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Alona [7]
3 years ago
15

3/50 The standard test to determine the maximum lateral acceleration of a car is to drive it around a 200-ft-diameter circle pai

nted on a level asphalt surface. The driver slowly increases the vehicle speed until he is no longer able to keep both wheel pairs straddling the line. If this maximum speed is 35 mi/hr for a 3000-lb car, determine its lateral acceleration capability an in g's and compute the magnitude F of the total friction force exerted by the pavement on the car tires.
Physics
1 answer:
Delicious77 [7]3 years ago
5 0

Answer:

(a): Its lateral acceleration is ac= 8.02 m/s² = 0.81 * g

(b): The magnitude F of the friction force exerted by the paviment is F= 1091.28 N.

Explanation:

r= 100 ft = 30.48 m

Vt= 35 miles/hr = 15.64 m/s

m= 3000 lb = 136.07 kg

w= Vt/r

w= 0.5131 rad/s

ac= w² * r

ac= 8.02 m/s² = 0.818 * g  (a)

F= m * ac

F= 1091.28 N (b)

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Please refer to the figure.

Explanation:

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Molecular formula of water molecule is H₂O.
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Find the equivalent resistance, current, and voltage across each resistor when the specified resistors are connected across a 20
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Answer:

Explanation:

The question is incomplete. Here is the complete question.

"Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified resistors are connected across a 20-V battery. Part (a) uses two resistors with resistance values that can be set with the animation sliders, and you can use the animation to verify your calculation. In part (b), three resistors are specified. (a) Two resistors are connected in series across a 20-V battery. Let R1 = 1 Ω and R2 = 2 Ω. Rea = (b) Add a third resistor to the circuit in series. Let R1 = 1 Ω, R2 = 2 Ω, and R3 = 3 Ω"

Using ohms law formula to solve the problem

E = IRt

E is the supply voltage

I is the total current

Rt is the total equivalent resistant.

a) Given two resistances

R1 = 1ohms and R2 = 2ohms

If the resistors are Connected in series across a 20V supply voltage,

-Equivalent resistance = R1+R2

= 1ohms + 2ohms

= 3ohms

- In a series connected circuit, same current flows through the resistors.

Using the formula E = IRt

I = E/Rt

I = 20/3

I = 6.67A

The current in both resistors is 6.67A

- Different voltage flows across a series connected circuit.

Using the formula V = IR

V is the voltage across each resistor

I is the current in each resistor

For 1ohms resistor,

V = 6.67×1

V = 6.67Volts

For 2ohms resistor

V = 6.67×2

V = 13.34Volts

b) If the resistors are three

R1 = 1ohms, R2 = 2ohms R3 = 3ohms

- Total equivalent resistance = 1+2+3

= 6ohms

- Current in each resistor I = E/Rt

I = 20/6

I = 3.33A

Since the same current flows through the resistors, the current across each of them is 3.33A

- Voltage across them is calculated as shown:

V = IR

For 1ohm resistor

V = 3.33×1

V = 3.33volts

For 2ohms resistor

V = 3.33×2

V = 6.66volts

For 3ohms resistor

V = 3.33×3

V = 9.99volts

3 0
3 years ago
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On a roller coaster ride the total mass of a cart - with two passengers included - is 319 kg. Peak K is at 43.6 m above the grou
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The mechanical energy is lost due to friction between the two peak is 78,458.688 J

<h3>What is mechanical energy?</h3>

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

Given is a roller coaster ride the total mass of a cart - with two passengers included - is 319 kg. Peak K is at 43.6 m above the ground and peak L is at 24.4 m. At location K the speed of the cart is 16.4 m/s, and at location L it is 12.4 m/s.

Total energy at peak K,

TE₁ = 1/2 mv₁² +mgh₁

Substitute the values, we get

TE₁ =  1/2 x319 x 16.4² +319 x 9.81 x 43.6

TE₁ = 179,340. 524 J

Total energy at peak L,

TE₂ = 1/2 mv₂² +mgh₂

Substitute the values, we get

TE₂ =  1/2 x319 x 12.4² +319 x 9.81 x 24.4

TE₂ = 100,881. 836 J

The mechanical energy lost is

M.E =  TE₁ -TE₂

M.E =  179,340. 524 J -  100,881. 836 J

M.E = 78,458.688 J

Thus, the mechanical energy is lost due to friction between the two peak i

Learn more about mechanical energy.

brainly.com/question/13552918

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7 0
2 years ago
Que fuerza será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s²
WARRIOR [948]

Answer:

La fuerza que será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s² es 80 N.

Explanation:

La segunda ley de Newton, llamada ley fundamental o principio fundamental de la dinámica, plantea que un cuerpo se acelera si se le aplica una fuerza.

De esta manera, esta ley establece que las aceleraciones que experimenta un cuerpo son proporcionales a las fuerzas que recibe. Dicho de otra forma, la aceleración de un cuerpo es proporcional a la fuerza neta que se le aplica. Cuanto mayor es la fuerza que se le aplica a  un objeto con una masa dada, mayor será su aceleración.

La segunda Ley de Newton se expresa matemáticamente como:

F = m*a

Donde:

  • F es la fuerza neta. Se expresa en Newton (N)
  • m es la masa del cuerpo. Se expresa en kilogramos (Kg.).
  • a es la aceleración que adquiere el cuerpo. Se expresa en metros sobre segundo al cuadrado (m/s²).

En este caso:

  • m= 20 kg
  • a= 4 m/s²

Reemplazando:

F= 20 kg* 4 m/s²

Resolviendo:

F= 80 N

<u><em>La fuerza que será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s² es 80 N.</em></u>

4 0
3 years ago
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