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3 years ago

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A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at

an angle 55º above the horizontal. What is the impulse delivered by the foot (magnitude and direction)? Ans: 43.6 kg*m/s @ 61.7° above the horizontal

1 answer:

pav-90 [236]3 years ago

7 0

Answer:

Explanation:

Given

mass of drop $m=2 kg$

height of fall $h=1 m$

ball leaves the foot with a speed of 18 m/s at an angle of $55^{\circ}$

Velocity of ball just before the collision with the floor

$u^=2gh$

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.8\times 1}=4.42 m/s$

Impulse delivered in Y direction

$J_y=m(v\sin (55)-(-u))$

$J_y=2(18\sin (55)+4.42)$

$J_y=38.32 kg-m/s$

Impulse in x direction

$J_x=m\times v\cos (55)$

$J_x=2\times 4.42\cos (55)=20.646$

$J_{net}=\sqrt{J_x^2+J_y^2}$

$J_{net}=\sqrt{(38.32)^2+(20.64)^2}$

$J_{net}=43.52 kg-m/s$

at an angle of $\tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}$

$\phi =tan^{-1}(1.856)$

$\phi =61.7^{\circ}$

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Answer:

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3 years ago

A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despi

aniked [119]

1) 4 forces

2) 165 N

3) 225 N

Explanation:

1)

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude $mg$ , where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

2)

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

$F_{net}=ma$

where

$F_{net}$ is the net force

m = 75.0 kg is the mass of the bicyclist

$a=2.20 m/s^2$ is its acceleration

Solving, we find the net force:

$F_{net}=(75.0)(2.20)=165 N$

3)

In this part, we basically want to find the forward force of push, F.

We can rewrite the net force acting on the bicyclist as

$F_{net}=F-R$

where:

F is the forward force of push

R is the air drag

We know that:

$F_{net}=165 N$ is the net force on the bicyclist

R = 60.0 N is the magnitude of the air drag

Therefore, by re-arranging the equation, we can find the force generated by the bicylicst by pedaling:

$F=F_{net}+R=165+60=225 N$

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4 years ago

Please answer my question!,

Karolina [17]

Answer:

Hope this helps! Mark as brainliest if liked thanks!

Explanation:

Your reasoning that the shadow is the shortest at mid-day is spot-on!

The wording of the question is the key to the answer. It says that the measurements were made in Summer. So this means that British Summer Time (BST) is being applied. BST is one hour ahead of Greenwich Mean Time and so what looks like 1pm is really 12 noon.

The safest sort of answer is to say that the shadow is shortest when the sun is at its highest point, and in this particular question that is at 1 pm because it is BST.

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3 years ago

A 1 000-V battery, a 3 000-Ω resistor, and a 0.50-μF capacitor are connected in series with a switch. The time constant for such

vampirchik [111]

Answer:

The current in the circuit at a time interval of τ seconds after the switch has been closed is 0.123 A

Explanation:

The time constant for an R and C in series circuit is given by τ = RC.

R = 3000 ohms, C = 0.5 × 10⁻⁶ F = 5.0 × 10⁻⁷ F

τ = 3000 × 5 × 10⁻⁷ = 0.015 s

The voltage across a capacitor as it charges is given be

V(t) = Vs (1 - e⁻ᵏᵗ)

where k = 1/τ

At the point when t = τ, the expassion becomes

V(t = τ) = 1000 (1 - e⁻¹) = 0.632 × 1000 = 632 V

Current flows as a result of potential difference,.

Current in the circuit at this time t = τ is given by

I = (Vs - Vc)/R

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R = 3000 ohms

I = (1000 - 632)/3000 = 0.123 A

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4 years ago

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

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Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

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4 years ago