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3 years ago

What is the main force that must be overcome in order to push an object

Physics

1 answer:

My name is Ann [436]3 years ago

4 0

Answer: Friction

Explanation:

Friction and the normal force would be the two initial forces to overcome.

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It’s for a cross word and mechanical doesn’t fit ☹️

VikaD [51]

Answer:

Circular wave

Explanation:

Circular waves are special types of mechanical waves. They all travel through a material medium or some times a vacuum.

An example of such wave is a ripple caused by dropping a stone in a tank of water.

A wave that propagates in circular form on the surface of water falls into this category.

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3 years ago

When playing a game of disc golf, each throw to your target is considered to be a O Point O Stroke O Hit O Toss

Hunter-Best [27]

Answer:

an o point i think

Explanation:

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3 years ago

imagine that a new planet is discovered with two moons of equal mass, moon A and moon B. The mass of the new planet is greater t

sdas [7]

The gravitational pull on moon A will be lower since it is further away from the new planet.

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4 years ago

Read 2 more answers

Which of the following is true about radiation from the sun?

Lelu [443]

><span>It can travel through vacuum.

The rays must travel in the vacuum of space between Earth's atmosphere and the sun.</span>

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4 years ago

Add a vector whose magnitude is 13 with angle 27 degrees to one whose magnitude is 11 with angle 45 degrees? Put the length firs

mafiozo [28]

Answer:

Magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

Explanation:

Magnitude of first vector = $|A| = 13\ \text{units}$

Angle = $\theta_1=27^{\circ}$

Magnitude of second vector = $|B| = 11\ \text{units}$

Angle = $\theta_2=45^{\circ}$

x component of first vector

$A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}$

y component of first vector

$A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}$

x component of second vector

$B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}$

y component of first vector

$B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}$

Adding the magnitudes

$C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}$

$C_y=A_y+B_y=5.9+7.8\\\Rightarrow C_y=13.7\ \text{units}$

Magnitude of the sum of the vectors would be

$|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}$

The direction would be

$\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}$

The magnitude of the vector is $23.75\ \text{units}$ and the direction is $35.23^{\circ}$

4 0

3 years ago