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3 years ago

How do i do this??????????

Physics

1 answer:

Sliva [168]3 years ago

4 0

Power = I^2 x R
Energy = Power x Time

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A stuntman drives a car with a mass of 1500 kg on a drawbridge. The car accelerates with a constant force of 10,000 N. While he

Viktor [21]

Answer:

1.77 m/s^2

Explanation:

There are two forces acting on the car along the direction parallel to the incline:

- The driving force of 10,000 N, which pushes forward

- The component of the weigth of the car parallel to the incline, which pulls backward

The component of the weight of the car parallel to the incline is:

$W_p = mg sin \theta=(1500 kg)(9.8 m/s^2)( sin 30^{\circ})=7350 N$

So now we can apply Newton's second law to find the acceleration of the car:

$F-W_p = ma\\a=\frac{F-W_p}{m}=\frac{10000 N-7350 N}{1500 kg}=1.77 m/s^2$

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3 years ago

2.65 In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75 kN. Knowing that n 5 0.30 an

Schach [20]

To solve this problem it is necessary to apply the concepts related to uniaxial deflection for which the training variable is applied, determined as

$\delta = \frac{Pl}{AE}$

Where,

P = Tensile Force

L= Length

A = Cross sectional Area

E = Young's modulus

PART A) The elongation of the bar in a length of 200 mm caliber, could be determined through the previous equation, then

$\delta = \frac{Pl}{AE}$

$\delta = \frac{(75*10^3)(200)}{(\pi/4*22^2)(200*10^3)}$

$\delta = 0.1973mm$

Therefore the elongaton of the rod in a 200mm gage length is $0.1973mm$

PART B) To know the change in the diameter, we apply the similar ratio of the change in length for which,

$\upsilon = \frac{l_{a}}{l_{s}}$

Where,

$\upsilon =$ Poission's ratio

$l_{a}$ = Lateral strain

$l_{s}$ = Linear strain

$\upsilon = \frac{\delta/d}{\delta/l}$

$0.3 = \frac{\delta/22}{0.1973/200}$

$0.3(\frac{0.1973}{200}) = \frac{\delta}{22}$

$\delta = 6.5109*10^{-3}mm$

Therefore the change in diameter of the rod is $6.5109*10^{-3}mm$

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3 years ago

48.36<br> g.<br> MgSO4 to motes

Julli [10]

Answer:120.3676

Explanation: using the molecular calculator and molar mass of MgSO4. hope this helps!

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3 years ago

What is the momentum of an object with 9.74 kg and 15 m/s?

dolphi86 [110]

Answer: 146.1 kg/m/s

Explanation: Momentum is equal to mass times velocity, so you do 9.74 kg times 15 m/s and get 146.1 kg/m/s.

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3 years ago

A block on the end of a spring is pulled to position x = a and released from rest. in one full cycle of its motion through what

xxMikexx [17]

A) 0.5 A......................

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3 years ago