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crimeas [40]
2 years ago
6

Observe: Up until now, all the problems you have solved have involved converting only one unit. However, some conversion problem

s require you to convert two or more units. Select Speed from the menu. What two units do you need to convert to solve this problem? meter and seconds Think about it: How do you think you can use conversion factors to solve this problem?
Physics
1 answer:
ipn [44]2 years ago
7 0

Answer:

t is appropriate to clarify that units such as time and angles the transformation is not in base ten, for example:

        60 s = 1 min

        60 min = 1 h

        24 h = 1 day

Therefore, for this transformation, you must be more careful

the length transformation is base 10

Explanation:

In many exercises the units used are transformed by equations into other units called derivatives, in general the transformation of derived units is the product of the transformation of the constituent units.

In the example of velocity, the derivative unit is m / s, which is why it works in the same way that you transform length and time if in the equation it is multiplying it is multiplied and if it is dividing it is divided.

It is appropriate to clarify that units such as time and angles the transformation is not in base ten, for example:

        60 s = 1 min

        60 min = 1 h

        24 h = 1 day

Therefore, for this transformation, you must be more careful

the length transformation is base 10

      1000 m = 1 km

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A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal
Setler [38]

Answer

given,

mass of the ball = 3 kg

swing in vertical circle with radius = 2 m

   work done by the gravity = ?          

   work done by the tension = ?            

Work done by the gravity = - m g Δh            

 Δ h = 2 + 2 = 4 m                                                                

Work done by the gravity =- 3 \times 9.8 \times 4

                                           = -117.6 J                  

work done by gravity is equal to -117.6 J            

Work done by tension will be equal to zero.        

Zero because tension is always perpendicular to velocity

work done by tension is equal to 0 J                          

7 0
2 years ago
Whale sharks swim forward while ascending or descending. They swim along a straight-line path at a shallow angle as they move fr
Aneli [31]
You did not include the quesetion, but I can help you to understand the problem and how to find the relevant information.

1) The angle of 13° with which the shark ascends meets this:

Vertical ascending velocity = 0.85m/s * sin(13°)

Horizontal velocity = 0.85m/s * cos(13°)

2) The length swan by the shark ascending meets this

Vertical ascending length = 50 m

Horizontal length, y:

\frac{y}{50} = \frac{0.85sin(13)}{0.85cos(13)}

From that y = 50 * tan(13°)

=> y = 11.54 m.

3) Conclusions:

1) The shark run 50 m vertically upward and 11.54 m horizontally.

2) The length of the path run by the shark may be calculated using Pythagoras' theorem:

hypotenuse^2 = (50m)^2 + (11.54m)^2 = 2633.25m^2

hypotenuse = 51.35m

So, the shark swan 51.35 m to reach the surface.

4 0
2 years ago
An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

4 0
2 years ago
A 10-meter long ramp has a mechanical advantage of 5. What is the height of the ramp?
denpristay [2]
<span><span>1.       </span>If the ramp has a length of 10 and has a mechanical advantage (MA) of 5. Then we need to find the height of the ramp.
Formula:
MA = L / H
Since we already have the mechanical advantage and length, this time we need to find the height .
MA 5 = 10 / h
h = 10 / 5
h = 2 meters

Therefore, the ramp has a length of 10 meters, a height of 2 meters with a mechanical advantage of 5.</span>



6 0
2 years ago
The speed of an object undergoing constant acceleration increased from 8.0 meters per second to 16.0 meters per second in 10. Se
saw5 [17]

v₀ = initial speed of the object = 8 meter/second

v = final speed of the object = 16 meter/second

t = time taken to increase the speed = 10 seconds

d = distance traveled by the object in the given time duration = ?

using the kinematics equation

d = (v + v₀) t/2

inserting the above values in the above equation

d = (16 + 8) (10)/2

d = 120 meter


6 0
2 years ago
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