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Vlad [161]
2 years ago
5

If the same force is applied to an object with a small mass it will have a

Physics
1 answer:
Rashid [163]2 years ago
8 0

Answer:

acceleration

Explanation:

i hope this helps

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If the speed of a wave doubles while the frequency remains the same, it means
sattari [20]

Answer:

Explanation:

Velocity of a wave is describe as

velocity =Frequency × Wavelength

Mathematically

v = fλ

Hence, Frequency, F = v / λ

Wavelength λ = v/f

So, if the frequency is kept constant, wavelength of the wave becomes directly proportional to velocity of the wave.

And this implies that, as the speed double, the wavelength is double.

5 0
3 years ago
You add 800 ml of water at 20c to 800 ml of water at 80c what is the most likely final temperature of the mixture ?
bekas [8.4K]

Answer:

d. 50 C

Explanation:

In this problem, we have to add 800 ml of water at 20 Celsius to 800 ml of water at 80 Celsius.

According to the 2nd law of thermodynamics, heat transfers from hot to cold temperature.

The quantity of both the different waters is equal so this makes it very easy. All we have to do is find the mean of both the temperatures:

Final temperature = (20 C + 80 C)/2

= 50 Celsius

3 0
3 years ago
Read 2 more answers
40ml of Liquid A are poured into a beaker, and 40.0ml of Liquid B are poured into an identical beaker. Stirrers in each beaker a
lakkis [162]

Answer:

d

Explanation:

there is not enough information about the liquid to know the force required for each. ie. stirring a cup of water is different than stirring a cup of pudding.

7 0
2 years ago
At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40° above horizontal. She
Citrus2011 [14]

Answer:

a

The height is   H  = 6.74 \  m

b

The horizontal distance is  D  = 23.74 \  m

Explanation:

From the question we are told that

  The speed is  v  =  15 \  m/s

  The angle is  \theta  =  40^o

   The height of the cannon from the ground is  h  =  2 m

  The distance of the net from the ground is k  =  1 m

 

Generally the maximum height she reaches is mathematically represented as  

     H  =  \frac{v^2 sin^2 \theta }{2 *  g }  +  h

=>    H  =  \frac{(15)^2 [sin (40)]^2 }{2 * 9.8}  +  2

=>    H  = 6.74 \  m

Generally from kinematic equation  

    s = ut + \frac{1}{2} at^2

Here s is the displacement which is mathematically represented as

         s  =  [-(h-k)]  

    =>  s =  -(2-1)

    =>  s  = -1 m

There reason why s =  -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half

     a =  -g = -9.8

    u  =  v sin (\theta)

So

    -1 = (vsin 40 )t + \frac{1}{2} * (-9.8) t^2

=>  -4.9t^2 + 9.6418t + 1 = 0

using  quadratic formula to solve the equation we have

    t  =  2.07 \  s

Generally distance covered along the horizontal is  

   D  =  v cos (40) *  2.07

=>   D  =  15 cos (40) *  2.07

=>   D  = 23.74 \  m

7 0
3 years ago
Please answer the number 5 6 and 7. I don't know what to do its our hw in physics. (new lesson as well)
Aleks04 [339]

From the picture, I see that you had no trouble at all with #4.
Well, #5, 6, and 7 are easily handled in exactly the same way.

Just as you did with #4, please sketch these on paper
as I walk you through the solutions.  That'll help you see
immediately what's going on.

#5.b).
Traveling east at 3 m/s for 4 seconds,
he covers  (3 m/s) x (4 sec) = 12 meters.

Traveling south at 5 m/s for 2 seconds,
he covers (5 m/s) x (2 sec) = 10 meters.

The total distance he covers is  (12m + 10m) = 22 meters.

#5.c).
Average speed (scalar)

                           = (distance covered)/(time to cover the distance)

                           = (22 meters)/(6 sec) = 3-2/3 m/s .

#5.d).
Displacement (vector)

                       = distance between the start-point and the end-point,
                          regardless of the route traveled,
                      
  in the direction from the start-point to the end-point.

Distance from the start-point to the end-point =

               √(12² + 10²) = √(144 + 100) = √(244) = 15.62 meters

in the direction of  arctan(10/12) south of east

                             =  39.8° south of east.
 
#5.e).
Average velocity (vector) =

             (displacement vector) / (time)

         =  15.62 meters directed 39.8° south of east / 6 seconds

         =  2.603 m/s directed 39.8° south of east.

 #6).
Magnitude = √(5.2² + 2.1²) = √(27.04 + 4.41) = √31.45 = 5.608 km.

Direction = arctan(5.2/2.1) south of east

                =   68° south of east  =  158° bearing . 

#7).
Magnitude = √(39² + 57²) = √(1521 + 3249) = √( 4770)

                                                                         =  69.07 m/s .

Direction = arctan (57/39) south of west

               =   55.6° south of west

                    Bearing = 214.4°

                    Compass: 0.65° past "southwest by south".  


I'm grateful for the privilege and opportunity to practice my math,
and I shall cherish the bounty of 5 points that came with it.

8 0
3 years ago
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