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bagirrra123 [75]
3 years ago
14

Find the magnitude & direction: 50N 25 N 35 N 10N

Physics
1 answer:
adoni [48]3 years ago
8 0

F1x + F2x = Rx

↓

Rx = F1x + F2x

↓

Rx = F1 cos45° + F2

↓

Rx = (50N)(cos45°) + 60N

↓

Rx = 95N

Similarly, if we sum all the y components, we will get the y component of the resultant force:

F1y + F2y = Ry

↓

Ry = F1y + F2y

↓

Ry = F1 sin45° + 0

↓

Ry = F1 sin45°

↓

Ry = (50N)(sin45°)

↓

Ry = 35N

At this point, we know the x and y components of R, which we can use to find the magnitude and direction of R:

Rx = 95N

Ry = 35N

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The metal wire in an incandescent lightbulb glows when the lights is switch on and stops glowing when switched off. This simple
lutik1710 [3]

Answer:

When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat.  

Explanation:

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tekilochka [14]

Answer:

changing the direction in which a force is exerted

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- What are (a) the x component, (b) the y component, and (c) the z component of r = a - b +c if a= 7.8 + 6.6 - 7.1 , b= -2.9 + 7
frez [133]

Answer:

Explanation:

a= 7.8i + 6.6j - 7.1k

b=   -2.9 i+ 7.4 j+ 3.9k , and

c = 7.6i + 8.8j + 2.2k

r = a - b +c

=7.8i + 6.6j - 7.1k - ( -2.9i + 7.4j+ 3.9k )+ ( 7.6i + 8.8j + 2.2k)

= 7.8i + 6.6j - 7.1k +2.9i - 7.4j- 3.9k )+  7.6i + 8.8j + 2.2k

= 18.3 i +18.3 j - k

the angle between r and the positive z axis.

cosθ = 18.3 / √18.3² +18.3² +1

the angle between r and the positive z axis.

= 18.3 / 25.75

cos θ= .71

45 degree

6 0
3 years ago
A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym
user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

                                     L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

                                    L_(p,i) = m*V_pi*R

                                    L_(p,i) = 3.6*3.3*0.44

                                    L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

                                    L_(c,i) = I*w_i

                                    L_(c,i) = 0.5*M*R^2*0

                                    L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

                                    L_i = L_(p,i) + L_(c,i)

                                    L_i = 5.2272 + 0

                                    L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

                                   L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

                                   L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

                                  L_f = L_(p,f) + L_(c,f)

                                  L_f = I_p*w_f + I_c*w_f

                                  L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

                                  L_i = L_f

                                  5.2272 = w_f*(I_p + I_c)

                                  w_f =  5.2272/ R^2*(m + 0.5M)

Plug in values:

                                  w_f =  5.2272/ 0.44^2*(3.6 + 0.5*45)

                                  w_f =  5.2272/ 5.05296

                                  w_f = 1.0345 rad/s

5 0
3 years ago
A cellist tunes the C string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.5
Dahasolnce [82]

Answer:

a

\lambda  = 1.18 \  m

b

v  =  77.172 \  m/s

c

T  = 151.41 \  N

Explanation:

From the question we are told that

   The frequency is  f =  65.4 \  Hz

   The  length of the vibrating string is  L  =  0.590 \  m

   The  mass is  m  =  15.0 \ g  =  0.015 \  kg

Generally the wavelength is mathematically represented as

           \lambda =  2 *  L

=>        \lambda  =  2 *   0.590

=>         \lambda  = 1.18 \  m

Generally the wave speed is  

          v  =  \lambda  *  f

=>       v  =  1.18 * 65.4

=>       v  =  77.172 \  m/s

Generally the tension on the wire is mathematically represented as

        T  =  v^2  *  \frac{ m }{L }

=>      T  =  77.172 ^2  *  \frac{  0.015  }{0.590}

=>      T  = 151.41 \  N

7 0
3 years ago
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