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zvonat [6]
1 year ago
11

Engineers are trying to improve a race car. Their goal is to increase the acceleration of the car using the same engine. Which c

hange shows the best application of Newton’s laws of motion to achieve this goal?
increasing the inertia of the car
decreasing the reaction force
increasing the action force
decreasing the mass of the car
Physics
1 answer:
Andru [333]1 year ago
4 0

To increase the acceleration of the car using the same engine, the   mass of the car must be decreased.

<h3>What is Newton's first law of motion</h3>

Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that path unless acted upon by an external force.

The first law is also called the law of inertia because it depends on the mass of the object. The greater the mass, the greater the inertia and more reluctant the object will be to move.

Thus, to increase the acceleration of the car using the same engine, the   mass of the car must be decreased.

a = F/m

Learn more about Newton's law here: brainly.com/question/25545050

#SPJ1

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The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting
White raven [17]

Answer:

\frac{\dot Q}{A} =20.129\ W.m^{-2}

T_1=27.58\ ^{\circ}C & T_2=2.41875\ ^{\circ}C

Explanation:

Given:

  • interior temperature of box, T_i=30^{\circ}C
  • height of the walls of box, h=3\ m
  • thickness of each layer of bi-layered plywood, x_p=1.25\ cm=0.0125\ m
  • thermal conductivity of plywood, k_p=0.104\ W.m^{-1}.K^{-1}
  • thickness of sandwiched Styrofoam, x_s=5\ cm=0.05\ m
  • thermal conductivity of Styrofoam, k_s=0.04\ W.m^{-1}.K^{-1}
  • exterior temperature, T_o=0^{\circ}C

<u>From the Fourier's law of conduction:</u>

\dot Q=\frac{dT}{(\frac{x}{kA}) }

\dot Q=\frac{dT}{R_{th} } ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

R_{th}=R_p+R_s+R_p

R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}

R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})

R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})

R_{th}=\frac{1.4904}{A} .....................(2)

Putting the value from (2) into (1):

\dot Q=\frac{30-0}{\frac{1.4904}{A} }

\dot Q=\frac{30\ A}{1.4904}

\frac{\dot Q}{A} =20.129\ W.m^{-2} is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}

20.129=0.104\times \frac{30-T_1}{0.0125}

T_1=27.58\ ^{\circ}C

&<u>For Styrofoam-plywood interface from inside:</u>

\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}

20.129=0.04\times \frac{27.58-T_2}{0.05}

T_2=2.41875\ ^{\circ}C

4 0
3 years ago
Mixing baking soda with vinegar is an example of an oxidation-reduction
34kurt

Answer:

B. False

Explanation:

The reaction of baking soda and vinegar produces carbon dioxide gas. It is an example of precipitation reactions.

4 0
3 years ago
if a tank filled with water contains a block and the height of the water above point A within the block is 0.6meter, what's the
Fittoniya [83]
The pressure exerted by a fluid solely relies on the depth or height of the fluid, its density, and the gravitational constant. These three are related in the equation:

Pressure = density x g x height

In the problem, point A is within the block inside the tank. The water above the block is assumed to be 0.6 meters. This gives a point A pressure of:

P = 1000 kg/m^3 * 9.81 m/s^2 * 0.6 m = 5,886 Pa or 5.88KPa
6 0
3 years ago
Read 2 more answers
A car starting from rest has a constant acceleration of 4 meters per second per second. How fast will it be going in 5 seconds?
irinina [24]

Answer:

18 m

Explanation:

Given : vo = 0 m/s ; t = 3 s; a = 4 m/s^2 ; d = ? m ; average velocity = ? m/s ; fonal velocity = ? m/s

solving for the final velocity, v

v = a * t

v = 4 m/s^2 * 3 s

v = 12 m / s

Solving for the average velocity. avg v

avg v = (vo + v) / 2

avg v = (0 m / s + 12 m/s) / 2

avg v = 6 m / s

Solving for the distance traveled after 3 s

d = avg v * t

d = 6 m / s * 3 s

d = 18 meters

In the first 3s the car travels 18 meters.

4 0
2 years ago
The pressure of a gas is reduced from 1200.0 mm hg to 850.0 mm hg as the volume of its container is increased by moving a piston
Kamila [148]
From gas laws:
\frac{PV}{T} = Constant

Therefore,
\frac{ P_{1}  V_{1} }{ T_{1} } =  \frac{ P_{2}  V_{2} }{ T_{2} }

P1 = 1200 mm
V1 = 85 ml
T1 = 90°C = 363.15 K
P2 = 850 mm
V2 = 350 ml
T2 = ?

Substituting;
T_{2} =  \frac{ P_{2}  V_{2}  T_{1} }{ P^{1}  V^{1} } =  \frac{850*350*363.15}{1200*85} = 1059.19 K = 786.04 °C
8 0
3 years ago
Read 2 more answers
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