Answer:
Explanation:
When a spring is compressed, the force exerted by the spring is given by:
where
k is the spring constant
x is the compression of the spring
In this problem we have:
k = 52 N/m is the spring constant
x = 43 cm = 0.43 m is the compression
Therefore, the force exerted by the spring on the dart is
Now we can apply Newton' second law of motion to calculate the acceleration of the dart:
where
F = 22.4 N is the force exerted on the dart by the spring
m = 75 g = 0.075 kg is the mass of the dart
a is its acceleration
Solving for a,
If a body p with a positive charge is placed in contact with a body q (initially uncharged), then the nature of charge gained by q must be positive, because rubbing an uncharged body with a charged body or placed in contact with a positive charged body, helps gain a charge to the uncharged body.
There are a variety of methods to charge an object. One method is known as induction. In the induction process, a charged object is brought near but not touched to a neutral conducting object.
Let's know, how a element gain positive charge?
A positive charge occurs when the number of protons exceeds the number of electrons. A positive charge may be created by adding protons to an atom or object with a neutral charge. A positive charge also can be created by removing electrons from a neutrally charged object.
To learn more about Positive charge here
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Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa
<u>Explanation:</u>
Given -
Stress Direction, A = [1 0 0 ]
Slip plane = [ 1 1 1]
Normal to slip plane, B = [ 1 1 1 ]
Critical stress, Sc = 2.92 MPa
Let the direction of slip on = [ 1 1 0 ]
Let Ф be the angle between A and B
cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3
cos Ф = 1/√3
σ = Sc / cosФ cosλ
For slip along [ 1 1 0 ]
cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1
cos λ = 1/√2
Therefore,
σ = 2.92 / 1/√3 1/√2
σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa
Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa
<em>Given that:</em>
mass of the ball (m) = 0.5 Kg ,
ball strikes the wall (v₁) = 5 m/s ,
rebounds in opposite direction (v₂) = 2 m/s,
time duration (t) = 0.01 s,
<em> Determine the force (F) = ?</em>
We know that from Newton's II law,
<em>F = m. a</em> Newtons
(velocity acting in opposite direction, so <em>a = ( (v₁ + v₂)/t</em>
= m × (v₁ + v₂)/t
= 0.5 × (5 + 2)/0.01
= 350 N
<em>The force acting up on the ball is 350 N</em>
15) C. The amount of each element that begins....