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lapo4ka [179]
3 years ago
6

A graph of velocity versus time (velocity displayed in m/s and time in seconds) displays a

Physics
1 answer:
xxMikexx [17]3 years ago
5 0

Answer: acceleration is 2m/s^2

initial velocity is 4m/s

Explanation:

Since slope is rise/ run in this graph it will be change in velocity/ change and time. Therefore the slope of the graph will tell you the acceleration is 2m/s^2. The y intercepts is the velocity then time is 0, therefore the initial velocity is 4 m/s

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A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
3 years ago
Read 2 more answers
Please help me to solve this and give me the summary of answer
Neko [114]
Using the principle of floatation.

u = w............(a)

Upthrust of fluid is equal to the weight of the object.

Let the volume of the wood be V.

The upthrust u, is related to the volume submerged in water, and that is 1/5 of it volume, that is (1/5)V = 0.2V

Formula for upthrust, u = vdg

where v = volume of fluid displaced
d = density of fluid
g = acceleration due to gravity

weight, w = mg
where m = mass
g = acceleration due to gravity

From (a)

                     u = w

                 vdg =  mg      Cancel out g

                   vd  =  m
 
The v  is equal to 0.2V, which is the submerged volume. Notice that the small letter v is volume of fluid displaced, and capital V is the volume of the solid.

d is density of fluid which is water in this case, 1000 kg/m³

         0.2V * 1000 =  m

           200V =  m

Hence the mass of the object is  200V  kg.

But Density of solid =  Mass of solid / Volume of solid

                                 =    200V / V

                                 =    200 kg/m³

Density of solid = 200 kg/m³     
7 0
3 years ago
It takes 9 sec for a 10 newton force to move an object 4 meters to the right. What is direction & magnitude of the force
lana66690 [7]

Answer:

the question is wrong

Explanation:

  1. M is not given
  2. after 9 second the acceleration multiply by the time divided by two then multiplied by the time is equal to 4 meter
  3. ((10/Mm/s *9s-)
3 0
3 years ago
A 6 m/s vector pointing North is added to a 2 m/s vector pointing East. What are the magnitude and direction of the resultant?
irina [24]

Answer:

A + B = C      Ax = 2   Ay = 0    Bx = 0  By = 6

Ax + Bx = Cx = 2

Ay + By = Cy = 6

C = (2^2 + 6^2)^1/2 = 6.32

Tan Cy / Cx = 6 / 2 = 3

Cy at 71.6 deg

6 0
3 years ago
Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19
Alisiya [41]

Answer:

The workdone is  W = 9.28 * 10^{3} J

Explanation:

From the question we are told that

   The height of the cylinder is  h = 0.588\ m

   The face Area is  A = 4.19 \ m^2

    The density of the cylinder is \rho  =  0.346 * \rho_w

     Where \rho_w is the density of freshwater which has a constant value

              \rho_w = 1000 kg/m^3

     

Now  

     Let the final height of the device under the water be  =  h_f

      Let  the initial volume underwater be = V_n

     Let the initial height under water be  = h_i

      Let the final volume under water be  = V_f

According to the rule of floatation

        The weight of the cylinder =  Upward thrust

This is mathematically represented as

          \rho_c g V_n = \rho_w gV_f

         \rho_c A h = \rho A h_f

So      \frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}

   =>     \frac{h_f}{h_c}  = 0.346

Now the work done is mathematically represented as  

          W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh

               =   \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.

              = \frac{g A \rho}{2}  [h^2 - h_f^2]

              = \frac{g A \rho}{2} (h^2)  [1  - \frac{h_f^2}{h^2} ]

Substituting values

        W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)

        W = 9.28 * 10^{3} J

4 0
2 years ago
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