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3 years ago

A 35 N force makes a 10 degree angle with the positive x-axis. What is the magnitude of the vertical component of the force?

Physics

2 answers:

Snowcat [4.5K]3 years ago

5 0

Answer:

6.07 N

Explanation:

Given that,

Force, F = 35 N

It makes 10 degree angle with the positive x-axis.

We need to find the magnitude of the vertical component of the force. It can be given by :

$F_y=F\sin\theta\\\\=35\times \sin(10)\\\\=6.07\ N$

So, the magnitude of the vertical component of the force is 6.07 N.

Mariulka [41]3 years ago

5 0

Answer:

Vertical component of force is equal to $6.067$ N

Explanation:

The magnitude of the vertical component on any force when α is the angle between the direction of force and the X -axis is

$F_y = F * Sin\alpha$

Substituting the given values in above equation, we get -

$F_y = 35 * Sin 10\\F_Y = 35 * 0.1736\\F_Y = 6.067$

Vertical component of force is equal to $6.067$ N

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un conductor de plata (p=1,6x10°-6ohm x m) tiene una seccion de 5x10°-6 m°2 y una longitud de 110 m . calcular la resistencia

Alexus [3.1K]

Responder:

35,2 ohm.

Explicación:

Dado:

La resistencia específica del conductor es,

La longitud del conductor es,

El área de la sección transversal del conductor es,

Sabemos que la resistencia de un conductor es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

Por lo tanto, la resistencia se puede expresar como:

$R=\frac{\rho\times l}{A}$

Ahora, conecte los valores dados y resuelva para 'R'. Esto da,

$R=\frac{1.6\times 10^{-6}\ ohm\cdot m\times 110\ m}{5\times 10^{-6}\ m^2}\\\\R=35.2\ ohm$

Por lo tanto, la resistencia del conductor es de 35,2 ohm.

4 0

3 years ago

A bicyclist starting at rest produces a constant angular acceleration of 1.30 rad/s2 for wheels that are 35.5 cm in radius.

Debora [2.8K]

Answer:

a) 0.462 m/s^2

b) 31.5 rad/s

c) 381 rad

d) 135m

Explanation:

the linear acceleration is given by:

$a=\alpha *r\\a=1.30rad/s^2*(35.5*10^{-2}m)\\a=0.462m/s^2$

the angular speed is given by:

$\omega=\frac{v}{r}\\\\\omega=\frac{11.2m/s}{35.5*10^{-2}m}\\\\\omega=31.5rad/s$

to calculate how many radians have the wheel turned we need the apply the following formula:

$\theta=\frac{1}{2}\alpha*t^2\\\\t=\frac{\omega}{\alpha}\\\\t=\frac{31.5rad/s}{1.30rad/s^2}\\\\t=24.2s\\\\\theta=\frac{1}{2}*1.30rad/s^2*(24.2s)^2\\\\\theta=381rad$