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2 years ago

If the x-component of a force vector is 5.17 newtons and its y-component is 8.80 newtons, then what is its magnitude?

Physics

1 answer:

olya-2409 [2.1K]2 years ago

3 0

Answer:

10.21 N

Explanation:

As the force is a vector, it can be decomposed in two components perpendicular each other, so there is no projection of one component in the direction of the other.

When divided in this way, the magnitude of the resultant vector can be found simply applying trigonometry, as follows:

F² = Fx² + Fy² ⇒ F = √(Fx)²+(Fy)²

Replacing by Fx= 5.17 N and Fy = 8.8 N, we get:

F = √(5.17)²+(8.8)² =10.21 N

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Starting at the same time, an arrow and a ball are shot horizontally with a speeds of 50 m/s and 44 m/s respectively from the to

Yuri [45]

Answer:

option D

Explanation:

given.

horizontal velocity of arrow and a ball given as 50 m/s and 44 m/s respectively from the top of a building over flat ground.

In vertical direction, they are both identical

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Their acceleration due to gravity is same for both arrow and a ball 9.8 m/s²

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2 years ago

A proton moving in the positive x direction with a speed of 9.9 105 m/s experiences zero magnetic force. When it moves in the po

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Answer:

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

Explanation:

In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:

$\vec{F_B}=q\vec{v}\ X\ \vec{B}$ (1)

v: speed of the proton = 9.9*10^5 m/s

q: charge of the proton = 1.6*10^-19C

B: magnetic field = ?

FB: magnetic force on the proton = 1.6*10^-13N

When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:

^j X (-^i) = -(-^k)=^k

To obtain the magnitude of the magnetic field you use:

$F_B=qvBsin90\°=qvB\\\\B=\frac{F_B}{qv}=\frac{1.6*10^{-13}N}{(1.6*10^{-19}C)(9.9*10^5m/s)}\\\\B=1.01T$

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

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3 years ago

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2 years ago

Rubidium forms the positive ion Rb+. Does this ion have more or fewer electrons than the neutral atom?

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Answer:

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Explanation:

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3 years ago

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