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
oksano4ka [1.4K]
3 years ago
13

While doing a lab, a student found the density of a piece of pure aluminum to be 2.85 g/cm3. The accepted value for the density

of aluminum is 2.70 g/cm3. What was the student's percent error?
Physics
1 answer:
bezimeni [28]3 years ago
3 0
The equation for percent error is

% Error = 100*|Experimental-Theoretical|/Theoretical

Our experimental is 2.85g/cm^3 and the accepted is 2.7g/cm^3

Thus our % Error = 5.555%
You might be interested in
10. Josie and Trey were working on their physics project and both built catapults. Trey's catapult shot a ball
SVEN [57.7K]

Answer:

Josie's ball faster than T

3 0
2 years ago
An astronaut is in space with a baseball and a bowling ball. The astronaut pushes both objects in the same direction. If both ba
Crazy boy [7]

Answer:

As a mass greater than that of baseball, at the moment of the bowling wave the moment of the baseball ball is also greater

Explanation:

This problem is an application of momentum and momentum. When the astronaut pushed balls, he needed more force to move the ball of greater mass (bowling). The expression for soul is

      p = m v

Besibol Blade

      p1 = m1 v

Bowling ball

      p2 = m2 v

As a mass greater than that of baseball, at the moment of the bowling wave the moment of the baseball ball is also greater

      p2 >> p1

3 0
3 years ago
. A stationary mass explodes into two parts of mass 4 kg and 40 kg . If the K.E of larger mass is 100 J . The K.E of small mass
alisha [4.7K]

Answer:

10J

Explanation:

KE = (1/2)mv²

100J = (.5)(40kg)v²

v²=(100J)/(20kg)

v²= 5

KE = 5(.5)(4kg)

KE = 10J

3 0
2 years ago
A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at
Readme [11.4K]

Answer:

 KE = 1.75 J

Explanation:

given,

mass of ball, m₁ = 300 g = 0.3 Kg

mass of ball 2, m₂ = 600 g = 0.6 Kg

length of the rod = 40 cm = 0.4 m

Angular speed = 100 rpm= 100\times \dfrac{2\pi}{60}

                         =10.47\ rad/s

now, finding the position of center of mass of the system

    r₁ + r₂ = 0.4 m.....(1)

 equating momentum about center of mass

  m₁r₁ = m₂ r₂

   0.3 x r₁ = 0.6 r₂

   r₁ = 2 r₂

Putting value in equation 1

2 r₂ + r₂ = 0.4

 r₂ = 0.4/3

 r₁ = 0.8/3

now, calculation of rotational energy

KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2

KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)

KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)

KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)

 KE = 1.75 J

the rotational kinetic energy is equal to 1.75 J

7 0
3 years ago
Fill in the blank
aliina [53]

Answer:

hypothesis

Explanation:

8 0
3 years ago
Other questions:
  • Which list shows the development of atomic models in chronological order?
    13·1 answer
  • What would happen to the apparent change in mass if the direction of the current is reversed?
    12·1 answer
  • In what changes of state do atoms lose energy? Check all that apply
    11·1 answer
  • what is the average speed of a car that travels 40mph for 1 hour and 60 mph in another hour PLEASE HELP WILL MARK BRAINLIEST
    7·1 answer
  • What is the voltage measured with a voltmeter across a wire in a circuit?
    6·2 answers
  • What influences the strength of an electric field?
    12·2 answers
  • Through which media did sound waves travel the fastest and the slowest?
    8·1 answer
  • Explain why a book placed on a table does not move?
    10·2 answers
  • I need the answer with steps
    6·1 answer
  • A metre rule is used to measure a length. Which reading is shown to the nearest millimetre? A 0.7m B 0.76m C 0.761m D 0.7614m
    5·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!