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Charra [1.4K]
3 years ago
14

Rosa records the distance that a toy car rolls and the time it takes to cover the distance. What scientific practice is this?

Physics
2 answers:
suter [353]3 years ago
8 0

collecting data...............................

Sergio039 [100]3 years ago
3 0
Rosa is conducting the scientific practice of observation.
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Write the mathemetical relation between work force and displacement​
il63 [147K]

Answer:

Work is measured as the product of force and the displacement in the direction of the force. Work = force × displacement in the direction of the force.

8 0
2 years ago
Help me please i dont understand
alexandr402 [8]
Option A is the correct answer.
5 0
2 years ago
A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 19
Hunter-Best [27]

The car will take 300 m before it stops due to applying break.

<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
  • As per Newton's equation of motion, V² - U² = 2aS
  • V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
  • Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
  • So, 0² - 60² = 2×6× S

=> -3600 = -12S

=> S = 3600/12 = 300 m

Thus, we can conclude that the distance covered by the car is 300 m before it stopped.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?

Learn more about the Newton's equation of motion here:

brainly.com/question/8898885

#SPJ1

7 0
2 years ago
What force is necessary to accelerate a 2500 kg care from rest to 20 m/s over 10 seconds?
EleoNora [17]
Force = mass x acceleration
force = 2500kg x (20m/s / 10m/s)
force = 2500kg x 2m/s^2
force = 5000kg m/s^2 = 5kN

i hope this is right (^^)
4 0
2 years ago
An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields a
Lapatulllka [165]

Answer:

The speed of the electron is 2.55\times 10^3\ m/s.

Explanation:

Given that,

The magnitude of electric field, E=1.32\ kV/m=1.32\times 10^3\ V/m

The magnitude of magnetic field, B = 0.516 T

Both the magnetic and electric fields are acting on the moving electron. Then,  the magnitude of electric field and magnetic field is balanced such that :

evB=eE\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.32\times 10^3}{0.516}\\\\v=2558.13\ m/s

or

v=2.55\times 10^3\ m/s

So, the speed of the electron is 2.55\times 10^3\ m/s. Hence, this is the required solution.

3 0
3 years ago
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