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Andrej [43]
3 years ago
15

During which moon phase do spring tides occur?

Physics
1 answer:
Murrr4er [49]3 years ago
6 0

Answer:

The Full Moon and New Moon

Explanation:

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A 70 kilogram hockey player skating east on an ice rink is hit by a 0.1 kilogram hockey puck moving toward the west. The puck ex
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Given:

70 Kilogram hockey

Hit by 0.1 kilogram

50 Newton

Description:

So in this case this bacially means that one object is experiencing a force then another object.  So the answer to the question will be 50N.

Answer: 50N (3rd Law)

Hope this helps.

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3 years ago
Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (Let |a| = 22 lb and |b| = 16 lb. Rou
SVEN [57.7K]

Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here

Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)  

Answer:

Resultant Force=33.8 lb

Angle=67.2°

Explanation:

Given data

Fa=22 lb

Fb=16 lb

Θ=55⁰

To find

(i) Resultant Force F

(ii)Angle α

Solution

First we need to represent the forces in vector form

\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j

Total Force

F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j

The Resultant Force is given as

|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb

For(ii) angle

We can find the angle bu using tanα=y/x

So

tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}

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3 years ago
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MissTica
The ground exerts an equal force on the golf ball.
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Two long, parallel wires are separated by 2.2 mm. Each wire has a 32-AA current, but the currents are in opposite directions. Pa
Alex

Answer:

B=1.1636*10^{-3}T

Explanation:

Given data

d_{wires}=2.2mm=0.022m\\ I_{current}=32A\\

To find

Magnitude of the net magnetic field B

Solution

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