1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
den301095 [7]
2 years ago
6

An object with a mass of 32 kg has an initial energy of 500 J. At the end of the experiment, the velocity of the object is recor

ded as 5.1 m/s. If the object travelled 50 m to get to this point, what was the average force of friction on object during the trip? Assume no potential energy. Show all work
Physics
1 answer:
Nezavi [6.7K]2 years ago
3 0

Explanation:
the concept of conservation of the mechanical nerve
initial
Em₀ = 500 J
The energy is totally kinetic
Em₀ = K = ½ m v₀²
v₀ =
v₀ = √ (2 500/32)
v₀ = 5.59 m / s
v² = v₀² - 2 a x
the negative sign is because its stopping
a =
a = (5.59² - 5.1²) / 2 50
a = 0.0524 m / s²
Newton's second law
F = ma
F = 32 0.0524
F = 1.68 N
You might be interested in
The astronomer who first worked out the mathematical description of black hole event horizons was?
vovangra [49]

Karl Schwarzschild devised the first general relativity model that would adequately describe a black hole in 1916.

What is Black Hole?

A black hole is an area of spacetime with such intense gravitational pull that nothing can escape from it, not even light or other electromagnetic waves. According to general relativity theory, a compact enough mass can bend spacetime into a black hole. The event horizon is the line beyond which there is no escape.

Black holes were once thought to be a mathematical curiosity, but theoretical research in the 1960s revealed that they were actually a general prediction of general relativity.

To know more about Black Hole refer:

brainly.com/question/7866362

#SPJ4

4 0
2 years ago
An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu
ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, qE \times S + mg \times S = 0.5 \times mv^{2}

     1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight  and the above equation will be as follows.

   (1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}

         v = sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}

           = 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0
2 years ago
A rocky space object of varying size
kow [346]
That would be an asteroid
3 0
3 years ago
Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s
Ne4ueva [31]
The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
F=qvB \sin \theta
where \theta is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is q=2.7 \mu C=2.7 \cdot 10^{-6}C. In this case, v and B are perpendicular, so \theta=90^{\circ}, therefore we have:
B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T

2) In this second case, the angle between v and B is \theta=55^{\circ}. The charge is now q=42.0 \mu C=42.0 \cdot 10^{-6}C, and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N
5 0
3 years ago
Read 2 more answers
Questions
KIM [24]

Answer:

Precisely, water has to absorb 4,184 Joules of heat (1 calorie) for the temperature of one kilogram of water to increase 1°C. For comparison sake, it only takes 385 Joules of heat to raise 1 kilogram of copper 1°C.

Explanation:

6 0
3 years ago
Other questions:
  • A 0.0502-kg pair of fuzzy dice is attached to the rearview mirror of a car by a short string. The car accelerates at constant ra
    5·1 answer
  • The unified atomic mass unit is defined to be 1 u = 1.6605 ✕ 10−27 kg. Verify that this amount of mass converted to energy yield
    7·2 answers
  • A horizontal spring attached to a wall has a force constant of k = 860 N/m. A block of mass m = 1.60 kg is attached to the sprin
    5·1 answer
  • A toy car runs off the edge of a table that is 2.225 m high. If the car lands 0.400 m away from the base of the table, how fast
    8·1 answer
  • _______________have a negative charge and are located on the outside of the nucleus.
    6·1 answer
  • The motion of a particle is given by the equation v = 8t3 – ½ t2 + 4t, where v is in meters and t is in seconds. Compute the val
    5·1 answer
  • Please help quick please
    5·1 answer
  • Billy drops a ball from a height of 1 m. The ball bounces back to a height of 0.8 m, then
    12·1 answer
  • Please answer the question in the picture
    12·1 answer
  • Lunar phases occur when the moon appears to change shape as seen from earth. What causes different phases of the moon?.
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!