The possible magnitude for the force of static friction on the stationary cart is 72.1 N.
The given parameters:
- <em>Applied force on the cart, F = 72.1 N</em>
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Based on Newton's second law of motion, the force applied to object is directly proportional to the product of mass and acceleration of the object.
F = ma
Static frictional force is the force resisting the motion of an object at rest.
where;
is the frictional force
Thus, the possible magnitude for the force of static friction on the stationary cart is 72.1 N.
Learn more about Newton's second law of motion: brainly.com/question/25307325
By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.
<h3>What are the limits?</h3>
First, we need to find the limits.
We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.
So if 58 in is the 100%, the 26% and 43% of that are:
- 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
- 43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.
But we know that the CG is found to be 45.5% MAC, then it measures:
(45.5%/100%)*58in = 0.455*58in = 26.39 in
We need to compare it with the largest limit, so we get:
26.39 in - 24.94 in = 1.45 in
This means that the CG is 1.45 inches out of limits.
If you want to learn more about percentages, you can read:
brainly.com/question/14345924
To create the shapes, stars are arranged on a piece of cardboard in the desired configuration. If the stars are placed in a smiley face pattern on the cardboard, for example, they will explode into a smiley face in the sky. In fact, you may see several smiley faces in the sky at one time.
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Answer:
The third drop is 0.26m
Explanation:
The drop 1 impacts at time T is given by:
T=sqrt(2h/g)
T= sqrt[(2×2.4)/9.8]
T= sqrt(4.8/9.8)
T= sqrt(0.4898)
T= 0.70seconds
4th drops starts at dT=0.70/3= 0.23seconds
The interval between the drops is 0.23seconds
Third drop will fall at t= 0.23
h=1/2gt^2
h= 1/2×9.81×(0.23)^2
h= 0.26m
Answer:
m1/m2 = 0.51
Explanation:
First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:
V = √F/u
This is the equation that describes the relation between speed of a pulse and a force exerted on it.
the value of "u" is:
u = m/L
Where m is the mass of the rod, and L the length.
Now, for the rod 1:
V1 = √F/u1 (1)
rod 2:
V2 = √F/u2 (2)
Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:
1.4V2 = √F/u1 (3)
Replacing (2) in (3):
1.4(√F/u2) = √F/u1 (4)
Now, let's solve the equation 4:
[1.4(√F/u2)]² = F/u1
1.96(F/u2) =F/u1
1.96F = F*u2/u1
1.96 = u2/u1 (5)
Now, replacing the expression of u into (5) we have the following:
1.96 = m2/L / m1/L
1.96 = m2/m1 (6)
But we need m1/m2 so:
1.96m1 = m2
m1/m2 = 1/1.96
m1/m2 = 0.51
