Answer:
Density = 1.1839 kg/m³
Mass = 227.3088 kg
Specific Gravity = 0.00118746 kg/m³
Explanation:
Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³
Now, from tables, density of air at 25°C is 1.1839 kg/m³
Now formula for density is;
ρ = mass(m)/volume(v)
Plugging in the relevant values to give;
1.1839 = m/192
m = 227.3088 kg
Formula for specific gravity of air is;
S.G_air = density of air/density of water
From tables, density of water at 25°C is 997 kg/m³
S.G_air = 1.1839/997 = 0.00118746 kg/m³
A real question, one with a question mark?
Answer:
The normal force the seat exerted on the driver is 125 N.
Explanation:
Given;
mass of the car, m = 2000 kg
speed of the car, u = 100 km/h = 27.78 m/s
radius of curvature of the hill, r = 100 m
mass of the driver, = 60 kg
The centripetal force of the driver at top of the hill is given as;
where;
Fc is the centripetal force
is downward force due to weight of the driver
is upward or normal force on the drive
Therefore, the normal force the seat exerted on the driver is 125 N.
<h2>
Answer:</h2>
0.126m
<h2>
Explanation:</h2>
According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;
F = k x e -------------------(i)
Where;
k = the spring's constant.
From the question, the force acting on the spring is the weight(W) of the mass. i.e
F = W -----------------------(ii)
<em>But;</em>
W = m x g;
where;
m = mass of the object
g = acceleration due to gravity [usually taken as 10m/s²]
<em>From equation (ii), it implies that;</em>
F = W = m x g
<em>Now substitute F = m x g into equation(i) as follows;</em>
F = k x e
m x g = k x e ------------------(iii)
<em>From the question;</em>
m = m1 = 3.5kg
k = 278N/m
<em>Substitute these values into equation (iii) as follows;</em>
3.5 x 10 = 278 x e
35 = 278e
<em>Now solve for e;</em>
e = 35/278
e = 0.126m
Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m
Answer:
a)
b)
Explanation:
Given:
mass of bullet, 
compression of the spring, 
force required for the given compression, 
(a)
We know
where:
a= acceleration
we have:
initial velocity,
Using the eq. of motion:
where:
v= final velocity after the separation of spring with the bullet.
(b)
Now, in vertical direction we take the above velocity as the initial velocity "u"
so,
∵At maximum height the final velocity will be zero
Using the equation of motion:
where:
h= height
g= acceleration due to gravity
is the height from the release position of the spring.
So, the height from the latched position be:
