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Pani-rosa [81]
2 years ago
10

The speedometer of your automobile shows that you are proceedig at a steady speed of 80 although your automobile is in accelerat

ed motion?
Physics
1 answer:
nika2105 [10]2 years ago
7 0

Answer:

<em>The engine must provide power to compensate friction</em>

Explanation:

<u>Accelerated Motion </u>

Newton's first law states that a body will keep its speed or state of rest while no external net force is applied. Our automobile is running through a road which surface exerts a frictional force on the wheels. If we didn't use fuel, the automobile will eventually stop, because the unbalanced friction force causes deceleration. To compensate that force, we must provide power to the engine. When our speedometer shows a constant speed, the net force is zero, but the car needs to accelerate some, just to compensate the friction force.

The total acceleration keeps being zero, but the engine is actually doing work against friction.

You might be interested in
Two boats start together and race across a 48-km-wide lake and back. boat a goes across at 48 km/h and returns at 48 km/h. boat
jolli1 [7]

Answer:

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

Explanation:

Case 1: Boat 1

Speed of boat = \frac{distance of river}{time}

time = \frac{distance of river}{speed of boat}

While going to another end

time = \frac{distance of river}{speed of boat}

time = \frac{48}{48}

time = 1 hour

While going back,

time = \frac{distance of river}{speed of boat}

time = \frac{48}{48}

time = 1 hour

Total time taken by boat 1 is,

Total time by boat 1 = 1 hour + 1 hour = 2 hour

Total time by boat 1 = 2 hour

Total time taken by boat 1 for the round trip is 2 hour.

Case 2: Boat 2

Speed of boat = \frac{distance of river}{time}

time = \frac{distance of river}{speed of boat}

While going to another end

time = \frac{distance of river}{speed of boat}

time = \frac{48}{24}

time = 2 hour

While going back,

time = \frac{distance of river}{speed of boat}

time = \frac{48}{72}

time = 0.66 hour

Total time taken by boat 2 is,

Total time by boat 1 = 2 hour + 0.66 hour

Total time by boat 1 = 2.66 hour

Total time taken by boat 2 for the round trip is 2.66 hour.

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

5 0
3 years ago
Which of these is an example of a chemical change?
xenn [34]

Hello!  B. would be an example of a chemical change.

5 0
3 years ago
Read 2 more answers
A milk truck carries milk with density 64.6 lb/ft3 in a horizontal cylindrical tank with diameter 12 ft. (a) Find the force exer
7nadin3 [17]

Answer:

F = 351×10³lb

Explanation:

Given the density

ρg = 64.6lb/ft³

Diameter d = 12ft

The tank is horizontally cylindrical. The vertical distance from the top to the bottom of the tank is h = 12ft

The pressure in the tank is

P = ρgh = 64.6 × 12 = 775.2lb/ft²

The force exerted on one end of the tank is therefore F = PA = 775.2 × πd² = 775.2π×12²

F = 351×10³lb.

4 0
3 years ago
traveling at about 30 mph, how many feet will the average driver cover from the time they see the danger until they hit the brak
barxatty [35]

Answer: 75 ft

Explanation:

Breaking distance = Speed²/ 20

= 30²/20

= 45 feet

Stopping distance = Speed + braking distance

= 30 + 45

= 75 ft

5 0
2 years ago
A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so sma
mel-nik [20]

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

              {} ⇩

[m = 20 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 0.5·R]

              {} ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

<em>1: m = 20 kg, r = 0.25·R</em>

<em>2: m = 10 kg, r = 1.0·R</em>

<em>3: m = 10 kg, r = 0.25·R</em>

<em>4: m = 15 kg, r = 0.75·R</em>

<em>5: m = 10 kg, r = 0.5·R</em>

<em>6: m = 40 kg, r = 0.25·R</em>

According to the principle of conservation of angular momentum, we have;

I_i \cdot \omega _i = I_f \cdot \omega _f

The moment of inertia of the merry-go-round, I_m = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²·\omega _i = (0.5·M·R² + m·r²)·\omega _f

Given that 0.5·M·R²·\omega _i is constant, as the value of  m·r² increases, the value of \omega _f decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[<u>m = 10 kg, r = 0.25·R</u>] > [<u>m = 20 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 0.5·R</u>] > [<u>m = </u>

<u>10 kg, r = 0.5·R</u>] = [<u>m = 40 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 1.0·R</u>].

Learn more here:

brainly.com/question/15188750

6 0
2 years ago
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