a_m_a [10]

Answer:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

#Where u is in meters and t in seconds.

Explanation:

Given that :m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s

From \omega=kL we have:

k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2

From F_d(t)=-\gamma u\prime(t) we have that:

\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m

Now,given that the initial value problem is given by:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

Hence,the position of u at time t is given by

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\, u in meters,t in seconds.

4 0
3 years ago
What is the speed of light in meters per second?
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The answer is approximately 2.998e+8
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