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ratelena [41]
2 years ago
9

A computational model predicts the maximum kinetic energy a roller coaster car can have given its mass, the speed at the highest

point of its journey, and the height of the highest point. The mass is 250 kg, and the highest point of the journey is at 30m, where the cart is at rest. What is the maximum kinetic energy?
A. 113500 J
B. 468750 J
C. 0 J
D. 73500 J
Physics
1 answer:
Ymorist [56]2 years ago
3 0

Answer:

D

Explanation:

The decrease in potential energy is equal to the increase in kinetic energy.

mgh

250 x 9.8 x 30

=73, 500

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A cat is sleeping on the floor in the middle of a 2.8-m-wide room when a barking dog enters with a speed of 1.40 m/s. As the dog
marysya [2.9K]

Answer:

the cat is  0.4238 m in front of the dog as it leaps through the window

Explanation:

Given that;

acceleration a = 0.85 m/s²

speed v = 1.40 m/s

the cat is at rest, so initial velocity u = 0  

we know that, since the cat is sleeping on the floor in the middle of a 2.8-m-wide room, it needs to cover (2.8 m / 2 ) distance to get to the window;

using the second equation equation of motion;

s = ut + 1/2 at²  

we substitute

2.8/2 = 0×t + 1/2 × 0.85 × t²

1.4 = 0.425t²

t = √( 1.4 / 0.425 )

t = 1.81497 sec

now, at acceleration 0.10 m/s²

the dog has to cover the distance;

s = ut + 1/2 at²  

s = ( 1.4 × 1.81497) - 1/2 × 0.10 × 1.81497²

s =  2.540958 - 0.1647

s = 2.3762 m  

The cant in front of the dog as it leaps through the window;

distance = 2.8 m - 2.3762 m

distance = 0.4238 m

Therefore, the cat is  0.4238 m in front of the dog as it leaps through the window

7 0
3 years ago
A boat traveling across a river has a resultant velocity of 10 km/h and travels 34 degrees with respect to the shore. A) What is
vodka [1.7K]

Answer:

a) 1.55 m/s

b) 2.3 m/s

Explanation:

We know that the boat travels across the river, if we define the river as the x-axis, then the velocity of the boat is only on the y-axis.

Then we can write the velocity of the boat in still water as:

S = (0, B)

Now, when the boat is on the river, the velocity of the boat will be equal to the velocity of the boat in still water plus the velocity of the river.

The velocity of the river is:

v = (R, 0).

Then the velocity of the boat in that river is:

V' = (0, B) + (R, 0) = (R, B)

Now, we know that the velocity of the boat is 10km/h, and it travels at an angle of 34° with respect to the shore.

We can use the Pythagoreans theorem to write the components of this velocity as:

x-axis component = 10km/h*cos(34°) = 8.29 km/h

y-axis component = 10km/h*sin(34°) =  5.59 km/h

Then the velocity of the boat can be written in components as:

velocity = ( 8.29 km/h,  5.59 km/h)

And we knew that the velocity of the boat was written as  (R, B)

Then we must have:

R = 8.29 km/h

B = 5.59 km/h

a) The speed of the boat in m/s:

We know that the speed of the boat is 5.59 km/h.

First, we know that:

1km = 1000m, then:

5.59 km/h = 5.59*(1000m)h = 5,590 m/h

And we know that:

1h = 3600s

Then we can write:

5,590 m/h = 5,590 m/(3600s) = 1.55 m/s

b) The speed of the river in m/s:

We know that the speed of the river is 8.29 km/h

Using the same reasoning as above, we can do the change of units as follows:

8.29 km/h = 8.29 (1000m)/(3600s) = 2.3 m/s

6 0
2 years ago
A car traveling with constant speed travels 150 km in 7200 s what is speed of the car
spayn [35]

The speed of the car is exactly 150/7200 km/sec, or 125/6 meters/sec. 

In more familiar units, that speed is equivalent to ...

-- (20 and 5/6) meters/sec

-- 75 km/hour

8 0
3 years ago
A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the
andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and  how far away from the overpass is the roadrunner when the coyote drops the net.

d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0
2 years ago
QUIZZ <br><br>12 + 14 + 66 - 87
svet-max [94.6K]

12 + 14 + 66 - 87

26+66-87

92-87

5

6 0
2 years ago
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