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jenyasd209 [6]
2 years ago
11

Consider a 2100-kg car cruising at constant speed of 70 km/h. Now the car starts to pass another car by accelerating to 110 km/h

in 5 s. Determine the additional power needed to achieve this acceleration. What would your answer be if the total mass of the car were only 700 kg
Physics
1 answer:
Liula [17]2 years ago
6 0

Answer: 51841.5 Watts

Explanation: Using the kinematic equation for the final velocity for a constant acceleration we have:

Vf=Vi+a*t

replacing the values the results is

a=(Vf-Vi)/t= (30.55 m/s-19.44 m/s)/5s= 2.22 m/s^2

Remenber that to convert the speed in Km/h to m/s we have to multiplier by the factor 0.277.

Finally to calculate the increment of power get the final velocity we have to use Neton second law to determine the Force applied to the car.

F=m* a=2100 Kg* 2.22 m/s^2= 4666.2 N

Then increment  power to accelerate is given by:

ΔPower= Force* Δ velocity= 4666.2 N* 11,11 m/s= 51841.5 Watts

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You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅  = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0
2 years ago
An inclined plane is made out of a short plank of wood. It is used to move a 300N box up onto a tabletop 1m above the floor. Wha
Ronch [10]

Answer:

<em>The purpose of an inclinded plane is to make easier to move objects to a certain height.</em>

The technology behind this is about the Work you need to use to move the object upwards. Basically, when we use an inclined plane, we are splitting the net force, making easier to move. All this means, the force needed to move the objecto up will be lower, due to the inclined plane.

So, if the force needed is lower, then the work is also lower, because the work done is defined as the product between the force applied and the distance traveled.

<em>In addition, if we have a longer inclined plane, that means the force needed is even lower,</em> beacuse the distance is increased, but the Work is the same, because it only depends on the initial and final point.

Therefore, in this case, the work would remain the same and the mechanical advantage would increase. As we said before, the work needed will be the same despite the force decreases, because the distance increases, remaining the work as a constant. And the mechanical advantage increases, because it's easier to move if the inclined plane is longer.

3 0
2 years ago
Read 2 more answers
Sebanyak 0,2 mol gas ideal berada dalam wadah yang volumenya 10 liter dan tekanan 1 atm . berapakah suhu gas tersebut
tamaranim1 [39]

The temperature of the gas is about 600 K

\texttt{ }

<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

\large {\boxed {PV = nRT} }

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

\texttt{ }

<u><em>Question (Translation):</em></u>

<em>A total of 0.2 moles of ideal gas are in containers with volume of 10 liters and a pressure of 1 atm. What is the temperature of the gas?</em>

<u>Given:</u>

number of moles = n = 0.2 moles

volume of gas = V = 10 liters

pressure of gas = P = 1 atm

gas contant = R = 0.0821 L.atm/mol.K

<u>Unknown:</u>

temperature of gas = T = ?

<u>Solution:</u>

PV = nRT

1 \times 10 = 0.2 \times 0.0821 \times T

10 = 0.01642 \times T

T = 10 \div 0.01642

T \approx 600 \texttt{ K}

\texttt{ }

<h3>Learn more</h3>
  • Minimum Coefficient of Static Friction : brainly.com/question/5884009
  • The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

\texttt{ }

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure

8 0
2 years ago
A heavy boy and a lightweight girl are balanced on a massless seesaw. If they both move forward so that they are one-half their
Yuki888 [10]

Answer:

The system is still balanced

Explanation:

If we suppose that the boy weights M and the girl m, and are balanced at distances L1 and L2 from the pivot point respectively, thy will be balanced if the resultant torque of all the farces from the pivot pint is cero:

(1)T=M*L1-m*L2=0

now they moved to distances (L1)/2 and (L2)/2, the resultant torque will be:

(2)T=M*(L1)/2-m*(L2)/2

1/2 can be taken out as a common factor:

(3)T=(M*L1-m*L2)*1/2

As the the inside of the parenthesis equals equation (1) that equals zero, hte whole equiation equals zero:

(3)T=(0)*1/2=0

So the system is still balanced

6 0
2 years ago
Two extremely large nonconducting horizontal sheets each carry uniform charge density on the surfaces facing each other. The upp
sashaice [31]

This problem refers to a parallel plate capacitor. There is an electric field between the two plates. The working equation to be used is the Gauss’s Law which is

Electric field = Surface charge density / ε0

The answer is -2.52 μC/m2.

8 0
2 years ago
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