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3 years ago

When low on gasoline, is it better to increase or decrease the speed of the vehicle? Explain.

1 answer:

5 0

Yeah!! It is better to decrease the speed of the vehicle so the ejection of gases would happen in slow manner.... then you can save your gasoline

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John flies directly east for 20km, then turns to the north and flies for another 10 km before dodging a flock of geese. what’s t

KIM [24]

The distance is 30 km and the displacement is 22.4 km North East

7 0

3 years ago

You have a bowling ball with a mass of 4kg. You throw it with an acceleration of 10 m/s/s. With how much force will it hit the p

Eddi Din [679]

Answer:

<h3>The answer is 40 N</h3>

Explanation:

The force acting on an object can be found by using the formula

<h3>force = mass × acceleration</h3>

From the question

mass = 4 kg

acceleration = 10 m/s²

So we have

force = 4 × 10

We have the final answer as

Hope this helps you

6 0

3 years ago

Which illustration represents low accuracy but high precision?

Misha Larkins [42]

Answer:

D i think

Explanation:

7 0

3 years ago

Read 2 more answers

Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d

alex41 [277]

Answer:

$v=18.89\frac{m}{s}$

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

$\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)$

Solving N from (2) and replacing in (1):

$N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Replacing and solving for v:

$gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}$

4 0

3 years ago

A resistor, inductor, and a battery are arranged in a circuit. The circuit has an inductance of L = 1 H and a resistance of 1.4

shtirl [24]

Answer:

$\tau \approx 7.14 \times 10^{-4}s \approx0.714ms$

Explanation:

In a LC circuit The time constant τ is the time necessary for 60% of the total current (maximum current), pass through the inductor after a direct voltage source has been connected to it. The time constant can be calculated as follows:

$\tau =\frac{L}{R}$

Therefore, the time needed for the current to reach a fraction f = 0.6(60%) of its maximum value is:

$\tau =\frac{1}{1.4\times 10^{3}} =7.142857143 \times 10^{-4} \approx7.14 \times 10^{-4}s$

8 0

3 years ago