Answer:
12J
Explanation:
Given parameters:
Mass = 1.5kg
Impulse = 6kgm/s
let us start first by find the velocity with which this body moves;
Impulse = mass x velocity
Velocity = Impulse / mass = 6/ 1.5 = 4m/s
Initial velocity = 0m/s
Unknown:
Resulting kinetic energy = ?
Solution:
To solve this problem use the formula below:
K.E = m (v - u)²
m is the mass
v is the final velocity
u is the initial velocity
So;
K.E = x 1.5 x (4 - 0)²
K.E = 1.5 x 8 = 12J
Answer:
B
Explanation:
It's because the line keeps going straight meaning it has a constant velocity from beginning to end.
Note that the density of mercury is
(i) 13545 kg/m^3 at 20 deg.C
(ii) 13472 kg/m^3 at 50 deg, C
For a given mass, M kg, by using the volume at 20 deg. C, obtain
M = (13545 kg/m^3)*(0.002 m^3) = 27.09 kg .
Let V m^3 = volume at 50 deg. C
Then
(13472 kg/m^3)*(V m^3) = 27.09 kg
V = 27.09/13472 = 0.002010837 m^3
Answer:
The closest answer is A. 0.002010812 m^3
Given Information:
Mass = m = 500 kg
Acceleration = a = 10 cm/s²
Required Information:
Magnitude of rightward net force = F = ?
Answer:
Magnitude of rightward net force = 50 N
Explanation:
From the Newton's second law of motion
F = ma
Where m is the mass and a is the acceleration
To get force in Newtons first convert 10 cm/s² into m/s²
10/100 = 0.1 m/s²
F = 500*0.1
F = 50 N
Therefore, the magnitude of rightward net force acting on it is 50 Newtons.
Answer:
Tension= 475N
Force= 225N
Explanation:
The question is not complete, here is the complete question
Also, see attached a free body diagram for your reference
<em>"A 20.0-kg uniform plank is supported by the floor at one end by a vertical rope at the other as shown in the figure. A 50.0-kg mass person stands on the plank a distance three-fourths of the length plank from the end on the floor.
</em>
<em>a. What is the tension in the rope?
</em>
<em>b. What is the magnitude of the force that the floor exerts on the plank?"</em>
<em />
given data
mass of man=50kg
mass of plank=20kg
length of plank=10m
<em />
let us make the lenght of the rope be d
The torque about the floor
That is taking moment about the floor
Force will be also zero
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