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1 year ago

A hopper jumps straight up to a height of 1.1 m. With what velocity did it leave the floor?

Physics

1 answer:

Amanda [17]1 year ago

4 0

Answer:

4.64m/s

Explanation:

We can use the formula [ v = √2gh ] to solve for this problem. We know that g is constant acceleration (9.8), and h is height (1.1).

v = √2(9.8)(1.1)

v ≈ 4.64m/s

Best of Luck!

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Is calculating the change of velocity the same as calculating acceleration?

koban [17]

Answer:

Yes! Thinking about it graphically a position vs time graph models meters per second in most cases, making every point on the line have the units m/s. If we want the find the slope we are finding the change between each point and those units would change to m/s/s or m/s^2 giving us the same units for acceleration. Simply put, slope of a velocity graph gives us acceleration.

Explanation:

4 0

2 years ago

A street light is on top of a 8 foot pole. Joe,

zubka84 [21]

8/4 = y/y-x

8y - 8x = 4y

y = 2x

y = 2 x 4

y = 8

Hope this helps

5 0

3 years ago

What is the final velocity of a car that starts at 22 m/s and accelerates at 3.78 m/s for distance of 45 m

Pepsi [2]

v^2 = v0^2 +2ad v^2 = 22^2 + 2*3.78*45 = 824.2 v= √824.2 = 28.7 m/s

5 0

3 years ago

A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s

Paraphin [41]

Answer:

$\omega = 0.016\,\frac{rad}{s}$

Explanation:

The rotation rate of the man is:

$\omega = \frac{v}{R}$

$\omega = \frac{0.80\,\frac{m}{s} }{5\,m}$

$\omega = 0.16\,\frac{rad}{s}$

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

$(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega$

The final angular speed is:

$\omega = 0.016\,\frac{rad}{s}$

3 0

2 years ago

A man paddles a canoe at 6 km per hour. If he paddles on a river with a current of 6 km per hour, what is the speed of the canoe

Romashka-Z-Leto [24]

Answer:

If the canoe heads upstream the speed is zero. And directly across the river is 8.48 [km/h] towards southeast

Explanation:

When the canoe moves upstream, it is moving in the opposite direction of the normal river current. Since the velocities are vector (magnitude and direction) we can sum each vector:

Vr = velocity of the river = 6[km/h}

Vc = velocity of the canoe = -6 [km/h]

We take the direction of the river as positive, therefore other velocity in the opposite direction will be negative.

Vt = Vr + Vc = 6 - 6 = 0 [km/h]

For the second question, we need to make a sketch of the canoe and we are watching this movement at a high elevation. So let's say that the canoe is located in point 0 where it is located one of the river's borders.

So we are having one movement to the right (x-direction). And the movement of the river to the south ( - y-direction).

Since the velocities are vector we can sum each vector, so using the Pythagoras theorem we have:

$Vt = \sqrt{(6)^{2} +(-6)^{2} } \\Vt=8.48[km/h]$

5 0

2 years ago