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ryzh [129]
2 years ago
10

The most useful unit of measurement for measuring the thickness of a side of a lab beaker would be a(n)

Physics
1 answer:
andrew11 [14]2 years ago
3 0
The answer is letter c

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A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at
Elenna [48]

The tension in the cable is 23.2 N

<h3>What is the tension in the string?</h3>

The tension in the cable can be resolved into horizontal and vertical forces Tcosθ and Tsinθ respectively.

Tcosθ, is acting perpendicularly, Tcosθ = 0

Taking moments about the pivot:

Tsinθ * 2.2 = 4 * 9.8 * 0.7

Solving for θ;

θ = tan⁻¹(1.4/2.2) = 32.5°

T = 27.44/(sin 32.5 * 2.2)

T = 23.2 N

In conclusion, the tension in the cable is determined by taking moments about the pivot.

Learn more about moments of forces at: brainly.com/question/23826701

#SPJ1

3 0
1 year ago
Read the scenario and solve these two problems.
Burka [1]

Answers:

a) 5400000 J

b) 45.92 m

Explanation:

a) The kinetic energy K of an object is given by:

K=\frac{1}{2}mV^{2}

Where:

m=12000 kg is the mass of the train

V=30 m/s is the speed of the train

Solving the equation:

K=\frac{1}{2}(12000 kg)(30 m/s)^{2}

K=5400000 J This is the train's kinetic energy at its top speed

b) Now, according to the Conservation of Energy Law, the total initial energy is equal to the total final energy:

E_{i}=E_{f}

K_{i}+P_{i}=K_{f}+P_{f}

Where:

K_{i}=5400000 J is the train's initial kinetic energy

P_{i}=0 J is the train's initial potential energy

K_{f}=0 J is the train's final kinetic energy

P_{f}=mgh is the train's final potential energy, where g=9.8 m/s^{2} is the acceleration due gravity and h is the height.

Rewriting the equation with the given values:

5400000 J=(12000 kg)(9.8 m/s^{2})h

Finding h:

h=45.918 m \approx 45.92 m

7 0
2 years ago
Read 2 more answers
When is the velocity of a mass on a spring at its maximum value?
ehidna [41]

Answer:

A.  when the mass has a displacement of zero

Explanation:

The velocity of a mass on a spring can be calculated by using the law of conservation of energy. In fact, the total energy of the mass-spring system is equal to the sum of the elastic potential energy (U) of the spring and the kinetic energy (K) of the mass:

E=U+K=\frac{1}{2}kx^2 + \frac{1}{2}mv^2

where

k is the spring constant

x is the displacement of the mass with respect to the equilibrium position of the spring

m is the mass

v is the velocity of the mass

Since the total energy E must remain constant, we can notice the following:

- When the displacement is zero (x=0), the velocity must be maximum, because U=0 so K is maximum

- When the displacement is maximum, the velocity must be minimum (zero), because U is maximum and K=0

Based on these observations, we can conclude that the velocity of the mass is at its maximum value when the displacement is zero, so the correct option is A.


8 0
3 years ago
Read 2 more answers
A sound is recorded at 19 decibels. What is the intensity of the sound?
sp2606 [1]

1 \times 10^{-10.1} \mathrm{Wm}^{-2} is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about 1 \times 10^{-12} \mathrm{Wm}^{-2}). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

                     \text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)

Where,

I = Intensity of the sound produced

I_{0} = Standard Intensity of sound of 60 decibels = 1 \times 10^{-12} \mathrm{Wm}^{-2}

So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

5 0
3 years ago
How many significant figures?<br> 5.0001<br> O None of these are correct<br> O 5<br> 02<br> 0 1
mezya [45]

5

if zero falls between two significant numbers it becomes significant.

6 0
3 years ago
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