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1 year ago

a force vector f has a magnitude of 12.0 n. it is oriented 60° to the left of the y ax what are its x and y components?

1 answer:

5 0

The force vector that has a magnitude of 12.0 N. and is oriented 60° to the left of the (y) has the followings components:

v(x) =6 N
v(y) = 10.39 N

To solve this exercise the formulas and procedures we will use are:

v(x) = v * cosine (angle)
v(y) = v * sine (angle).

Where:

v= magnitude of the vector
v(x) = component of the vector on the (x) axis
v(y) = component of the vector on the (y) axis
angle = angle

Information about the problem:

angle = 60º
v = 12.0 N
v(x)= ?
v(y)= ?

Applying the formula of the component of the vector in the (x) axis we have:

v(x) = v * cosine (angle).

v(x) = 12.0 N * cosine (60º)

v(x) =6 N

Applying the formula of the component of the vector in the (y) axis we have:

v(y) = v * sine (angle)

v(y) = 12.0 N * sine (60º)

v(y) = 10.39 N

<h3>What is a vector?</h3>

It can be said to be a straight line described by a point (a) and (b) that has direction and sense.

Learn more about vector at: brainly.com/question/2094736

#SPJ4

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Power is measured in unit of Joules per second or

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3 years ago

What metric unit would you use to estimate the actual distance between Boston and New York?

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3 years ago

The half-life of cobalt-60 is 5.26 years. If 50 g are left after 15.8 years, how many

PtichkaEL [24]

Answer:

400 g

Explanation:

The computation of the number of grams in the original sample is shown below:

Given that

half-life = 5.26 years

total time of decay = 15.8 years

final amount = 50.0 g

Now based on the above information

number of half-lives past is

= 15.8 ÷ 5.26

= 3 half-lives

Now

3 half-lives = 1 ÷ 8 remains = 50.0 g

So, the number of grams would be

= 50.0 g × 8

= 400 g

4 0

2 years ago

Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while a

wel

Answer:

a) $v=7.32m/s$

b) $\alpha =-35$ º

c) $ΔK=-1094.62J$

Explanation:

From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

So, If we analyze the collision we got

$p_{1x}=p_{2x}$

To simplify the problem, lets name D for Daniel and R for Rebecca

a) $p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}$

Since Daniel's initial velocity is 0

$m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}$

$v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}$

$v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s$

Now, lets analyze the movement in the vertical direction

$p_{1y}=p_{2y}$

Since $p_{1y}=0$

$0=m_{D}v_{D2y}+m_{R}v_{R2y}$

$v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s$

Now, we can find the magnitude of Daniel's velocity after de collision

$v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s$

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle

$\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35$ º

c) The change in the total kinetic energy is:

ΔK= $K_{2}-K_{1}$

ΔK= $\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J$

That means that the kinetic energy decreases

5 0

3 years ago

A 1520 kg car is rolling<br> down a frictionless<br> 14.4° slope. What is<br> its acceleration?

Umnica [9.8K]

Answer:

a = 2.44 m/s²

Explanation:

The Net force acting on the car downwards the plane will be equal to the component of the weight that is parallel to the plane.

Thus;

mg(sin θ) = ma

Thus;. g(sin θ) = a

g is acceleration due to gravity = 9.81 m/s²

θ = 14.4°

Thus;

9.81 × sin 14.4 = a

a = 2.44 m/s²

7 0

2 years ago