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Margarita [4]
2 years ago
11

What force does it take to accelerate a 7.2 kg object 3.0 m/s^2.

Physics
1 answer:
grin007 [14]2 years ago
3 0

Answer:

<h2>21.6 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 7.2 × 3 = 21.6

We have the final answer as

<h3>21.6 N</h3>

Hope this helps you

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I think the answer would be D.
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The same pipe is used to carry both air and water. For the same fluid velocity and friction factor for the air and water flows:
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Answer:

the pressure drop for the water flow is greater than that for the air flow.

Explanation:

Detailed analysis of the problem is show below.

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3 years ago
A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of 384,000 km. Sup
Tema [17]

Answer:

a) v = 19,149.6 m/s

b) f = 95%

c) t = 346.5min

Explanation:

First put all values in metric units:

15.8 min*\frac{60s}{1min}=948s

The equation of motion you need is:

v_f = a*t+v_0

where v_f is the final velocity, a is acceleration and t is time in hours.

Since the spaceship starts from 0 velocity:

v_f = a*t = 20.2*948 = 19,149.6 m/s

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

x=\frac{v_0+v_f}{2}t

solving for first and last interval:

Since the spaceship starts and finish with 0 velocity:

x=\frac{v}{2}t=\frac{19,149.6}{2}948=9,076,910.4m=9,076.9104km

Then the ship traveled 384,000-9,076.9104*2 = 361,846.1792km at constant speed, which means that it traveled:

f_{constant_speed} =\frac{ x_{constant_speed}}{x_total} =\frac{361,846.1792}{380,000} =0.95

Which in percentage is 95% of the trip.

to calculate total time you need to calculate the time used during constant speed:

t = \frac{361,846,179.2}{19,149.6} = 18,895.75s = 314min

That added to the other interval times:

t_{total} = t_1+t_2+t_3=15.8+314.93+15.8=346.5min

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2 years ago
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8 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
3 years ago
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