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erica [24]
3 years ago
5

An automobile starts from rest and travels down a straight section of road. The distance s (in feet) of the car from the startin

g position after t seconds is given by s(t)=5t3 (a) Find the instantaneous velocity (in feet per second) at t=4 seconds.
Physics
1 answer:
Vladimir [108]3 years ago
4 0

Answer: 240 ft/sec

Explanation:

If the distance s, of the car regarding his starting position, is defined as a function of time, by definition, the instantaneous velocity at any time, is given by the derivative of the displacement (distance from origin) respect the time, as follows;

v = ds / dt ⇒ v= d/dt (5 t³) = 15 t² = 15 t²

Replacing by t=4 s, we get:

v = 15. 4² = 15.16 = 240 ft/sec

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The two main states of mechanical energy are ___________ and potential energy.
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<span>Potential energy and Kinetic energy</span>
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A yellow train of mass 100 kg is moving at 8 m/s towards an orange train of mass 200 kg traveling on the opposite direction on t
vladimir1956 [14]
Mass of yellow train, my = 100 kg

Initial Velocity of yellow train, = 8 m/s

mass of orange train = 200 kg

Initial Velocity of orange train = -1 m/s (since it moves opposite direction to the yellow train, we will put negative to show the opposite direction)

To calculate the initial momentum of both trains, we will use the principle of conservation of momentum which

The sum of initial momentum = the sum of final momentum


Since the question only wants the sum of initial momentum,

(100)(8) + (200)(-1) = 600 m/s

8 0
2 years ago
A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the
Genrish500 [490]

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

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

in this case we know the speed of the runner

v =8.00\ m/s

The radius "r" will be the distance from the runner to the center of the track

r = 28.2\ m

a_c = \frac{8^2}{28.2}\ m/s^2

a_c = 2.27\ m/s^2

The answer is the last option

3 0
3 years ago
A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia
Musya8 [376]

Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

When the tiger reaches the ground, y = 0:

0 = 6.00 m - 1/2 · 9.81 m/s² · t²

2 · (-6.00 m) / -9.81 m/s² = t²

t = 1.11 s

We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:

x = x0 + v0 · t

At t = 1.11 s, x = 2.00 m

x0 = 0

2.00 m = v0 · 1.11 s

2.00 m / 1.11 s = v0

v0 = 1.80 m/s

The initial speed of the tiger is 1.80 m/s

4 0
3 years ago
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