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3 years ago

5

A soccer ball is kicked with an initial speed of 23 m/s at an angle of 21Â° with respect to the horizontal. (a) find the maximum

height reached by the ball. m (b) find the speed of the ball when it is at the highest point on its trajectory. m/s (c) where does the ball land

1 answer:

Archy [21]3 years ago

3 0

Just try your best best friend everyone

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a straight line on a graph means there is a _________ relationship between the dependent variable and the independent variable

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X-rays travel in vacuum at a higher speed than the visible light. (A) True (B) False

AleksandrR [38]

Answer:

(B) False

Explanation:

The visible light and the x-rays are part of the electromagnetic spectrum, all in the spectrum travel in the vacuum at the same speed, at the speed of light (c = 299,792,458 m/s ). If the visible light ant the x-rays are traveling inside a medium, their velocity will change due to the properties of the medium but in the vacuum, their speed is the same.

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4 years ago

A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi

bogdanovich [222]

Answer:

1) $W_{object} = 400 N$

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

$\sum F_{Tot} = 0$

In this case we just have y-direction forces.

$\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0$

Now, let's solve the equation for W(object).

$W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N$

2 ) To find the position of the heaviest weight we need to use the torque definition.

$\sum \tau = 0$

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

$L = 1.50 m$ ; total length of the bar.

$D_{AP} = x$ ; distance between Tension A and P point.

$D_{BP} = L-x$ ; distance between Tension B and P point.

$D_{W_{bar}P} = \frac{L}{2}-x$ ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

$\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0$

Finally we just need to solve it for x.

$x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}$

$x = 0.28 m$

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

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4 years ago

A man standing on a bus remains still when the bus is at rest. When the bus moves forward and then slows down the man continues

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When the bus starts moving forward, the man remains at rest,
causing him to lean back.

When the bus slows down, the man continues to move forward,
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Both events are examples of the effect of inertia.

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