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Amanda
15 days ago
9

An electric field of 4.0 μV/m is induced at a point 2.0 cm from the axis of a long solenoid (radius = 3.0 cm, 800 turns/m). At w

hat rate is the current in the solenoid changing at this instant?
Physics
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a hippopotamus produces a pressure of 250000 pa when it is standing on all four feet if the weight of the hippo is 40000 N what
Maru [3345]

0.04m²

Explanation:

Known values:

Pressure = 250000Pa

Weight = 40000N

Unknown:

Area of each foot =?

Solution:

Pressure is defined as the force applied per unit area of an object

  Pressure = \frac{force}{area}

To determine the area;

        Area = \frac{force }{pressure}

    Area = \frac{40000}{250000} = 0.16m²

The force exerted by all four feet amounts to 0.16m²

thus, the area for each foot is \frac{0.16}{4} = 0.04m²

Learn more:

Pressure

8 0
2 months ago
Two of the types of ultraviolet light, uva and uvb, are both components of sunlight. their wavelengths range from 320 to 400 nm
Sav [3153]

In terms of light energy, a higher frequency corresponds to increased energy within the light.

We establish that frequency is essentially the inverse of wavelength:

frequency = 1 / wavelength

Calculating frequencies:

f UVA = 1/320 to 1/400

f UVA = 0.0031 to 0.0025

 

f UVB = 1/290 to 1/320

f UVB = 0.0034 to 0.0031

Since UVB occupies a higher frequency range, it consequently possesses greater energy than UVA.

7 0
2 months ago
Read 2 more answers
13–82. the 8-kg sack slides down the smooth ramp. if it has a speed of 1.5 m> s when y = 0.2 m, determine the normal reaction
Softa [3030]
The second question necessitates a figure to provide an answer. For the initial question
The acceleration of the sack is
1.5² - 0² = 2a(0.2)
a = 5.63 m/s²
The ramp's reaction force is
F = 8 kg (5.63 m/s²)
F = 45 N
Differentiate the kinematic equation with respect to time to find the velocity's rate of increase.
4 0
2 months ago
A particle leaves the origin with an initial velocity v⃗ =(2.40 m/s)xˆv→=(2.40 m/s)x^ , and moves with constant acceleration a⃗
Maru [3345]

Answer:

The distance before stopping is 1.52 m,

velocity is 4.0 m/s on the y-axis

Explanation:

The particle’s motion is two-dimensional due to acceleration along both the x and y axes; each axis can be addressed independently for calculations.

a) At the moment the particle starts to reverse, its velocity should be zero (Vfx = 0)

     Vfₓ = V₀ₓ + aₓ t  

     t = -  V₀ₓ/aₓ

     t = - 2.4/(-1.9)

     t=  1.26 s

When the particle stops, we calculate its position

     X1 = V₀ₓ t + ½ aₓ t²

     X1= 2.4 1.26 + ½ (-1.9) 1.26²

     X1= 1.52 m

At this point, the particle begins returning.

b) The velocity comprises both x and y components.

   For the x section, Vₓ = 0 m/s indicates a halt, but the y component retains a velocity

    Vfy= Voy + ay t

    Vfy= 0 + 3.2 1.26

    Vfy = 4.0 m/s

Thus, the velocity reads as

    V = (0 x^ + 4.0 y^) m/s

c) To graph the motion, we create a table listing position x and y at given time intervals; let's begin the calculations for equations

      X = V₀ₓ t+ ½  aₓ t²

      Y = Voy t + ½  ay t²

      X= 2.4 t + ½ (-1.9) t²

      Y= 0 + ½ 3.2 t²

      X= 2.4 t – 0.95 t²

      Y=   1.6 t²

With these equations, we construct two graphs for position against time, one for the x-axis and another for the y-axis

                       Chart for graphing

              Time (s)     x(m)            y(m)

                 0                0               0

                 0.5             0.960       0.4

          1       1.45          1.6

                 1.50      1.46      3.6

                2.00      1.00      6.4

7 0
2 months ago
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