density = mass/volume = 100kg/10ml = 10kg/ml
voluime = mass/density = 50g/2 g/ml = 25 ml
mass = density x volume = 2x55 = 110 kg
To the Earth in less than ten minutes.
Before coming into conclusion first we have to understand both scalar and vector .
A scalar quantity is a physical quantity which has only magnitude for it's complete specification.
A vector quantity is that physical quantity which not only requires magnitude but also possesses direction for it's complete specification.
So the most important factor that differentiate vector from scalar is the direction.
As per the question the student is doing an experiment where he is recording the data obtained during the process.
In order to arrange them in data table, he should ask about the direction of the quantity under consideration.
Hence the correct option is the third option(C)i.e does the measurement include direction?
Answer:
The load has a mass of 2636.8 kg
Explanation:
Step 1 : Data given
Mass of the truck = 7100 kg
Angle = 15°
velocity = 15m/s
Acceleration = 1.5 m/s²
Mass of truck = m1 kg
Mass of load = m2 kg
Thrust from engine = T
Step 2:
⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:
T = (m1+m2)*g*sinθ
⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes m1*gsinθ .
Resultant force on truck is F = T – m1*gsinθ
F causes the acceleration of the truck: F= m*a
This gives the equation:
T – m1*gsinθ = m1*a
T = m1(a + gsinθ)
Combining both equations gives:
(m1+m2)*g*sinθ = m1*(a + gsinθ)
m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ
m2*g*sinθ = m1*a
Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:
m2*g*sinθ = (7100 – m2)*a
m2*g*sinθ = 7100a – m2a
m2*gsinθ + m2*a = 7100a
m2* (gsinθ + a) = 7100a
m2 = 7100a/(gsinθ + a)
m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)
m2 = 2636.8 kg
The load has a mass of 2636.8 kg
Answer:
Answered
Explanation:
x= 0.02 m
E_p= 10.0 J
E_p= 0.5kx^2
10= 0.5k(0.02)^2
solving we get
K= 50.0 N/m
Now
E'_p= 0.5kx'^2
E'_p= 0.5×50×(0.04)^2
E'_p=40 J
b) potential energy is a scalar quantity and it only depends magnitude and not direction so it will remain same in compression and expansion both
c) 20 J = 0.5×50,000×x^2
solving
x= 0.028 m
d) k is 50.0 N/m from above calculation
