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djverab [1.8K]
3 years ago
11

If a plane is traveling at a velocity of 100 km/hr East, and if the wind velocity is 25 km/hr to the East, then what is the velo

city of the plane relative to an
observer on the ground below?
100 km/hr East
75 km/hr West
75 km/hr East
125 km/hr East
Physics
1 answer:
kkurt [141]3 years ago
5 0

Answer:

75 west

Explanation:

I think not sure lucky guess

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Learning Goal:To understand and be able to use the rules for determining allowable orbital angular momentum states.Several numbe
Free_Kalibri [48]

Answer:

Explanation:

1) for a given n value the l value can be from 0 to n-1

So if n= 5 it can take 0,1,2,3,4

i.e it can take 5 values

2)for an electron with l =3

it can be from -3 -2 -1 0 1 2 3

i.e it can take 7 values

3) n = 3 !!

l = 0 , 1 , 2

for l=0 , m = 0 total = 1

for l= 1 ,m = -1,0,1 total = 3

for l = 2, m=-2,-1,0,1,2 total = 5

5+3+1 = 9

total possible states = 9 * 2 = 18

Answer is 168

4)given l=3 and n=3

orbital quantum number cannot be equal to principal quantum number

its max value is l-1 only

5)L = sqrt(l(l+1))x h'

for it to be max l should be max

for n = 3 max l value is 2

therfore it is sqrt(2(2+1)) x h'

\sqrt(6) \times h'

this is the answer

5 0
3 years ago
A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.
kondaur [170]

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

F_k = \mu_k N = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

3 0
3 years ago
Explain the relationship between the current output of the power supply and the current through each component in the parallel c
baherus [9]

Explanation:

Current output at the battery will be current of entire circuit, while the current through each bulb in the parallel circuit is the total current circuit.

So, current output through power supply is i and current through each component be i_1, i_2 , i_3 considering only three component.

Then in a parallel circuit

i = i_1+i_2+i_3

4 0
2 years ago
A student is given a sample of matter and asked to determine whether it is an element.
Pepsi [2]
I think the correct answer is C
5 0
2 years ago
Read 2 more answers
Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
Tatiana [17]

Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

Explanation:

Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m

x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
3 years ago
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