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larisa86 [58]
3 years ago
12

A turntable has an angular velocity of 3.5 rad/s. A dust bunny is on the disk of the turntable at a distance of 0.2m from the ce

nter of rotation. The coefficient of static friction required to keep the dust bunny from getting slung off is at leastA. 0.51B. 0.45C. 0.33D. 0.12E. 0.25
Physics
1 answer:
diamong [38]3 years ago
3 0

Answer:

E. 0.25

Explanation:

Given that

Angular speed ω = 3.5 rad/s

Distance ,r= 0.2 m

Lets take mass of dust bunny = m

We know that

Radial force F = m ω² r

The friction force on the dust bunny Fr

Fr= μ m g

To getting slung off  dust bunny from disk

m ω² r =  μ m g

ω² r =  μ  g

3.5²  x 0.2 =  μ x 10                        ( take g =10 m/s²)

μ = 0.245

μ = 0.25

Therefore answer is E

E. 0.25

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Diagnostic ultrasound of frequency 3.45 MHz is used to examine tumors in soft tissue.
Mekhanik [1.2K]

Answer:

4.35×10⁻⁴ m

Explanation:

(a)

From wave,

v = λf...................... Equation 1

Where v = speed of sound in air, λ = wavelength of sound, f = frequency of sound.

Make λ the subject of formula in  equation 1

λ = v/f................. Equation 2

Given: v = 343, f = 3.45 MHz = 3.45×10⁶ Hz

Substitute into equation 2

λ = 343/(3.45×10⁶)

λ = 99.42×10⁻⁶ m

λ = 9.942×10⁻⁵ m

(b)

using,

v' = λ'f............... Equation 3

Where v' = speed of sound in tissue, λ' = wavelength of sound in tissue.

make λ' the subject of the equation

λ' = v'/f......................Equation 4

Given: v' = 1500 m/s, f = 3.45 MHz = 3.45×10⁶ Hz

Substitute into equation 4

λ' = 1500/(3.45×10⁶)

λ' = 434.783×10⁻⁶

λ' ≈ 4.35×10⁻⁴ m.

6 0
3 years ago
TIME REMAINING
dmitriy555 [2]

Answer: d

Explanation:

I took the test

4 0
3 years ago
A car is driving away from a crosswalk. The formula d = t 2 + 2 t expresses the car's distance from the crosswalk in feet, d , i
Ede4ka [16]

Answer:

1) No, the car does not travel at constant speed.

2) V = 9 ft/s

3) No, the car does not travel at constant speed.

4) V = 5.9 ft/s

Explanation:

In order to know if the car is traveling at constant speed we need to derive the given formula. That way we get speed as a function of time:

V(t) = 2*t + 2   Since the speed depends on time, the speed is not constant at any time.

For the average speed we evaluate the formula for t=2 and t=5:

d(2) = 8 ft     and      d(5) = 35 ft

V_{2-5}=\frac{d(5)-d(2)}{5-2}=9 ft/s

Again, for the average speed we evaluate the formula for t=1.8 and t=2.1:

d(1.8) = 6.84 ft     and      d(2.1) = 8.61 ft

V_{1.8-2.1}=\frac{d(2.1)-d(1.8)}{2.1-1.8}=5.9 ft/s

4 0
4 years ago
A straight segment of wire 35.0 cm long carrying a current of 2.60 A is in a uniform magnetic field. The segment makes an angle
den301095 [7]

Answer:

Magnetic field, B = 0.275 T

Explanation:

Given that,

Length of the wire, L = 35 cm = 0.35 m

Current carried in the wire, I = 2.6 A

The segment makes an angle of 53∘ with the direction of the magnetic field, \theta=53^{\circ}

Magnetic force, F = 0.2 N

To find,

The magnitude of the magnetic field.

Solution,

The magnetic force acting on the wire is given by :

F=ILBsin\theta

\theta is the angle between the length of wire and the magnetic field.

0.2=2.6\times 0.35\times B\times sin(53)

B = 0.275 T

Therefore, the magnitude of the magnetic field is 0.275 T. Hence, this is the required solution.

5 0
3 years ago
Moving a greater ______ with a smaller _______ is the purpose of most machines. (fill the blanks asap please)
8_murik_8 [283]
Greater mass with a smaller force
5 0
4 years ago
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