Using only one management style with all people is the most effective leadership technique for any organization.

What is the month of winter

Tanya [424]

Period of months where the weather is the coldest and the days are the shortest.

5 0

3 years ago

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What’s the difference between gravitational force & gravitational field strength?

Zepler [3.9K]

Answer:

Gravitational field strength is the force experienced by a unit mass. Gravitational force is the amount of force acting on a body. It is the product of field strength times the mass under consideration. Gravitational pull is just a more colloquial name for gravitational force.

Explanation:

hope it helps u

5 0

3 years ago

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Please help me with this question, especially the angle part. I have solved the initial velocity already. I don't know what to d

Helga [31]

I think you almost got it.

At the top, the velocity only has horizontal component, so v=12 m/s is v_x, which is v*cos(theta), because v_x is constant, so the same when it was launched or now.

With the value of the initial speed (28 m/s, which is the total speed), you can set

v_x = v * cos( theta ) ---> 12 = 28*cos(theta) --> cos(theta)=12/28=3/7

or theta = 64.62 deg, it is D. Think about it. I hope you see it.

5 0

3 years ago

Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1

harkovskaia [24]

Answer: 14.1 m/s

Explanation:

We can solve this with the Conservation of Linear Momentum principle, which states the initial momentum $p_{i}$ (before the elastic collision) must be equal to the final momentum $p_{f}$ (after the elastic collision):

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=m_{1}V_{i} + m_{2}U_{i}$

$p_{f}=m_{1}V_{f} + m_{2}U_{f}$

Where:

$m_{1}=200 kg +100 kg=300 kg$ is the combined mass of Tubby and Libby with the car

$V_{i}=10 m/s$ is the velocity of Tubby and Libby with the car before the collision

$m_{2}=25 kg + 100 kg=125 kg$ is the combined mass of Flubby with its car

$U_{i}=0 m/s$ is the velocity of Flubby with the car before the collision

$V_{f}=4.12 m/s$ is the velocity of Tubby and Libby with the car after the collision

$U_{f}$ is the velocity of Flubby with the car after the collision

So, we have the following:

$m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f}$ (2)

Finding $U_{f}$ :

$U_{f}=\frac{m_{1}(V_{i}-V_{f})}{m_{2}}$ (3)

$U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg}$ (4)

Finally:

$U_{f}=14.1 m/s$

8 0

3 years ago

A metallic wire has a diameter of 4.12mm. when the current in the wire is 8.00a, the drift velocity is 5.40×10−5m/s.

Darina [25.2K]

R = 2.06 mm = 2.06 x 10^(-3) m
Q = 1.6 x 10^(-19) C
v = 2.5 x 10^(-5) m/s
I = 8 A = 8 C/s
A = r² π = ( 2.06 x 10^(-3) ) ² x 3.14 = 13.325 x 10^(-6 ) m² =
= 1.3325 x 10^(-5) m²
I = n Q v A
n = I / (Q v A)
n = 8 C/s / ( 1.6 x 10 ^(-19) * 5.4 x 10^(-5) * 1.3325 x 10^(-5) ) =
= 0.694 x 10^(29) m^(-3)
n = 6.94 x 10^(28) m^(-3)

8 0

3 years ago

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