Answer:
The graph appears to be in error.
The actual figure appears to be a rhombus with sides of 5 and 15 with a height of 5
The work done (F * S) is the area of the rhombus
1/2 * (5 +15) * 5 = 50 J
Answer:
Explanation:
Given
Frequency of SHM is 
Amplitude of SHM is 
Cup begins to slip when it overcomes the friction force
Friction force 
Applied force 
and maximum acceleration during SHM is
D. They are heterotrophs that digest food internally.
(a) The net force on the shopping cart is zero.
(b) The the force of friction on the shopping cart is 25 N.
(c) When same force is applied to the shopping cart on a wet surface, it will move faster.
<h3>Net force on the shopping cart</h3>
The net force on the shopping cart is calculated as follows;
F(net) = F - Ff
where;
- F is the applied force
- Ff is the frictional force
ma = F - Ff
where;
- a is acceleration of the cart
- m is mass of the cart
at a constant velocity, a = 0
0 = F - Ff
F(net) = 0
F = Ff = 25 N
Net force is zero, and frictional force is equal to applied force.
<h3>On wet surface</h3>
Coefficient of kinetic friction of solid surface is greater than that of wet surface.
Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.
Thus, the cart will move faster on a wet surface due to decrease in friction.
Learn more about frictional force here: brainly.com/question/24386803
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Answer:
a = 17.68 m/s²
Explanation:
given,
length of the string, L = 0.8 m
angle made with vertical, θ = 61°
time to complete 1 rev, t = 1.25 s
radial acceleration = ?
first we have to calculate the radius of the circle
R = L sin θ
R = 0.8 x sin 61°
R = 0.7 m
now, calculating at the angular velocity
ω = 5.026 rad/s
now, radial acceleration
a = r ω²
a = 0.7 x 5.026²
a = 17.68 m/s²
hence, the radial acceleration of the ball is equal to 17.68 rad/s²
