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Oksi-84 [34.3K]

2 years ago

9

A small mass m is tied to a string of length L and is whirled in vertical circular motion. The speed of the mass v is such that

the ratio of the string tension at the top of the circle to that at the bottom of the circle is FtopT/FbotT = 0.5. Derive an expression for the speed v.

1 answer:

adell [148]2 years ago

7 0

Answer:

(mv^2/R)/(mg)=1/2

v^2=R/2g

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What is the effect of the following change on the volume of 1 mol of an ideal gas? The initial pressure is 722 torr, the final p

Ray Of Light [21]

Answer:

A. Volume is unchanged

Explanation:

$P_{i}$ = initial pressure of the gas = 722 torr = 96258.7 pa

$P_{f}$ = final pressure of the gas = 0.950 atm = 96258.75 pa

$T_{i}$ = initial temperature = 32 °F = 272.15 K

$T_{f}$ = final temperature = 273 K

$V_{i}$ = initial volume

$V_{f}$ = final volume

Using the Equation

$\frac{P_{i} V_{i}}{T_{i}} = \frac{P_{f} V_{f}}{T_{f}}$

Inserting the values

$\frac{(96958.7) V_{i}}{272.15} = \frac{(96958.75) V_{f}}{273}$

$V_{f} = (1.00312) V_{i}$

Hence the volume is unchanged.

4 0

2 years ago

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

$v=18.9 m/s$

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0

2 years ago

Physics Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

djverab [1.8K]

Good work on solving part a).
b) may look complicated, but it's not too bad.

It says that the body is 25% efficient in converting fat to mechanical energy.
In other words, only 25% of the energy we get from our stored fat shows up
in the physical, mechanical moving around that we do. (The rest becomes
heat, which dissipates into the environment as we keep our bodies warm,
breathe hot air out,and perspire.)

You already know how much mechanical energy the climber needed to lift
himself to the top of the mountain... 2.4x10⁶ joules.
That's 25% of what he needs to convert in order to accomplish the climb.
He needs to pull 4 times as much energy out of fat.

-- Fat energy required = 4 x (2.4 x 10⁶) = 9.6 x 10⁶ joules.

-- Amount stored in 1kg of fat = 3.8 x 10⁷ joules

-- Portion of a kilogram he needs to use = (9.6 x 10⁶) / (3.8 x 10⁷)

Note:
That much of a kilogram weighs about 8.9 ounces ... which shows why it's so
hard to lose weight with physical exercise alone. It also helps you appreciate
that fat is much more efficient at storing energy than batteries are ... that one
kilogram of fat stores the amount of energy used by a 100-watt light bulb, to
burn for 105 hours (more than 4-1/2 days ! ! !)

5 0

2 years ago

Read 2 more answers

1) My 14V car battery could be used to charge my laptop, but I need to use an inverter to first convert it to a standard 120V. T

LenaWriter [7]

Answer:

1) Charge chord resistance is 75 Ω

2) Charge chord resistance is 6.33 Ω

Explanation:

1) To answer the question, we note that the the formula voltage is found as follows;

V = IR

Therefore,

$R = \frac{V}{I} = \frac{120}{1.6} = 75 \, \Omega$

2) Where the voltage, V = 19.5 V and the current, I = 3.33 A, we have;

Initial resistance R₁ = 19.5 V/(3.33 A) = 5.86 Ω

However, to reduce the current to 1.6 A, we have;

$R_T = \frac{19.5}{1.6} = 12.1875 \ \Omega$

Therefore, where the resistance is found by the sum of the total resistance we have;

$R_T$ = R₁ + Charge chord resistance

∴ 12.1875 = 5.86 + Charge chord resistance

Hence, charge chord resistance = 6.33 Ω

8 0

2 years ago

When a pendulum is pulled back from its equilibrium position by 10∘, the restoring force is 1.0 N. When it is pulled back to 30∘

jarptica [38.1K]

Answer: B

Explanation: I said B because if you pull something back what is going to be more of a force pulling back or letting it go for a rubier band yes it will have more force if you let it go

5 0

2 years ago

Read 2 more answers