All categories

4 years ago

How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and

Physics

1 answer:

aev [14]4 years ago

4 0

Answer:

h= 46.66 m

Explanation:

Given that

Initial speed of the car ,u = 110 km/h

We know that

1 km/h= 0.277 m/s

u= 30.55 m/s

lets height gain by car is h.

The final speed of the car will be zero at height h.

v²=u²- 2 g h

v= 0 m/s

0²=30.55²- 2 x 10 x h ( g = 10 m/s²)

h= 46.66 m

You might be interested in

a Ferrari with an initial velocity of 10 m/s, comes to a complete stop in 5 seconds, what will its acceleration be?

11111nata11111 [884]

10 m/s divided by 5 s

-2m/s/s

minus - deceleration

6 0

4 years ago

Read 2 more answers

Three small objects are located in the x-y plane as shown in the figure. The objects have the following masses: mA = 1.09 kg, mB

aliina [53]

The moment of inertia of this set of objects with respect to the axis perpendicular to the the x-y plane passing through location x = 4.00 m and y = 3.00 m is 144.97 Kg.m^2.

<h3>What is moment of inertia?</h3>

The moment of inertia is the amount of rotation obtained by an object when it is in state of motion or rest.

Three small objects are located in the x-y plane as shown in the figure. The objects have the following masses: mA = 1.09 kg, mB = 2.83 kg, mC = 3.39 kg.

The coordinates of Ball A :(2,1) Ball B :(8,2) and Ball C: (5,8) and axis is at (4,3)

Here, for the moment of inertia of each ball

For ball A

IA = 1.09 x ((4 - 2)^2 + (3-1)^2)

IA = 8.72 Kg.m^2

for the ball B

IB = 2.83 * ((4 - 8)^2 + (3 - 2)^2)

IB = 48.11 Kg.m^2

for the ball C

IC = 3.39 * ((4 - 5)^2 + (3 - 8)^2)

IC = 88.14 Kg.m^2

Total moment of inertia = 8.72 + 48.11 + 88.14

Total moment of inertia = 144.97 Kg.m^2

Thus, the moment of inertia of this set of objects with respect to the axis perpendicular to the the x-y plane passing through location x = 4.00 m and y = 3.00 m is 144.97 Kg.m^2.

Learn more about moment of inertia.

brainly.com/question/15246709

#SPJ1

3 0

2 years ago

where would a bowling ball and a napkin fall with the same acceleration. Check all that apply. A. Dropped off of the Leaning Tow

Kazeer [188]

Answer: here ya go

A Bowling ball and a napkin would fall with the same acceleration in space where there is vacuum. Basically where there is no drag force and the two objects are freely falling under gravity. The two objects would fall with same acceleration due to gravity.

Explanation:

7 0

3 years ago

A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. Wh

mixas84 [53]

Answer:

the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Explanation:

Given;

diameter of the ceiling fan, d = 90 cm = 0.9 m

angular speed of the fan, ω = 64 rpm

time taken for the fan to stop, t = 28 s

The distance traveled by the ceiling fan when it comes to a stop is calculated as;

$d = vt = \omega r\times t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m$