Answer: If x + y = a, xxy = b and x • a = 1 , then 2 (a~ - l)a- a x b (b2 ... xy-plane, then the vector in the same plane having projections
<u>We are given:</u>
Mass of load = 140 kg
Acceleration = 0.4 m/s² downward (from Q24)
<u>24. Finding the tension in the cable:</u>
Ma = Mg - T
where 'a' is the acceleration of the load and T is the tension in the cable
replacing the variables, we get:
(140)(0.4) = (140)(9.8) - T
T = (140)(9.8) - (140)(0.4)
T = 1372 - 56
T = 1316N
<u>25. Finding the distance covered when the load accelerates for 20 s:</u>
in this case, the load will start from rest
so, initial velocity = 0 m/s
acceleration = 0.4 m/s²
from the second equation of motion:
s = ut + (1/2)at²
where s is the distance covered, u is the initial velocity and t is the time taken
s = (0)(20) + (1/2)(0.4)(20)²
s = 0 + 0.2*(400)
s = 80 m
here we didn't take the acceleration due to gravity of the mass into consideration because it is connected to the crane, which would act against any forces and since we are given that the total acceleration is 0.4 m/s², we will just assume that it's the final acceleration of the mass
Answer:
a) 20.54 J
b) 30.97 J
c) 10.43 J
d) -20.54 J
Explanation:
m = Mass
g = Acceleration due to gravity = 9.81 m/s²
h = Height
![W=mgh\\\Rightarrow W=0.537\times 9.81\times (5.88-1.98)\\\Rightarrow W=20.54\ J](https://tex.z-dn.net/?f=W%3Dmgh%5C%5C%5CRightarrow%20W%3D0.537%5Ctimes%209.81%5Ctimes%20%285.88-1.98%29%5C%5C%5CRightarrow%20W%3D20.54%5C%20J)
Work done by the ball's weight is 20.54 J
![W=mgh\\\Rightarrow W=0.537\times 9.81\times (5.88)\\\Rightarrow W=30.97\ J](https://tex.z-dn.net/?f=W%3Dmgh%5C%5C%5CRightarrow%20W%3D0.537%5Ctimes%209.81%5Ctimes%20%285.88%29%5C%5C%5CRightarrow%20W%3D30.97%5C%20J)
Gravitational potential energy of the basketball, relative to the ground, when it is released is 30.97 J
![W=mgh\\\Rightarrow W=0.537\times 9.81\times (1.98)\\\Rightarrow W=10.43\ J](https://tex.z-dn.net/?f=W%3Dmgh%5C%5C%5CRightarrow%20W%3D0.537%5Ctimes%209.81%5Ctimes%20%281.98%29%5C%5C%5CRightarrow%20W%3D10.43%5C%20J)
Gravitational potential energy of the basketball, relative to the ground, when it is released is 10.43 J
Change in gravitational potential energy
![\Delta U=10.43-30.97=-20.54\ J](https://tex.z-dn.net/?f=%5CDelta%20U%3D10.43-30.97%3D-20.54%5C%20J)
Change in gravitational potential energy is given by -20.54 J
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:
The resistor has a resistance of 10.667 ohms.
Explanation:
By Ohm's Law, voltage (
), in volts, is directly proportional to the current (
), in amperes, and by definition of power (
), in watts, we have the following formula:
(1)
Where
is the resistance, in ohms.
If we know that
and
, then the resistance of the resistor is:
![R = \frac{\dot W}{i^{2}}](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7B%5Cdot%20W%7D%7Bi%5E%7B2%7D%7D)
![R = 10.667\,\Omega](https://tex.z-dn.net/?f=R%20%3D%2010.667%5C%2C%5COmega)
The resistor has a resistance of 10.667 ohms.