Hello, I want to ask. . anyone knows the answer.

A slanted vector has a magnitude of 41 N and is at an angle of 23 degrees north of east. What are the magnitude and direction of

jenyasd209 [6]

Answer:

The horizontal component is 37.74 N to east

The vertical component is 16.02 N to North

Explanation:

Any slant vector has two components

→ Horizontal component = R cos Ф

→ Vertical component = R sin Ф

→ R is the magnitude of the vector

→ Ф is the direction of the vector with positive part of the horizontal axis

A slanted vector has a magnitude of 41 N and is at an angle of 23

degrees north of east

23° north of east means the angle between the vector and the

east direction is 23° (east is the positive horizontal direction)

That means R is 41 N and Ф is 23°

→ R = 41 N , Ф = 23°

→ The horizontal component = 41 × cos(23) = 37.74 N east

→ The vertical component = 41 × sin(23) = 16.02 N North

<em>The horizontal component is 37.74 N to east</em>

<em>The vertical component is 16.02 N to North</em>

7 0

3 years ago

a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet

ANTONII [103]

Answer:

Explanation:

Given

radius of circular path $r=4\ in.$

Position is given by

$\theta =\cos 2t---1$

Differentiate 1 to angular velocity we get

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2$

Differentiate 2 to get angular acceleration

$\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3$

Net acceleration is the vector summation of tangential and centripetal force

$a_t=\alpha \times r$

$a_t=-4\cos 2t\times 4=-16\cos 2t$

$a_r=\omega ^2\cdot r$

$a_r=(-2\sin 2t)^2\cdot 4$

$a_r=16\sin^2(2t)$

$a_{net}=\sqrt{a_r^2+a_t^2}$

$a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}$

$a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}$

6 0

2 years ago

If you need 40.0 Nm of torque in order to loosen a nut on a wn

KonstantinChe [14]

Answer:

0.301 m

Explanation:

Torque = Force × Radius

τ = Fr

40.0 Nm = 133 N × r

r = 0.301 m

The mechanic must apply the force 0.301 m from the nut.

6 0

2 years ago

How does wood and plastic relate to forces

WARRIOR [948]

Answer:

There are six kinds of forces that act on objects when they come into contact with one another: Normal force, applied force, frictional force, tension force, spring force and resisting force. These forces make objects change their motion or movement , the act of going from one place to another.

6 0

3 years ago

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$

$h = \frac{3}{4g}(v_1^2 -v_f^2)$

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

$h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m$

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0

2 years ago