How close does the proton get to the line of charge?

Copper has a specific heat of 0.386 J/g°C. How much heat is required to increase 5.00 g of copper from 0.0°C to 10.0°C?

Leto [7]

The answer is 19.3 j

3 0

3 years ago

A small boat coasts at constant speed under a bridge. A heavy sack of sand is dropped from the bridge onto the boat. The speed o

Tju [1.3M]

Answer:

d. decreases

Explanation:

The law of conservation of momentum tells us that the sum of momenta before the collision is equal to the sum of momenta after the collision. The bag has no momentum as it falls onto the boat because its velocity is zero in the horizontal direction. But after it hits the boat, it's momentum increases while the momentum of the system remains the same. That means a component of the system must decrease somewhere else. And that component is the velocity, not the mass, of the boat.

7 0

3 years ago

We want to find how much charge is on the electrons in a nickel coin. Follow this method. A nickel coin has a mass of about 4.2

monitta

Answer:

The number of atoms is $N = 4.37*10^{22} \ atoms$

Explanation:

From the question we are told that

The mass of coin is $m_n = 4.2g$

Number of atom in one mole = $n =6.02*10^{23} \ atoms$

Molar mass of nickel $M = 57.8g$

Now the relation to obtain the number of atom in the nickel coin is

$N = \frac{Mass \ of Nickel\ coin}{Molar \ mass\ of nickel } * No\ of\atoms \ in \ \ one\ mole\ of\ nickel$

$= \frac{4.2}{57.8}* 6.02*10^{23}$

$=4.37 *10^{22} atoms$

8 0

3 years ago

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5× $10^{-6}$ m.

Tension = 95.6N, Δx = 15.4× $10^{-5}$ m

Explanation: A speed of wave on a string under a tension force can be calculated as:

$|v| = \sqrt{\frac{F_{T}}{\mu} }$

$F_{T}$ is tension force (N)

μ is linear density (kg/m)

Determining velocity:

$|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{0.00874 }$

$|v| =$ 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

$\Delta x = |v|.t$

$\Delta x = 9.35.10^{-2}*1.23.10^{-3}$

Δx = 11.5× $10^{-6}$

With tension of 47.8N, a pulse will travel Δx = 11.5× $10^{-6}$ m.

Doubling Tension:

$|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{2.0.00874 }$

$|v| = \sqrt{0.01568}$

|v| = 0.1252 m/s

Displacement for same time:

$\Delta x = |v|.t$

$\Delta x = 12.52.10^{-2}*1.23.10^{-3}$

$\Delta x =$ 15.4× $10^{-5}$

With doubled tension, it travels $\Delta x =$ 15.4× $10^{-5}$ m

4 0

3 years ago

A golf ball (m = 59.2 g) is struck a blow that makes an angle of 50.5 ◦ with the horizontal. The drive lands 130 m away on a fla

fomenos

Answer:

The force is calculated as 338.66 N

Explanation:

We know that force is given by

$\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{m(v_{f}-v_{o})}{\Delta t}$

We know that range of a projectile is given by

$R=\frac{v_{o}^{2}sin(2\theta )}{g}$

it is given that R=130 m applying values in the above equation we get

$R=\frac{v_{o}^{2}sin(2\theta )}{g}\\\\v_{0}=\sqrt{\frac{Rg}{sin(2\theta )}}\\\\\therefore v_{o}=\sqrt{\frac{130\times 9.81}{sin(2\times 50.5_{o}} )}\\\\v_{o}=36.04m/s$

Thus the force is obtained as

$\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{59.2\times 10^{}-3(36.04-0)}{6.3\times 10^{-3}}$

Thus force equals $F=338.66N$

4 0

3 years ago

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