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sp2606 [1]
3 years ago
10

FORCE AND DISPLACEMENT AT AN ANGLE A sailor pulls a boat a distance of 30.0 m along a dock using a rope that makes a 25.0° angle

with the horizontal. How much work does the rope do on the boat if its tension is 255 N?​
Physics
1 answer:
Bingel [31]3 years ago
7 0

Answer: 6117.58 J

Explanation:

We know that W=Fd*cos(theta) where theta is the angle between the displacement and the force.

In this case, we are given that F=225 N, d=30 m, and theta=25 degrees.

Plugging all this in we get

W=225*30*cos(25)=6117.58 J

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The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3
viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;

\Rightarrow mg\sin{\theta} = F

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at \theta. Solving for the angle, we get

\sin{\theta} = \dfrac{F}{mg}

or

\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)

\;\;\;=  \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]

\;\;\;=42.9°

6 0
2 years ago
If you want to play a tune on wine glasses, you’ll need to adjust the oscillation frequencies by adding water to the glasses. Th
jonny [76]

Answer:Reducing mass i.e. water

Explanation:

Frequency For given mass in glass is given by

f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}

where k =stiffness of the glass

m=mass of water in glass

from the above expression we can see that if mass is inversely Proportional to frequency

thus reducing mass we can increase frequency  

6 0
4 years ago
A large fake cookie sliding on a horizontal surface isattached to one end of a horizontal spring with spring constantk = 425 N/m
omeli [17]

Answer:

Explanation:

spring constant k = 425 N/m

a ) At the point of equilibrium

restoring force = frictional force

= kx = 10 N

425 x = 10

x = 2.35 cm

b )

Work done by frictional force

= -10 x 2.35 x 10⁻² x 2  J ( Distance is twice of 2.35 cm )

= - 0.47 J

= Kinetic  energy remaining with the cookie  as it slides back through the position where the spring is unstretched .

= 425 - 0.47

= 424.53 J

=

4 0
3 years ago
Https://phet.colorado.edu/sims/html/balloons-and-static-electricity/latest/balloons-and-static-electricity_en.html
Katarina [22]

Answer:

there is not pic

Explanation:

5 0
3 years ago
An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect
Dvinal [7]

Answer:

The distance is d =1.66*10^{-9}m

Explanation:

From the question we are told that

         The initial speed of the  electron is v_i  = 3.2 *10^5 m/s

         The mass of electron is m = 9.11*10^{-31}kg

         Let d be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

         Let KE_i be the initial kinetic energy of the electron \

          Let KE_d be the kinetic energy of the electron at the distance d from the proton

  Considering that energy is conserved,

  The energy at the initial position of the electron = The energy at the final position of the electron

      i.e

             KE_i +U_1 = KE_d + U_2

U_1 \ and \ U_2 are the potential energy at the initial  position of the electron and at distance d of the electron to the proton

                Here U_1 = 0

So the equation becomes

                   \frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2  + \frac{kq_1 q_2}{d}

Here q_1 \ and  \ q_2 are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

 k is electrostatic constant with value 8.99*10^9 N \cdot m^2 /C^2

i.e q = 1.602 *10^{-19}C

           v_d is the velocity at distance d from the proton = 2v_i

  So the equation becomes

             \frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}

            \frac{1}{2} mv_i^2  = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}

           3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}

Making d the subject of the formula

           d = \frac{2k(q)^2}{3mv_i^2}

              = \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}

              =1.66*10^{-9}m

             

           

         

                 

   

6 0
3 years ago
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