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weqwewe [10]
3 years ago
7

Calculate the work done by applying a 125 N force to the right on a 65.0 kg sled to displace

Physics
1 answer:
Elanso [62]3 years ago
5 0
W=fd
W= (125N) (3.5m)
W= 437.5 J
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An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height. The bullet remai
Vsevolod [243]
<span>79.75m/s  .................................</span>
5 0
3 years ago
Describe how the mass, luminosity, surface temperature, and radius of main-sequence stars change in value going from the “bottom
Sladkaya [172]

Answer:

1. Least massive stars are the coolest and least luminous, lower right of main sequence, on HR diagram.  

2. Most massive are the hottest and most luminous, upper left of main sequence on Hr Diagram.  

3. The radius of stars are related to their sprectral type. having the O being the hottest upper left and M being the coolest bottom right.

4 0
3 years ago
Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.
AVprozaik [17]

The speed of a proton after it accelerates from rest through a potential difference of 350 V is 25.86 \times 10^4 ~m/s.

Initial velocity of the proton u = 0

Given potential difference \Delta V = 350V

let's assume that the speed of the proton is v,

Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge q when accelerated with a potential difference \Delta V is,

    U = q \Delta V

Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy  P.E must be equal to gain in Kinetic Energy K.E</em> i.e

\Delta K = \Delta V

If the initial and final velocity of the proton is u and v respectively then,

change in Kinetic Energy  \implies  \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0

change in Potential Energy \implies \Delta U = q\Delta V

from conservation of energy,

             v= \sqrt{\frac{2q\Delta V}{m}}

so,         v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}

                = 25.86 \times 10^4 ~m/s

To read more about the conservation of energy, please go to brainly.com/question/14668053

7 0
2 years ago
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
What is a element that tends to be shiny, easy shaped, and a good conductor of heat?
ozzi
The answer to your question is Metal
8 0
3 years ago
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