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andrezito [222]
3 years ago
9

The blades in a blender rotate at a rate of 6800 rpm . When the motor is turned off during operation, the blades slow to rest in

2.8 s .
Physics
1 answer:
tangare [24]3 years ago
5 0
The angular acceleration of the blade when it's switched off is (-6800 rev/min) divided by (2.8 sec) = -2,428.6 rev/(min-sec) = -40.5 rev/sec^2 .
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Which of the following accurately shows how to calculate the weight of a 20kg object
forsale [732]
The answer to your question is "20kgx9.8m/s" because weight is the force an object is exerting on another object, and the formula used to calculate force is <em>Force = Mass * Acceleration</em>.
4 0
3 years ago
Read 2 more answers
A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li
Alexeev081 [22]

Note: if the professor wants the distance between the m = 0 and m = 1 maxima to be 25 cm

Answer:

d = 1.0128×10⁻⁵m

Explanation:

given:

length L = 4.0m

maximum distance between m = 0 and m = 1 , y = 25cm = 0.25m

wavelength λ = 633nm = 633×10⁻⁹m

note:

dsinθ = mλ (constructive interference)

where d is slit seperation, θ is angle of seperation , m is order of interference , and λ is wavelength

for small angle

sinθ ≈ tanθ

d (\frac{y}{L})  = mλ

d (\frac{y}{L}) = (1)(633nm)

d(\frac{0.25}{4} ) = (1)(633nm)

d = 1.0128×10⁻⁵m

6 0
3 years ago
Richard is driving home to visit his parents. 150 mi of the trip are on the interstate highway where the speed limit is 65 mph .
Serggg [28]

Answer:

t = 25.5 min

Explanation:

To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.

t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h

Next, you calculate the difference between both times t1 and t2:

\Delta t=t_1-t_2=2.30h-1.875h=0.425h

This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:

0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s

hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph

3 0
3 years ago
4 . A negative charge -0.450 exerts an upward 0.150N force on an unknown charge 0.250m directly below it . ( a ) what is the unk
Elodia [21]

(a) The unknown charge kept at distance 0.250m directly below  -0.450 charge is  2.315 x 10⁻¹² C.

(b) The magnitude of the force is 29.16 x 10¹² N.

<h3>What is electrostatic force?</h3>

The electrostatic force F between two charged objects placed distance apart is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

F = kq₁q₂/r²

where k = 9 x 10⁹ N.m²/C²

Given is a negative charge -0.450 exerts an upward 0.150N force on an unknown charge 0.250m directly below it .

In equation of the magnitude of force, only the magnitude of charges are taken.

Substituting the values into the expression ,we get the charge

0.150 = (9 x 10⁹ x 0.450x q)/ (0.25)²

q = 2.315 x 10⁻¹² C

Thus, the magnitude of the unknown charge is 2.315 x 10⁻¹² C.

b)Let the unknown charge be  0.450C to exert on -450C.

Substituting the values into the expression ,we get the electrostatic force

F = kq₁q₂/r²

F = (9 x 10⁹ x 0.450x 0.450)/ (0.25)²

q = 2.315 x 10⁻¹² C

Thus, the magnitude of the force is 29.16 x 10¹² N.

Learn more about electrostatic force.

brainly.com/question/9774180

#SPJ1

8 0
2 years ago
A solid sphere with a radius of 50.0 cm has a total positive charge of 40.0 μC uniformly distributed in its volume. Calculate th
il63 [147K]

Answer:

E=2.88\times 10^5\ N/C                              

Explanation:

It is given that,

The radius of the solid sphere, R = 50 cm = 0.5 m

Charge on the sphere, q=40\ \mu C=40\times 10^{-6}\ C

We need to find the magnitude of the electric field r = 10.0 cm away from the center of the sphere. The electric field at point r away form the center of the sphere is given by :

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

E=\dfrac{9\times 10^9\times 40\times 10^{-6}\times 0.1}{0.5^3}

E=2.88\times 10^5\ N/C

So, the electric field 10.0 cm away from the center of the sphere is 2.88\times 10^5\ N/C. Hence, this is the required solution.              

7 0
3 years ago
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