Ask question

Ask question

All categories

2 years ago

11

A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed

is doubled, what happens to the tension in the string?

1 answer:

Liono4ka [1.6K]2 years ago

8 0

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

$F = \dfrac{mv^2}{r}$

m is the mass, v is the velocity and r is the radius.

It follows that $F \propto v^2$ , provided m and r are constant.

When v is doubled, the new force, $F_1$ , is

$F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F$

Hence, the tension in the string is quadrupled.

You might be interested in

A boat moves through the sea.

sergiy2304 [10]

Answer:

dont you have to times it

Explanation:

4 0

2 years ago

The energy stored in a wooden log transforms when the log is burned. Which of the following explanations BEST describes how the

Makovka662 [10]

Answer:

D. the amount of chemical energy equals the amount of heat and light energy.

Explanation:

Given that the first law of thermodynamics affirmed that energy is neither created nor destroyed however, it can be transformed from one form to another. In other words, while, during the transformation of energy, no energy is lost, the input energy is also equal to output energy.

Hence, the chemical energy stored in the log is EQUAL to the heat and light energy produced by burning.

5 0

2 years ago

The position of an object that is oscillating on an ideal spring is given by the equation x=(12.3cm)cos[(1.26s−1)t]. (a) at time

Natali5045456 [20]

<span>x=((12.3/100)m)cos[(1.26s^−1)t]
v= dx/dt = -</span><span>((12.3/100)*1.26)sin[(1.26s^−1)t]
v=</span>-((12.3/100)*1.26)sin[(1.26s^−1)t]=-((12.3/100)*1.26)sin[(1.26s^−1)*(0.815)]
v=<span> <span>-0.13261622 m/s
</span></span>the object moving at 0.13 m/s <span>at time t=0.815 s</span>

6 0

2 years ago

Large and small numbers are often best entered in exponential notation. For example, the mass of the earth is 5.98x1024 kg and t

Doss [256]

Answer:

The charge to mass ratio is $-1.76\times10^{11}$

Explanation:

$Mass\ of\ electron=m_e=9.11\times10^{-31} kg\\ Charge\ of\ the\ electron=q_e=-1.60\times10^{-19} C\\$

We need to find how much charge is contained in the electron per unit of mass, to do this we divide the charge in an electron and the mass of an electron:

$\frac{Charge\ of\ electron}{Mass\ of\ electron}=\frac{q_e}{m_e}\frac{C}{kg}=\frac{-1.60\times10^{-19} C}{9.11\times10^{-31} kg}=-1.76\times10^{11} \frac{C}{kg}$

4 0

2 years ago

Read 2 more answers

A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik

Lady bird [3.3K]

Answer:

The options are approximations of the exact answers:

A) $1\times10^6N/C$

B) $2\times10^{-13}N$

C) $1\times10^{-2}V$

D) Toward the inner wall

E) $3\times10^{-17}J$

Explanation:

A) The electric field in a parallel plate capacitor is given by the formula $E=\frac{\sigma}{k\epsilon_0}$ , where $\epsilon_0=8.85\times10^{-12}C/Vm$ and in our case $\sigma=10^5C/m^2$ and, for air, $k=1.00059$ , so we have:

$E=\frac{10^5C/m^2}{(1.00059)(8.85\times10^{-12}C/Vm)}=1.13\times10^6N/C$

B) The K+ ion has one elemental charge excess, so its charge is $q=1.6\times10^{-19}C$ , and the force a charge experiments under an electric field E is given by F=qE, so we have:

$F=(1.6\times10^{-19}C)(1.13\times10^6N/C)=1.81\times10^{-13}N$

C) The potential difference between two points separated a distance d under an uniform electric potential E is given by $\Delta V=dE$ , so we have:

$\Delta V=(10\times10^{-9}m)(1.13\times10^6N/C)=1.13\times10^{-2}V$

D) The electic field goes from positive to negative charges, so it goes towards the inner wall.

E) The work done by an electric field through a potential difference $\Delta V$ on a charge Q is $W=Q\Delta V$ , and is equal to the kinetic energy imparted on it, so we have:

$K=(3\times10^{-15}C)(1.13\times10^{-2}V)=3.39\times10^{-17}J$

5 0

2 years ago