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2 years ago

What is the momentum of a 35–kilogram cart moving at a speed of 1.2 meters/second?

Physics

1 answer:

Alexus [3.1K]2 years ago

6 0

Given:-

Mass of the cart (m) = 35 kg
Speed (consider Velocity) = 1.2 m/s

To Find: Momentum of the cart.

We know,

p = mv

where,

p = Momentum,
m = Mass &
v = Velocity.

Thus,

p = (35 kg)(1.2 m/s)

→ p = 42 kg m/s (Ans.)

Conclusion:-

A. ☑️ 42 kilogram - metre per second.

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A wave emitted from a source has a frequency of 10 Hz and wavelength 2.5 m. How much time will it take to reach a person located

const2013 [10]

Answer:

time taken by the wave to reach the person is 0.2 s

Explanation:

As we know that the speed of the wave is given as

$v = \lambda f$

here we know that the wavelength of the wave is

$\lambda = 2.5 m$

$f = 10 Hz$

now speed of the wave is given as

$v = 10(2.5)$

$v = 25 m/s$

Now time taken by the wave to reach 5 m distance is

$t = \frac{L}{v}$

$t = \frac{5}{25}$

$t = 0.2 s$

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2 years ago

What speed would a fly with a mass of 0.55g need in order to have a kinetic energy of 7.6 •10^4 j?

masya89 [10]

Answer:

16613 m/s

Explanation:

Given that

mass of the fly, m = 0.55 g = 0.55*10^-3 kg

Kinetic Energy of the fly, E = 7.6*10^4 J

Speed of the fly, v = ? m/s

We know that the Kinetic Energy is that energy that an object, in this case, the fly, possesses due to its motion.

The Kinetic Energy, KE of any object is represented by the formula

KE = 1/2 * m * v²

If we substitute the values in the relation, we have,

7.6*10^4 = 1/2 * 0.55*10^-3 * v²

v² = (15.2*10^4) / 0.55*10^-3

v² = 2.76*10^8

v = √2.76*10^8

v = 16613 m/s

Thus, the fly would need a speed of 16.6 km/s in order to have a Kinetic Energy of 7.6*10^4 J

7 0

3 years ago

How much energy to be removed to drop the temperature of 5.7kg of wood from 20degrees to 7degrees. #100points

Neporo4naja [7]

see i was trying to figure out the answer but i didn't understand it so i took the time to research and work it out but i still didn't understand i found one that was close to it and i got the same one as the other person which is D but idk if it is that type of question if it is than it is d if not then idk

5 0

3 years ago

Read 2 more answers

2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo

seropon [69]

Answer:

The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times60$

$\omega=376.9\ rad/s$

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

$B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}$

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

$B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}$