Ask question

Ask question

All categories

2 years ago

14

A student is bouncing on a trampoline. at her highest point, her feet are 65 cm above the trampoline. when she lands, the trampo

line sags 15 cm before propelling her back up. part a for how long is she in contact with the trampoline?

1 answer:

Ede4ka [16]2 years ago

3 0

<span>she is in contact with the trampoline for less than one second.</span>

You might be interested in

A 0.10 kg piece of copper at an initial temperature of 95°c is dropped into 0.20 kg of water contained in a 0.28 kg aluminum cal

DedPeter [7]

<span>(cp of Copper = 387J / kg times degrees C; cp of Aluminum = 899 J / kg times degrees C; cp of Water = 4186J / kg times degrees C)
</span> Use the law of conservation of energy and assuming no heat loss to the surroundings, then
<span>Heat given up by copper = heat absorbed by water + heat absorbed by calorimeter
</span><span> Working formula is
</span> <span>Q = heat = MCp(delta T)
</span><span> where
</span><span> M = mass of the substance
</span><span> Cp = specific heat of the substance
</span><span> delta T = change in temperature
</span> Heat given up by copper = 0.10(387)(95 - T)
<span> Heat absorbed by water = 0.20(4186)(T - 15)
</span><span> Heat absorbed by calorimeter = 0.28(899)(T - 15)
</span> where
<span> T = final temperature of the system
</span><span> Substituting appropriate values,

</span> 0.10(387)(95 - T) = 0.20(4186)(T - 15) + 0.28(899)(T - 15)
<span> 38.7(95 - T) = 1088.92(T - 15)
</span><span> 3676.50 - 38.7T = 1088.92T - 16333.8
</span><span>1127.62T = 20010.3
</span><span> T = 17.75 C </span>

8 0

2 years ago

Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

AysviL [449]

The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
$E= \frac{\sigma}{2\epsilon_0}$
where
$\sigma$ is the charge density
$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

7 0

2 years ago

A typical human contains 5.00 l of blood, and it takes 1.00 min for all of it to pass through the heart when the person is resti

Mrrafil [7]

<span>(a) 0.0676 l (b) 67.6 cc So we've been told that 5.00 L of blood flows through the heart every minute and that the heart beats 74.0 times per minute. So that means that for every beat of the heart, 5.00 L / 74.0 = 0.067567568 L of blood flows through the heart. Rounding to 3 significant figures gives 0.0676 l. Converting from liters to cubic centimeters simply require a multiplication by 1000, so we have 67.6 cc of blood pumped per beat.</span>

5 0

2 years ago

Read 2 more answers

Umar has two copper pans, each containing 500cm3 of water. Pan A has a mass of 750g and pan B has a mass of 1.5kg. Which pan wil

Olin [163]

Answer:

heat required in pan B is more than pan A

Explanation:

Heat required to raise the temperature of the substance is given by the formula

$Q = ms\Delta T$

now we know that both pan contains same volume of water while the mass of pan is different

So here heat required to raise the temperature of water in Pan A is given as

$Q_1 = (m_w s_w + m_ps_p)\delta T$

$Q_1 = (0.5(4186) + 0.750(s))\Delta T$

Now similarly for other pan we have

$Q_2 = (m_w s_w + m_ps_p)\delta T$

$Q_2 = (0.5(4186) + 1.50(s))\Delta T$

So here by comparing the two equations we can say that heat required in pan B is more than pan A

3 0

2 years ago

A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t

Mashutka [201]

As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."

Now the general formula of the Doppler's effect is:
$f = (\frac{g + v_{r}}{g + v_{s}})f_{o}$ -- (A)

Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
g = Speed of sound = 340m/s.
$v_{r}$ = Velocity of the receiver/observer relative to the medium = ?.
$v_{s}$ = Velocity of the source with respect to medium = 0 m/s.
$f_{o}$ = Frequency emitted from source = 400 Hz.
$f$ = Observed frequency = 408Hz.

Plug-in the above values in the equation (A), you would get:

$408 = ( \frac{340 + v_{r}}{340 + 0})*400$

$\frac{408}{400} = \frac{340 + v_{r}}{340}$

Solving above would give you,
$v_{r}$ = 6.8 m/s

The correct answer = 6.8m/s

7 0

2 years ago