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Alexeev081 [22]
2 years ago
9

Date: 12-3-21 4. The momentum of a 5-kilogram object moving at 6 meters per second is

Physics
2 answers:
Dafna11 [192]2 years ago
6 0
30 kg•m/s

p = mv
p = 5kg • 6m/s
p = 30 kg•m/s
ASHA 777 [7]2 years ago
5 0

Answer

30 kg . m/sec

Explanation:

mark me as brainliest!

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If the mass of the object is doubled and the speed is halved then kinetic energy will change by a factor of:
Orlov [11]
I could be wrong but I believe it’s 1/2
6 0
3 years ago
During strengthening heat treatment, the _______ step traps the material in an unstable crystalline structure. a)-Quenching, b)-
ZanzabumX [31]

Answer: A) Quenching

Hope this helps

4 0
3 years ago
Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce
lutik1710 [3]

Answer:

speed of electrons = 3.25 × 10^{7} m/s

acceleration in term g is 3.9 × 10^{17} g.

radius of circular orbit is 2.76 × 10^{-4} m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = \sqrt{\frac{2qV}{m}}

v = \sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}

v = 3.25 × 10^{7} m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = \frac{Bqv}{m}  

a =  \frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}

a = 3.82 × 10^{18} m/s²

and acceleration in term g

a = \frac{3.82\times 10^{18}}{9.81}  

a = 3.9 × 10^{17} g

acceleration in term g is 3.9 × 10^{17} g.

and

electron moving in circular orbit has centripetal force

F = \frac{mv^2}{r}  

Bqv = \frac{mv^2}{r}  

r = \frac{mv}{Bq}  

r = \frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}  

r = 2.76 × 10^{-4} m

radius of circular orbit is 2.76 × 10^{-4} m

8 0
3 years ago
If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
Bas_tet [7]

Answer:

  W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

              F = G m₁ M / r²

              W = ∫ F. dr

              W = G m₁ M ∫ dr / r²

we integrate

             W = G m₁ M (-1 / r)

                 

We evaluate between the limits, lower r = R_ Moon and r = ∞

           W = -G m₁ M (1 /∞ - 1 / R_moon)

            W = G m1 M / r_moon

Body weight is

             W = mg

             m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

            m = 1000/32 = 165 / g_moon

            g_moom = 165 32/1000

            .g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system  

           W_capsule = 1000 pound (1 kg / 2.20 pounds)

           W_capsule = 454 N

           W = m_capsule g

           m_capsule = W / g

           m = 454 /9.8

           m_capsule = 46,327 kg

Let's calculate

          W = 6.67 10⁻¹¹ 46,327   7.36 10²² / 1.74 10⁶

          W = 1,307 10⁶ J

7 0
3 years ago
A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightl
USPshnik [31]

Given Information:  

Wavelength =  λ = 39.1 cm = 0.391 m

speed of sound = v = 344 m/s

linear density = μ = 0.660 g/m = 0.00066 kg/m

tension = T = 160 N

Required Information:

Length of the vibrating string = L = ?

Answer:

Length of the vibrating string = 0.28 m

Explanation:

The frequency of beautiful note is

f = v/λ

f = 344/0.391

f = 879.79 Hz

As we know, the speed of the wave is

v = √T/μ

v = √160/0.00066

v = 492.36 m/s

The wavelength of the string is

λ = v/f

λ = 492.36/879.79

λ = 0.5596 m

and finally the length of the vibrating string is

λ = 2L

L = λ/2

L = 0.5596/2

L = 0.28 m

Therefore, the vibrating section of the violin string is 0.28 m long.

3 0
2 years ago
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