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laila [671]
3 years ago
13

How much work is done ON THE BOOK witih mass of 10 kg that is being pushed across a 2m table at 5m/s?

Physics
1 answer:
ruslelena [56]3 years ago
5 0
I don’t know that topic
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What is the definition of density?​
Luba_88 [7]

Answer:

Density is the amount of mass in a specified space. It is a way to measure how compact an object is

Explanation:

6 0
3 years ago
A bumblebee darts past at 3 m/s. The frequency of the hum made by its wings is 152 Hz. Assume the speed of sound to be 342 m/s.
Veronika [31]

It'll be 152 Hz at the exact instant the bumblebee
is right at the tip of your nose, on his way past you.

Before he gets there, while he's coming at you,
he sounds like a frequency higher than 152 Hz.

After he passes by, and is going away from you,
he sounds like a frequency lower than 152 Hz.

8 0
4 years ago
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Which depends on your location weight or mass
Zepler [3.9K]
I believe it would be weight. mass never changes.
6 0
3 years ago
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The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t
jek_recluse [69]

Answer:

e=3367.2J

%e=1.43%

Explanation:

From the exercise we know two information. The real speed and the experimental measured by the speedometer

v_{r}=10km/h=2.77m/s

Since the speedometer is only accurate to within 0.1km/h the experimental speed is

v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s

Knowing that we can calculate Kinetic energy for the real and experimental speed

E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J

E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J

Now, the potential error in her calculated kinetic energy is:

e=E_{r}-E_{e}=(234023-230656)J=3367.2J

%e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43%

4 0
4 years ago
A layer of oil (n = 1.38) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid.
Vinil7 [7]

Answer:

Refractive index of unknown liquid = 1.56

Explanation:

Using Snell's law as:

n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}

Where,  

{\theta_i}  is the angle of incidence  ( 65.0° )

{\theta_r} is the angle of refraction  ( 53.0° )

{n_r} is the refractive index of the refraction medium  (unknown liquid, n=?)

{n_i} is the refractive index of the incidence medium (oil, n=1.38)

Hence,  

1.38\times {sin65.0^0}={n_r}\times{sin53.0^0}

Solving for {n_r},

Refractive index of unknown liquid = 1.56

4 0
4 years ago
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