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UkoKoshka [18]
3 years ago
12

A 2.50-kg solid, uniform disk rolls without slipping across a level surface, translating at 3.75 m/s. If the disk’s radius is 0.

100 m, find its (a) translational kinetic energy and (b) rotational kinetic energy.
Physics
1 answer:
Gennadij [26K]3 years ago
4 0

Answer

given,

mass of the solid = 2.50 Kg

speed of translating disk = 3.75 m/s

radius of the disk = 0.1 m

a) transnational Kinetic energy

    KE = \dfrac{1}{2}mv^2

    KE = \dfrac{1}{2}\times 2.5 \times 3.75^2

    KE =17.58\ J

b)  rotational kinetic energy.

    KE = \dfrac{1}{2}I\omega^2

     for disk

    I = \dfrac{1}{2}mr^2       and v = rω

    KE = \dfrac{1}{2}( \dfrac{1}{2}mr^2)(\dfrac{v}{r})^2

    KE = \dfrac{1}{4}mv^2

    KE = \dfrac{1}{4}\times 2.5 \times 3.75^2

    KE =8.79\ J

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Answer:

A.) 42.7 m/s

B.) 0.33 m/s^2

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Explanation:

A.) If Justin races his Chevy S-10 down highway 37 north for 2,560 meters in 60 seconds, what is his velocity? 

Velocity = displacement/time

Velocity = 2560/60

Velocity = 42.67 m/s

B.) The Chevy S-10 started rounding at 10 meters per hour. What is the acceleration at 30 seconds on the highway?

Acceleration = velocity/time

Acceleration = 10/30

Acceleration = 0.33 m/s^2

C.) The S-10 has a force of 30 N. What is the mass of the car?

Force = mass × acceleration

30 = mass × 0.33

Mass = 30/ 0.33

Mass = 90 kg

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NEED HELP ASAPTsunamis are often called tidal waves because they are caused by tides.
garik1379 [7]

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false

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A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
3 years ago
An airplane weighing 11,000 N climbs to a
Gennadij [26K]

The power in horsepower is 40.1 hp

Explanation:

We start by calculating the work done by the airplane during the climb, which is equal to its change in gravitational potential energy:

W=(mg)\Delta h

where

mg = 11,000 N is the weight of the airplane

\Delta h = 1.6 km = 1600 m is the change in height

Substituting,

W=(11,000)(1600)=17.6\cdot 10^6 J

Now we can calculate the power delivered, which is given by

P=\frac{W}{t}

where

W=17.6\cdot 10^6 J is the work done

t=9.8 min \cdot 60 = 588 s is the time taken

Substituting,

P=\frac{17.6\cdot 10^6 J}{588}=2.99\cdot 10^4 W

Finally, we can convert the power into horsepower (hp), keeping in mind that

1 hp = 746 W

Therefore,

P=\frac{2.99\cdot 10^4}{746}=40.1 hp

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

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