The specific gravity is how the density of the object compares to the density of water. Water's density is 1gram per milliliter. We just need to figure out the density of the object.
The object is .8 kg and it displaces 500mL of water, so the density is the mass divided by the volume. Since the density of water is given in grams, we have to convert the objects mass from kg to g and then we can get the density.
.8kg * 1000g/kg = 800 grams
So
800g/500ml = 1.6grams/mL this is the density.
So divide the density of your object by the density of water, which is 1g/mL, you get 1.6 as the specific gravity. This means the object is 1.6 times more dense than water.
Answer:
2.1J
Explanation:
Given parameters:
Spring constant = 128N/m
Compression = 0.18m
Unknown:
Potential energy of the spring = ?
Solution
The potential energy of the spring is the elastic potential energy within the spring.
To solve this;
Elastic potential energy = k e²
k is the spring constant
e is the compression
Now;
Elastic potential energy = x 128 x 0.18² = 2.1J
The answer is substance A
Answer:
P=1.53 i +1.92 j kg.m/s
P=2.45 kg.m/s
α = 51.34
Explanation:
Given that
m=123 g = 0.123 Kg
U= 25 m/s
θ=30°
t= 0.6 s
This is the case of projectile motion
So the horizontal component of velocity U = u cosθ
u = 25 cosθ
u = 25 cos 30°
u=21.65 m/s
The vertical component of velocity U = U sinθ
Vo= U sinθ
Vo= 25 sin 30°
Vo = 12.5 m/s
We know that horizontal component of velocity of ball will remain same.So the horizontal component of momentum
Px= m u
Px= 0.143 x 12.5 kg.m/s
Px=1.53 kg.m/s
The vertical component of ball after 0.6 s
V= Vo- g t
V= 21.65 - 10 x 0.6 m/s
V= 15.65 m/s
Py= m V
Py= 0.123 x 15.65 kg.m/s
Py=1.92 kg.m/s
P=1.53 i +1.92 j kg.m/s
Magnitude P
P=2.45 kg.m/s
Direction
α = 51.34° (measured from x direction)
Question:
A roller-coaster vehicle has a mass of 500 kg when fully loaded with passengers. If the vehicle has a speed of 20.0 m/s at point A, what is the force exerted by the track on the vehicle at this point?
<em>See attachment</em>
Answer:
33700 Newton
Explanation:
Given
First, we determine the forces acting on mass m.
They are: the force exerted by the track (Fn) and the weight of the vehicle (W)
So, the net force is:
So:
Make Fn the subject
Fnet is calculated as:
--- i.e. the centripetal force
So: