Answer:
a)5.925
ii)149.5 or383.5
Explanation:
ii)3.9×105-2.5×104
409.5-260
the zero in 260 may or may not be significant since it is after two significant figures thus we have two options
Answer:
His speed, in metres per second, is approximately 10m/s. The correct option is B.
Explanation:
Speed is defined as the rate in which an object cover a distance in a given time. It is measured in meters per second ( m/s). From the question, the jogger covered a distance of 3Km which when converted to meter is 3000meters in a given time 5 minutes which is 300seconds. The calculation of his speed in meter per second is shown below:
Distance= 3Km to meters ( as 1000meters = 1Km). Therefore 3× 1000 = 3000m
Time= 5 minutes to seconds ( as 1 minute = 60 Seconds). Therefore 5× 60= 300.
Speed= distance/ time
= 3000/300
= 10m/s
Therefore his speed, in metres per second, is approximately 10.
To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,
Here
n = Number of node
T = Tension
= Linear density
L = Length
Replacing the values in the frequency and value of n is one for fundamental overtone
Similarly plug in 2 for n for first overtone and determine the value of frequency
Similarly plug in 3 for n for first overtone and determine the value of frequency
Energy consumed in doing the work = 300 Joules
Force applied on the object = 75 N
Let the distance moved by the object be d.
Work done by the force is determined by the force applied and the displacement happened in the direction of the force applied.
Work done = Force x displacement
300 = 75 x d
d = 4 m
Hence, the maximum distance moved by the object = 4 meters
Answer:
a)Taking into consideration Newton’s second law, we know that
Net_Force = mass * acceleration
Since the box is pulled at constant speed, the acceleration is equal to zero.
This means that
Net_Force = 0 N
b) Force of friction
The net force is equal to the sum of all forces,
Net_Force = Force_applied - Friction
We found that Net_Force = 0, which means
Friction = Force_applied = 48 N
c) If the box comes to a stop. And the applied force becomes zero, the friction force becomes also zero.
Friction = 0 N
