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hammer [34]
3 years ago
11

Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular

acceleration of 0.311 rad/s2 for 4.13 s. What is the drill's angular displacement during that time interval?
Physics
1 answer:
RUDIKE [14]3 years ago
6 0

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, \omega _i = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

You might be interested in
g Determine the magnitude of the electric field at the surface of a lead-208 nucleus, which contains 82 protons and 126 neutrons
Nookie1986 [14]

Answer:

Explanation:

Electric field at the surface of the the lead 208 = KQ/ R²

where K = 8.99 × 10⁹ Nm² /C²

Q ( total charge inside the nucleus) and e is the charge of a proton = Ne = 82 × 1.6 × 10⁻¹⁹ C = 1.312 × 10⁻¹⁷ C

V of the lead = 208 v of a proton assuming they both are sphere

4/3 πR³ =208( 4/3 πr³) where R is the radius of the sphere and r is the radius of the proton

R³ = 208 r³

R = ∛( 208 r³) = 5.92r

replace r with 1.20 x 10-15 m

R = 5.92 ×1.20 x 10-15 m = 7.11 × 10⁻¹⁵ m

E = ( 8.99 × 10⁹ Nm² /C² × 1.312 × 10⁻¹⁷ C ) / (7.11 × 10⁻¹⁵ m)² =  0.233 × 10²² N/C = 2.33 × 10²¹ N/C

4 0
3 years ago
Interestingly, there have been several studies using cadavers to determine the moment of inertia of human body parts by letting
loris [4]

Answer:

0.08735 kgm²

Explanation:

m = Mass of lower leg = 5 kg

L = Length of leg = 18 cm

g = Acceleration due to gravity = 9.81 m/s²

f = Frequency = 1.6 Hz

I = Moment of inertia

Time period is given by

T=2\pi\sqrt{\dfrac{I}{mgL}}

Also

T=\dfrac{1}{f}

So,

I=\dfrac{mgL}{(2\pi f)^2}\\\Rightarrow I=\dfrac{5\times 9.81\times 0.18}{(2\pi 1.6)^2}\\\Rightarrow I=0.08735\ kgm^2

The moment of inertia of the lower leg is 0.08735 kgm²

8 0
4 years ago
A car moving at 10.0 m/s encounters a depression in the road that has a circular cross-section with a radius of 30.0 m. What is
Dennis_Churaev [7]

Answer:

F = 789 Newton

Explanation:

Given that,

Speed of the car, v = 10 m/s

Radius of circular path, r = 30 m

Mass of the passenger, m = 60 kg

To find :

The normal force exerted by the seat of the car when the it is at the bottom of the depression.

Solution,

Normal force acting on the car at the bottom of the depression is the sum of centripetal force and its weight.

N=mg+\dfrac{mv^2}{r}

N=m(g+\dfrac{v^2}{r})

N=60\times (9.81+\dfrac{(10)^2}{30})

N = 788.6 Newton

N = 789 Newton

So, the normal force exerted by the seat of the car is 789 Newton.

6 0
3 years ago
Need some help with homework but where doing it in class rn
Brilliant_brown [7]

Answer:

now it lets me. B

Explanation:

hope that helps

6 0
3 years ago
Calculate the solubility (in m units) of ammonia gas in water at 298 k and a partial pressure of 2.50 bar . the henry's law cons
Eduardwww [97]
Henry's Law (formulated in 1803 by William Henry) states that aa constant temperature, the amount of gas dissolved in a liquid is directly proportional to the partial pressure exerted by that gas on the liquid.
 Mathematically it can be formulated as
 C = H⨯P
 being:
 C: the molar concentration of dissolved gas A,
 P: the partial pressure of it
 H: Henry's constant
 Substituting:
 C = P * H
 C = (2.50 * 0.9869) * 58.0
 C = 143.1
 Answer:
 the solubility (in m units) is
 C = 143.1
8 0
3 years ago
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