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3 years ago

How would Newton's Three laws apply to a Ferris Wheel?

1 answer:

5 0

The Central pedal force of the wheels center is opposed by an equal force from the objects being spun around by the wheel

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A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The

Alex787 [66]

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

K= 1300 N/m

x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

$\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2$

$Kx^2=mv^2$

$v=\sqrt{\dfrac{kx^2}{m}}$

By putting the values

$v=\sqrt{\dfrac{kx^2}{m}}$

$v=\sqrt{\dfrac{1300\times 0.264^2}{25}}$

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

$V_f^2=V_i^2-2gh$

By putting the values

$0^2=1.9^2-2\times 10\times h$

h=0.1805 m

h=18.05 cm

4 0

3 years ago

The force of attraction between a -130.0 C and +180.0 C charge is 8.00 N. What is the separation between these two charges in me

kondaur [170]

Answer with Explanation:

The force of attraction between 2 charges of magnitude $q_1,q_2$ separated by a distance 'r' is given by

$F=\frac{1}{4\pi \epsilon _o}\frac{q_1\times q_2}{r^2}=k\frac{q_1\times q_2}{r^2}$

where

$\epsilon _o$ is a constant known as permitivity of free space

$k=9\times 10^{9}Nm^2/C^2$

Applying the given values in the above relation we get

$8=9\times 10^{9}\times \frac{130\times 180}{r^{2}}\\\\r^{2}=\frac{9\times 10^{9}\times 130\times 180}{8}\\\\r^{2}=26325\times 10^{9}\\\\\therefore r=\sqrt{26325\times 10^{9}}=5.131\times 10^{6}meters$

5 0

3 years ago

Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deflecting plates

Alina [70]

Answer:

The capacitance of the deflecting plates is $C=1.59 pF$ .

Explanation:

The expression for the capacitance of the capacitor in terms of area and distance is as follows;

$C=\frac{\varepsilon _{0}A}{d}$

Here, C is the capacitance, A is the area, d is the distance and $\varepsilon _{0}$ is the absolute permittivity.

Convert the side of the square from cm to m.

s= 3.0 cm

s= 0.030 m

Calculate the area of the square.

$A= s_^{2}$

Put s= 0.030 m.

$A=(0.030)_^{2}$

$A=9\times10^{-4}m^{-2}$

Convert distance from mm to m.

d= 5.0 mm

$d=5\times10^{-3}m$

Calculate the capacitance of the deflecting plates.

$C=\frac{\varepsilon _{0}A}{d}$

Put $d=5\times10^{-3}m$ , $A=9\times10^{-4}m^{-2}$ and $\varepsilon _{0}=8.85\times 10^{-12}Fm^{-1}$ .

$C=\frac{(8.85\times 10^{-12})(9\times10^{-4})}{5\times10^{-3}}$

$C=1.59\times 10^{-12}F$

$C=1.59 pF$

Therefore, the capacitance of the deflecting plates is $C=1.59 pF$ .

4 0

3 years ago

Which of the following statements best describes the object's motion between the time period of 23s to 30s?

inessss [21]

Constant acceleration

8 0

3 years ago

What is centripetal force?

daser333 [38]

The centripetal force is force acting on a body in circular motion. In circular motion, velocity is always on tangent and if we took 2 different positions on a circle, the change on velocity is a vector pointing in the middle of circle. In circular motion velocity is constant, and acceleration lies on radius of circle pointing to te middle. This acceleration is called centripetal acceleration, and the force is centripetal.

8 0

3 years ago