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PtichkaEL [24]
3 years ago
8

a motorcycle begins at rest and accelerates uniformly at 7.9 m/s2. We want to find the time it takes the motorcycle to reach a s

peed of 100 km/h. Which kinetic formula would be most useful to solve for the target unknown?
Physics
2 answers:
gladu [14]3 years ago
6 0

Answer:

It takes the motorcycle about 12.65 seconds to reach 100mph

Explanation:

AURORKA [14]3 years ago
3 0

Answer:3.5 seconds

Explanation:

initial velocity=u=0

Final velocity=v=100km/h

Final velocity=v=(100x5)/18=500/18=27.8 m/s

Acceleration=a=7.9m/s^2

t=(v-u)/a

t=(27.8-0)/7.9

t=27.8/7.9

t=3.5 seconds

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How many neutrons are contained in 2 kg? Mass of one neutron is 1.67x10-27 kg.
liraira [26]

Answer:

1.2 × 10^27 neutrons

Explanation:

If one neutron = 1.67 × 10^-27 kg

then in 2kg...the number of neutrons

; 2 ÷ 1.67 × 10^-27

There are.... 1.2 × 10^27 neutrons

4 0
3 years ago
What type of system uses a liquid to transmit a force
tankabanditka [31]
It can be transmit in hydraulic machine, exerting a small cross-sectional area can lead to pressure being transmitted
3 0
3 years ago
A spacecraft at rest has moment of inertia of 100 kg-m^2 about an axis of interest. If a 1 newton thruster with a 1 meter moment
snow_lady [41]

Answer:

18 radians

Explanation:

The computation is shown below:

As we know that

Torque = Force × Moment arm

= 1N × 1M

= 1N-M

Torque = I\alpha

\alpha = \frac{torque}{I}\\\\= \frac{1}{100}\\\\= 0.01 rad/s^2

Now

\theta = w_ot + \frac{1}{2} \alpha t^2\\\\w_o = 0\\\\\theta = 0 \times 60 + \frac{1}{2} \times 0.01 \times 60^2\\\\= 18\ radians

Here t = 1 minutes = 60 seconds

3 0
3 years ago
An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field
lawyer [7]

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = \frac{1}{2} mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = \frac{1}{2}  \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             K_e = K_p

              K_p = ½ m v_e²

              K_p = \frac{1}{2}  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = \frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = \frac{7 \ 10^4 }{ 3 \ 10^8}

        v/c= 2.33 10⁻⁴

8 0
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