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3 years ago

15

More people end up in U.S. emergency rooms because of fall-related injuries than from any other cause. At what speed v would som

eone hit the ground who accidentally steps off the top rung of a 7.00 ft tall stepladder

1 answer:

miss Akunina [59]3 years ago

6 0

Answer:

$v_{f}=6.47m/s$

Explanation:

Given data

Distance d=7.00 ft= 7*(1/3.281) =2.1336m

Initial velocity vi=0m/s

To find

Final velocity

Solution

From Kinematic equation we know that:

$v_{f}^{2} =v_{i}^{2}+2gd\\v_{f}^{2}=0+2(9.81m/s^{2} ) (2.1336m)\\v_{f}^{2}=41.86\\v_{f}=\sqrt{41.86}\\v_{f}=6.47m/s$

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A small wooden block with mass 0.750 kg is suspended from the lower end of a light cord that is 1.72 m long. The block is initia

Ierofanga [76]

Answer:

$v_{0}=319.2 m/s$

Explanation:

We need to use the momentum and energy conservation.

$p_{0}}=p_{f}$

$mv_{0}=(m+M)V_{1}$

Where:

m is the mass of bullet (m=0.01 kg)
M is the mass of the wooden (M=0.75 kg)
v(0) initial velocity of bullet
V(1) is the velocity of the combined object in the moment the bullet hist the block

Conservation of energy.

We have kinetic energy at first and kinetic and potential energy at the end.

$(1/2)(m+M)V_{1}^{2}=(1/2)(m+M)V_{2}^{2}+(m+M)gh$

Here:

V(1) is the velocity of the combined at the initial position
h is the vertical height ( h = 0.800 m)

We can find V(2) using the definition of force at this point:

$\Sigma F=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)$

$T-(m+M)gcos(\theta)=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)$

$cos(\theta) =(L-h)/L=(1.72-0.8)/1.72$

$\theta =57.66$

Now, we can solve the equation to find V(2)

$V_{2}=\sqrt{\frac{R*(T-(m+M)*g*cos(\theta))}{(m+M)}}$

$V_{2}=\sqrt{\frac{1.72*(4.86-(0.01+0.75)*9.81*cos(57.66))}{(0.01+0.75)}}$

$V_{2}=1.40 m/s$

Now we can find V(1) using the conservation of energy equation

$(1/2)V_{1}^{2}=(1/2)V_{2}^{2}+gh$

$V_{1}=\sqrt{V_{2}^{2}+2gh}$

$V_{1}=\sqrt{1.40^{2}+2*9.81*0.8}$

$V_{1}=4.20 m/s$

Finally, using the momentum equation we find v(0)

$v_{0}=\frac{(m+M)V_{1}}{m}$

$v_{0}=\frac{(0.01+0.75)*4.20}{0.01}$

$v_{0}=319.2 m/s$

I hope it helps you!

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4 years ago

What is similar between metals and suspensions

dimulka [17.4K]

<em>They both are heterogenous mixtures. </em>

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3 years ago

a bowling ball and a bag of potato chips are dropped at the same time from a tall on Earth , which hits the ground first ?

malfutka [58]

Isn’t it a bowling ball cause it’s the heaviest so it likely to go faster and hit the ground .

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3 years ago

Read 2 more answers

1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of

Masja [62]

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = $\sqrt{3^{2} +2^{2} }$ = $\sqrt{13}$ = 3.61cm = 0.036m

r₂ = $\sqrt{4^{2} + 3^{2} }$ = $\sqrt{25}$ = 5cm = 0.05m

electric potential V = $\frac{kq}{r}$

change in potential ΔV = $V_{1} - V_{2}$

ΔV = $\frac{2kq_{1} }{r_{1}} - \frac{2kq_{2} }{r_{2} }$ , where $q_{1} = q_{2}=$ 2.00μC

ΔV = $2kq(\frac{1}{r_{1}} - \frac{1}{r_{2} })$

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × $(\frac{1}{0.036} - \frac{1}{0.05} )$

ΔV= 2.789×10⁵

$\frac{1}{2}mv^{2}$ = ΔV × q₃

$\frac{1}{2}$ ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

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liberstina [14]

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