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horrorfan [7]
3 years ago
14

A suspension bridge is 60.0 m above the level base of a gorge. A stone is thrown or dropped from the bridge. Ignore air resistan

ce. At the location of the bridge g has been measured to be 9.83 m/s2. (a) If you drop the stone, how long does it take for it to fall to the base of the gorge? (b) If you throw the stone straight down with a speed of 20.0 m/s, how long before it hits the ground? (c) If you throw the stone with a velocity of 20.0 m/s at 30.0° above the horizontal, how far from the point directly below the bridge will it hit the level ground?
Physics
1 answer:
Alexxx [7]3 years ago
3 0

Answer:

a)t_{1}=3.49s

b)t_{2}=2.00s

c)Xmax=80.71m

Explanation:

<u>a)Kinematics equation for the Stone, dropped:</u>

v(t)=v_{o}-g*t

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

y_{o}=h=60m       initial position is bridge height

v_{o}=0m/s       the stone is dropped

The ball reaches the ground, y=0, at t=t1:

0=h-1/2*g*t_{1}^{2}

t_{1}=\sqrt{2h/g}=\sqrt{2*60/9.83}=3.49s

<u>b)Kinematics equation for the Stone, with a initial speed of 20m/s:</u>

v(t)=v_{o}-g*t

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

y_{o}=h=60m       initial position is bridge height

v_{o}=-20m/s       the stone is thrown straight down

The ball reaches the ground, y=0, at t=t1:

0=h+v_{o}t_{2}-1/2*g*t_{2}^{2}

0=60-20t_{2}-1/2*9.83*t_{2}^{2}

t2=-6.01       this solution does not have physical sense

t2=2.00

<u>c)Kinematics equation for the Stone, with a initial speed of 20m/s with an angle of 30° above the horizontal:</u>

v(t)=v_{o}-g*t

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

y_{o}=h=60m       initial position is bridge height

v_{o}=20sin(30)=10m/s       the stone is thrown with an angle of 30° above the horizontal

The ball reaches the ground, y=0, at t=t3:

0=h+v_{o}t_{3}-1/2*g*t_{3}^{2}

0=60+10t_{3}-1/2*9.83*t_{3}^{2}

t3=-2.62       this solution does not have physical sense

t3=4.66

the movement in x:

v=constant=20cos(30)m/s

x(t)=v*t

Xmax=v*t3=20cos(30)*4.66=80.71m

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Explanation:

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A 2.00 kg block hangs from a spring. A 300 g body hung below the block stretches the spring 2.00 cm farther. (a) What is the spr
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The spring constant is 147 N/m

Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm

We need to find the spring constant

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.

The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring

We know that F = kx

300(9.8)= k (0.02)

k = 147.15 N/m

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2 years ago
Find the quantity of heat needed
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Approximately 3.99\times 10^{4}\; \rm J (assuming that the melting point of ice is 0\; \rm ^\circ C.)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}

The energy required comes in three parts:

  • Energy required to raise the temperature of that 0.100\; \rm kg of ice from (-10\; \rm ^\circ C) to 0\; \rm ^\circ C (the melting point of ice.)
  • Energy required to turn 0.100\; \rm kg of ice into water while temperature stayed constant.
  • Energy required to raise the temperature of that newly-formed 0.100\; \rm kg of water from 0\; \rm ^\circ C to 10\;\ rm ^\circ C.

The following equation gives the amount of energy Q required to raise the temperature of a sample of mass m and specific heat capacity c by \Delta T:

Q = c \cdot m \cdot \Delta T,

where

  • c is the specific heat capacity of the material,
  • m is the mass of the sample, and
  • \Delta T is the change in the temperature of this sample.

For the first part of energy input, c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}.

Similarly, for the third part of energy input, c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}.

The second part of energy input requires a different equation. The energy Q required to melt a sample of mass m and latent heat of fusion L_\text{f} is:

Q = m \cdot L_\text{f}.

Apply this equation to find the size of the second part of energy input:

\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}.

Find the sum of these three parts of energy:

\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}.

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dimension = 30.0 m ✕ 15.0 m ✕ 5.0 m.

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  • The definition of Density is the amount of matter in a given space, or volume
  • Density = mass/volume
  • units for density kg/m^3
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Answer:

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