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Liula [17]
3 years ago
12

a truck is moving around a circular curve @ a uniform velocity of 13m/s. if the centifrical force on truck is 3,300N with truck

mass of 1600kg. What is curve radius?
Physics
1 answer:
VMariaS [17]3 years ago
8 0

<u>Answer</u>

81.94 m


<u>Explanation</u>

The centripetal force of an object moving in a circular path is given by:

F = mv²/r  Where m is the mass of the object, v is the constant velocity and r is the radius of the curve.

F = mv²/r

3,300 = (1600×13²)/r

3,300 = 270,400/r

r = 270,400/3,300

  = 81.94 m

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If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re
jenyasd209 [6]

Answer:

The amount of work done required to stretch spring by additional 4 cm is 64 J.

Explanation:

The energy used for stretching spring is given by the relation :

E = \frac{1}{2}kx^{2}           .......(1)

Here k is spring constant and x is the displacement of spring from its equilibrium position.

For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).

8 = \frac{1}{2}\timesk\times k \times(0.02)^{2}

k = 4 x 10⁴ N/m

Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).

Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).

E = \frac{1}{2}\times4\times10^{4}\times (0.06)^{2}

E = 72 J

But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :

E = ( 72 - 8 ) J = 64 J

7 0
3 years ago
The humber bridge in england has the world's longest single span, 1410 m . calculate the change in length of the steel deck of t
Pepsi [2]
Applicable linear expansion equation:
ΔL = αΔTL
In which
ΔL = change in length, α = Linear expansion coefficient of steel, ΔT = change in temperature, L = original length

Therefore,
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3 0
2 years ago
A 120-V rms voltage at 1000 Hz is applied to an inductor, a 2.00-μF capacitor and a 100-Ω resistor, all in series. If the rms va
natima [27]

Answer:

The inductance of the inductor is 35.8 mH

Explanation:

Given that,

Voltage = 120-V

Frequency = 1000 Hz

Capacitor C= 2.00\mu F

Current = 0.680 A

We need to calculate the inductance of the inductor

Using formula of current

I = \dfrac{V}{Z}

Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}

Put the value of Z into the formula

I=\dfrac{V}{\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}}

Put the value into the formula

0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}

L=35.8\ mH

Hence, The inductance of the inductor is 35.8 mH

4 0
3 years ago
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