Answer:
Work done = 0.3142 Nm
Explanation:
Mass of Object is 50 g
Circular path of radius is 10 cm ⇒ 0.1 m
Work done = Force × Distance = ?
*Distance moved (circular path) ⇒ Circumference of the circular path
2πr = 2 × 3.142 × 0.1 ⇒ 0.6284 m
*Force that is enough to move a 50 g must be equal or more than its weight.
therefore convert 50 grams to newton = 0.5 N
Recall that; work done is force times distance
∴ 0.5 N × 0.6284 m
Work done = 0.3142 Nm
11 grams were released.think it help
Answer:
Please see below as the answers are self-explanatory.
Explanation:
1) The resultant force is along the line that joins both charges or both masses (assuming both objects can be represented as points)
2) Both type of forces obey Newton's 3rd law.
3) Both are proportional to the product of the property that is affected by the force (charges and masses)
4) Both obey an inverse - square law (consequence of our universe being three-dimensional)
1) Main difference, is that while the gravitational force is always attractive, the electrostatic force can be attractive or repulsive, as there are two types of charges, which attract each other being of different type, and repel each other if they are of the same type.
2) It is possible, artificially, to block the influence of the electrostatic force, shielding a room, for instance, which is not possible for the gravitational force.
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
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<h2>
Answer:442758.96N</h2>
Explanation:
This problem is solved using Bernoulli's equation.
Let 
be the pressure at a point.
Let 
be the density fluid at a point.
Let 
be the velocity of fluid at a point.
Bernoulli's equation states that 
for all points.
Lets apply the equation of a point just above the wing and to point just below the wing.
Let 
be the pressure of a point just above the wing.
Let 
be the pressure of a point just below the wing.
Since the aeroplane wing is flat,the heights of both the points are same.
So,
Force is given by the product of pressure difference and area.
Given that area is 
.
So,lifting force is 
