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Natasha_Volkova [10]
3 years ago
8

Bodies weighing 1 kilogram and 5 kilograms lie on a smooth horizontal surface. If a traction force of 0.6 N acts on another 5 kg

body, determine the acceleration of these bodies.​
Physics
1 answer:
natima [27]3 years ago
6 0

0.6/5,1,5

so calculate it

not so sure though

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A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30° angle. The block’s initial speed is 10 m/s. What vert
Romashka [77]

Answer:

Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s

Explanation:

Given Data:

V is the overall velocity vector, Vi and Ui are its initial vertical and horizontal components

R = 10 m/s\\ Projection Angle (theta) = 30 degrees\\Vi   = 10*sin(30) = 5 m/s\\Ui  = 10*cos(30) = 8.66 m/s

To find:

Max Height h achieved

Calculation:

1) Using the 3^{rd} equation of motion, we know

2*a*s = Vf^{2}  - Vi^{2}

2) In terms of gravity g height h and  the vertical component of Velocity Vf , Vi.

3) As Vf = 0 as at maximum height the vertical component of velocity is zero maximum height achieved

2*g*h = Vf^{2}  -Vi^{2}

putting values

4) h = 0.784 m/s

5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)

3 0
4 years ago
A rectangular block measures 8cm by 5cm by 4cm, a mass of 1.25kg what is the volume of the block
Burka [1]
To calculate the volume of a rectangular, we have to multiply the value of height, length and width.

8x5x4 = 160 cm ^3
3 0
2 years ago
Two manned satellites approaching one another at a relative speed of 0.550 m/s intend to dock. The first has a mass of 2.50 ✕ 10
NNADVOKAT [17]

Answer: Their final relative velocity is -0.412 m/s.

Explanation:

According to the law of conservation,

      m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v

Putting the given values into the above formula as follows.

      m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v

     2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v

           -4.12 \times 10^{3} kg m/s = (10^{4} kg) v

                   v = \frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}

                      = -0.412 m/s

Thus, we can conclude that their final relative velocity is -0.412 m/s.

8 0
4 years ago
At an accident scene on a level road, investigators measure a car's skid mark to be 93 m long. It was a rainy day and the coeffi
USPshnik [31]

Answer:

u = 25 m/s

Explanation:

given,                                

length of skid = 93 m          

coefficient of friction = 0.35

final velocity = 0 m/s              

initial velocity = ?                        

force here is friction  f = μ mg

F = ma                                                

now com paring                      

-μ mg = m a                      

a = - μ g                    

a = - 0.35 x 9.8              

a = -3.43 m/s²

we know,              

v² = u² + 2 a s                        

0 = u² - 2 x 3.43 x 93                

u² = 637.98                    

u = 25.26 m/s                      

u = 25 m/s (two significant figure)

6 0
3 years ago
Why are some animals and plants can become pests when introduced into Australia
Yanka [14]
<span>Plants there are a lot. Australian plants are extremely hardy an have adapted to grow in conditions others have not. So when taken outside of Australia they do very, very well. Animals, less so. But there are a few very interesting cases.

</span>
8 0
4 years ago
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