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Igoryamba
2 years ago
3

Read each scenario and then answer the question. Scenario A: A 7 N bucket is raised a distance of 6 m. Scenario B: A 10 N bucket

is raised a distance of 4 m. Scenario C: A 12 N bucket is raised a distance of 3 m. Which scenario requires the most work?
Physics
2 answers:
Bad White [126]2 years ago
8 0
The work done is equal to the weight (mg) of the bucket times the altitude at which it is raised, h:
W=mgh
Let's calculate the work for the 3 scenarios:
W_A = (7 N)(6 m)=42 J
W_B=(10 N)(4 m)=40 J
W_C=(12 N)(3 m)=36 J

So, the scenario which requires more work is scenario A.
zepelin [54]2 years ago
3 0

Answer: Scenario A

Explanation: i got it correct on edge

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When a car is 100 meters from its starting position traveling at 60.0 m/s., it starts braking and comes to a stop 350 meters fro
NISA [10]
Remember your kinematic equations for constant acceleration. One of the equations is x_{f} =  x_{i} +  v_{i}(t) + \frac{1}{2} at^{2}, where x_{f} = final position, x_{i} = initial position, v_{i} = initial velocity, t = time, and a = acceleration. 

Your initial position is where you initially were before you braked. That means x_{i} = 100m. You final position is where you ended up after t seconds passed, so x_{f} = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was v_{i} = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
350\:m = 100\:m + (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = 498\:m +34.445\:s^{2}(a)\\
-248\:m = 34.445\:s^{2}(a)\\
a \approx -7.2 \: m/s^{2}

Your acceleration is approximately -7.2 \: m/s^{2}.
4 0
2 years ago
A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end
Triss [41]

Answer: m= 35.6 kg

Explanation:

For finding the mass of the stone we have the formula

v= \sqrt{\frac{Tension}{Linear. Mass. density} }

Here, Tension= m*g = m*9.81

and linear mass density= \frac{8.25 g}{65 cm}

Linear mass density= \frac{8.25*10^-3}{65*10^-2}

Linear mass density= 0.0127 kg/m

Velocity= 2*\frac{l}{t}

Velocity= 2 * \frac{65*10^-2}{7.84}

Velocity= 165.8 m/s

So putting all these values in equation we get

v= \sqrt{\frac{Tension}{Linear. Mass. density} }

165.8= \sqrt{\frac{m*9.81}{0.0127} }

Solving we get

m= 35.58 kg

or m= 35.6 kg

3 0
2 years ago
Two wires are stretched between two fixed supports and have the same length. One wire A there is a second-harmonic standing wave
lina2011 [118]

(a) Greater

The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, f_1:

f_n = n f_1

So we have:

- On wire A, the second-harmonic has frequency of f_2 = 660 Hz, so the fundamental frequency is:

f_1 = \frac{f_2}{2}=\frac{660 Hz}{2}=330 Hz

- On wire B, the third-harmonic has frequency of f_3 = 660 Hz, so the fundamental frequency is

f_1 = \frac{f_3}{3}=\frac{660 Hz}{3}=220 Hz

So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

(b) f_1 = \frac{v}{2L}

For standing waves on a string, the fundamental frequency is given by the formula:

f_1 = \frac{v}{2L}

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

v=2Lf_1

We know that the two wires have same length L. For wire A, f_1 = 330 Hz, while for wave B, f_B = 220 Hz, so we can write the ratio between the speeds of the waves in the two wires:

\frac{v_A}{v_B}=\frac{2L(330 Hz)}{2L(220 Hz)}=\frac{3}{2}

So, the waves travel faster on wire A.

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2 years ago
What is the advantage of the SI unit over CGS units?
MatroZZZ [7]

Answer:

The advantage of the SI unit over CGS unit are:

  • S.I has broader base. It has seven base units and two supplementary units.
  • S.I is metric.
  • S.I is coherent.
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List some reasons why growth characteristics are more useful on agar plates than on agar slants
SpyIntel [72]
Usually, in culturing of the bacteria we have a slant and then portion f it is transferred to the agar plate. The growth characteristics are more useful in the agar plates because it is where we really do the observation because bacteria in slants are still to be transferred in the agar plates. 
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2 years ago
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