Https://www.slideshare.net/mobile/jan_parker/life-cycle-of-stars-3196871
the answer to your question!
Curved line
Explanation:
Acceleration of motion is represented by a curved line on a non-linear distance-time graph.
The acceleration of a non-linear motion is depicted using a parabola which is a curve. This implies that the velocity is constantly changing and the distance covered by the body is also changing with equal amount of time.
- A plot of this will give a parabola. This can be further established using one of the equations of motion below:
x = u + 
at ²
This is a quadratic function where:
x is the distance
u is the initial velocity
t is the time
a is acceleration
A quadratic function gives a curved line which is a parabola.
Learn more:
Acceleration brainly.com/question/10932946
#learnwithBrainly
Answer:
F = 39.36 N
Explanation:
given,
initial speed, u = 38 m/s
final speed, v = 0 m/s
mass of ball = 0.145 Kg
time, t = 0.14 s
Force = ?
using impulse formula
J = change in momentum
J = F x t
m(v - u) = F x t
0.145 x (0 - (-38)) = F x 0.14
F x 0.14 = 5.51
F = 39.36 N
force exerted by the ball is equal to 39.36 N.
Answer:
the correct affirmation is the 3
Explanation:
Let's analyze the problem with Newton's second law before looking at the claims.
X axis parallel to the plane, positive down
F -fr + Wₓ = ma
Y Axis perpendicular to the plane
N -Wy = 0
With trigonometry
Wₓ = W sin θ
Wy = w cos θ
Let's multiply by the displacement along the plane, to relate to the work, which has as expression W = F d
F d -fr d + Wx d = ma d
Push W₁ = Fd
frictional force W₂ = -fr d
gravity W₃ = Wx d
W₁ + W₃ -W₂ = m a d
Analysis affirmations:
R1) false. The work of gravity is the subtraction
R2) false. Each force contributes according to its magnitude
R3) true. In the equation we see that, if the acceleration is zero, W2 = W1 + W3
R4) False. It equals the difference
the correct affirmation is the 3
What's the weight of the pear ?
Weight = (mass) x (gravity) = (1 kg) x (9.8 m/s²) = 9.8 Newtons.
OK. We know there's a force of 9.8 Newtons acting downwards on the pear.
Is the pear accelerating ? No ! It's just laying there on the table.
If it's not accelerating, then we know that the net force on it must be zero.
So there must be ANOTHER force acting UPWARDS on it, to exactly
cancel out the downward force of its weight. THAT's the "normal" force ...
the upward force that the table exerts on the pear. It must also be 9.8N,
but UPwards, so that if you add it to the weight, the sum is zero.
