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jolli1 [7]
4 years ago
13

The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact wit

h the ground. Determine the weight of the automobile.
Physics
1 answer:
Stels [109]4 years ago
7 0

Answer:

Force, F = 20240 N

Explanation:

It is given that,

Pressure exerted by the four tires of an automobile, P=2.2\times 10^5\ Pa

Area of each tire, A=0.023\ m^2

Area of 4 tires,  A=0.092\ m^2

We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :

P=\dfrac{F}{A}

F=P\times A

F=2.2\times 10^5\times 0.092

F = 20240 N

So, the weight of the automobile is 20240 N. Hence, this is the required solution.

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3 years ago
A mass is placed at the end of a spring. It has starting velocity of V & allowed to oscillate freely. If the mass has a star
LiRa [457]

Answer:

Equation for SHM can be written

V = w A cos w t        where w is the angular frequency and the velocity is a                                         maximum at t = 0

V1 = w1  A cos w1 t

V2 = w2 A cos w2 t

V2 / V1 = w2 / w1     since cos X t = 1 if t = zero

V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2

If the velocity is twice as large the period will be 1/2 long

8 0
3 years ago
A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find
FrozenT [24]

Answer:

The time rate of change of flux is 1.34 \times 10^{10} \frac{V}{s}

Explanation:

Given :

Current I = 0.106 A

Area of plate A = 36 \times 10^{-4} m^{2}

Plate separation d = 4 \times 10^{-3} m

(A)

First find the capacitance of capacitor,

   C = \frac{\epsilon _{o} A }{d}

Where \epsilon _{o} = 8.85 \times 10^{-12}

   C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4}  }{4 \times 10^{-3} }

   C = 7.9 \times 10^{-12} F

But   C = \frac{Q}{V}

Where Q = It

  C = \frac{It}{V}

  V = \frac{It}{C}

Now differentiate above equation wrt. time,

  \frac{dV}{dt} = \frac{I}{C}

       = \frac{0.106}{7.9 \times 10^{-12} }

       = 1.34 \times 10^{10} \frac{V}{s}

Therefore, the time rate of change of flux is 1.34 \times 10^{10} \frac{V}{s}

8 0
3 years ago
Carlos gives a grocery cart a 60-N push. The cart has a mass of 40 kg. What is the cart's acceleration?
andre [41]
<h3>Answer:</h3>

1.5 m/s²

<h3>Explanation:</h3>

We are given;

Force as 60 N

Mass of the Cart as 40 kg

We are required to calculate the acceleration of the cart.

  • From the newton's second law of motion, the rate of change in momentum is directly proportional to the resultant force.
  • That is, F = ma , where m is the mass and a is the acceleration

Rearranging the formula we can calculate acceleration, a

a = F ÷ m

  = 60 N ÷ 40 kg

  = 1.5 m/s²

Therefore, the acceleration of the cart is 1.5 m/s²

3 0
3 years ago
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