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3241004551 [841]
3 years ago
13

A car manufacturer wants to change its car's design to increase the car's acceleration. Which changes should the engineers

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
5 0

Increase the Force provided by the Engine.

Decrease the Mass of the Car.

Increase the top velocity the Car can travel.

Explanation:

A design targeted at increasing the engine force, decreasing the mass of the car and increasing the top velocity will increase the acceleration of the car.

From newton's second law, we know that force is the product of mass and acceleration.

 Force = mass x acceleration

  Acceleration = \frac{force }{mass}

Acceleration is the change in the velocity of a body with time.

  • We see that acceleration is directly related to force. As the force of the engine increases, its acceleration increases.
  • But acceleration is inversely related to mass. The more the mass of the car, the lesser the acceleration. The car should be made lighter.
  • The top velocity is the maximum velocity the car can go.
  • The higher the top level velocity, the more the acceleration a car can have.

Learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

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It's being planned to launch in the 2020's
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3 years ago
A stone is dropped into water from a bridge 52 m above the water's
dusya [7]

Answer:

Its final velocity and how much time it takes to reach the water

Explanation:

The motion of the stone is a uniformly accelerated motion, so we can use the following suvat equation to determine its final velocity:

v^2-u^2=2as

where

v is the final velocity

u = 0 is the initial velocity

a=g=9.8 m/s^2 is the acceleration of gravity

s = 52 m is the distance covered during the fall

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0^2+2(9.8)(52)}=31.9 m/s

We can also find how much time it takes to reach the water, using the equation

v=u+at

where

v = 31.9 m/s is the final velocity

u = 0 is the initial velocity

a=g=9.8 m/s^2

t is the time

And solving for t,

t=\frac{v-u}{a}=\frac{31.9-0}{9.8}=3.26 s

3 0
3 years ago
A 100-g toy car moves along a curved frictionless track. At first, the car runs along a flat horizontal segment with an initial
Sliva [168]

Answer:

The final velocity of the car is 1.85 m/s

Explanation:

Hi there!

The initial kinetic energy of the toy car can be calculated as follows:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass.

v = velocity.

KE = 1/2 · 0.100 kg · (2.66 m/s)² = 0.354 J

The gain in altitude produces a gain in potential energy. This gain in potential energy is equal to the loss in kinetic energy. So let´s calculate the potential energy of the toy car after gaining an altitude of 0.186 m.

PE = m · g · h

Where:

PE = potential energy.

m = mass.

g = acceleration due to gravity.

h = height.

PE = 0.100 kg · 9.8 m/s² · 0.186 m = 0.182 J

The final kinetic energy will be: 0.354 J - 0.182 J = 0.172.

Using the equation of kinetic energy, we can obtain the velocity of the toy car after running up the slope:

KE = 1/2 · m · v²

0.172 J = 1/2 · 0.100 kg · v²

2 · 0.172 J / 0.100 kg = v²

v = 1.85 m/s

The final velocity of the car is 1.85 m/s

3 0
3 years ago
A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35m x 0.55m. The magnetic field has a
Cloud [144]

Answer:

E = 0.38V

Explanation:

See attachment below.

4 0
3 years ago
Light travels 300 000 000 m/s and one year has approximately 32 000 000 second a light year is the distance light travels in one
Lelu [443]

Explanation:

It is given that,

Speed of light, v=300 000 000\ m/s=3\times 10^8\ m/s

Seconds in 1 year, t=32 000 000=32\times 10^6\ s

We need to find the distance traveled by light in one year. Speed of an object is given by :

v=\dfrac{d}{t}

So,

d=v\times t\\\\d=3\times 10^8\times 32\times 10^6\\\\d=9.6\times 10^{15}\ m

Since,

1\ \text{light year}=9.46\times 10^{15}\ m\\\\1\ m=\dfrac{1}{9.46\times 10^{15}}\ \text{ly}\\\\9.6\times 10^{15}\ m=\dfrac{9.6\times 10^{15}}{9.46\times 10^{15}}\\\\d=1.01\ \text{ly}

So, the distance covered by light is 1.01 light years.

8 0
3 years ago
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