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2 years ago

Car B is being pushed by a force of 22000 N. If it has a mass of 1375 kg.,

Physics

1 answer:

Burka [1]2 years ago

5 0

Answer:

a = 16 m/s²

General Formulas and Concepts:

Dynamics

Newton's Law of Motions

Newton's 1st Law of Motion: An object at rest remains at rest and an object in motion stays in motion
Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration)
Newton's 3rd Law of Motion: For every action, there is an equal and opposite reaction

Explanation:

Step 1: Define

Identify variables

[Given] F = 22000 N

[Given] m = 1375 kg

[Solve] a

Step 2: Find Acceleration

Substitute in variables [Newton's 2nd Law of Motion]: 22000 N = (1375 kg)a
Isolate a: 16 m/s² = a
Rewrite: a = 16 m/s²

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Citrus2011 [14]

Answer:

(a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

Explanation:

Given that,

Speed v₁= 60 mph

Speed v₂ = 225 mph

Speed v₃ = 235 mph

(a). We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{1}=P+\dfrac{1}{2}\rho v^2+\rho gh$

there, no vertical movement

So, on neglect of height term

$P_{1}=P+\dfrac{1}{2}\rho v_{1}^2$

Where, P= atmospheric pressure

$\rho$ = air density

v = speed

Put the value in the equation

$P_{1}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(60\times1.4667)^2)$

$P_{1}=2126.0\ lb/ft^2$

$P_{1}=\dfrac{2126.0}{144}$

$P_{1}= 14.76\ psi$

(b). Speed v₂ = 225 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{2}=P+\dfrac{1}{2}\rho v_{2}^2$

Put the value in the equation

$P_{2}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(225\times1.4667)^2)$

$P_{2}=2246.17\ lb/ft^2$

$P_{2}=\dfrac{2246.17}{144}$

$P_{2}= 15.59\ psi$

(c). Speed v₃ = 235 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{3}=P+\dfrac{1}{2}\rho v_{3}^2$

Put the value in the equation

$P_{3}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(235\times1.4667)^2)$

$P_{3}=2257.93\ lb/ft^2$

$P_{3}=\dfrac{2257.93}{144}$

$P_{3}= 15.68\ psi$

According to bernoulli's equation,

If the car increases the velocity the the pressure on the surface of the driver's hand increases.

The pressure from P₁ to P₃ are all near the value of one atmosphere.

So, the pressure difference of one atmosphere is not enough to break the driver's hand.

Hence, (a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

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2 years ago

If a steel containing 1.88 wt%C is cooled relatively slowly to room temperature, what is the expected weight fraction of pearlit

Oxana [17]

Answer:

The answer is %pearlite = 0.06%

Explanation:

according to the exercise we have that the percentage is 1.88% C, therefore, the percentage of perlite is equal to:

%pearlite = (B*C)/(A*C) = (2-1.88)/(2-0) = 0.06%

The percentage of cementite is equal to:

%cementite = (1.88-0)/(2-0) = 0.94%

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2 years ago

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

Talja [164]

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

$F_{g}$ = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

$T$ = Tension force in upward direction

$F_{d}$ = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

$F_{g}$ - $F_{d}$ - $T$ = 0

7000 - 1800 - $T$ = 0

$T$ = 5200 N

$T$ = 5.2 x 10³ N

Part B)

$F_{g}$ = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

$T$ = Tension force in upward direction

$F_{d}$ = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

$T$ - $F_{g}$ - $F_{d}$ = 0

$T$ - 7000 - 1800 = 0

$T$ = 8800 N

$T$ = 8.8 x 10³ N

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3 years ago

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Answer:

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Explanation:

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