Answer:
311,850 N
Explanation:
We can solve the problem by using Newton's second law:
where
F is the net force applied on an object
m is the mass of the object
a is its acceleration
For the object in this problem,
m = 27 kg
Substituting, we find the force required:
I'll assume that the chair has four legs.
Since the chair weights 3.7 kg by itself, it will weigh (79+3.7)=82.7 kg with the person sitting on it. And each of the chair's legs will take about (82.7/4)=20.675 kg.
Each leg touches the floor in a circle with 1.3cm diameter. The area of that circle is about (3.14*(1.3/2)^2)=1.327 cm^2.
Pressure is measured by force per area. So, the pressure from each leg is about 20.675kg / 1.327cm^2. That simplifies to 15.58 kg/cm^2.
Answer:
Vf= 7.29 m/s
Explanation:
Two force act on the object:
1) Gravity
2) Air resistance
Upward motion:
Initial velocity = Vi= 10 m/s
Final velocity = Vf= 0 m/s
Gravity acting downward = g = -9.8 m/s²
Air resistance acting downward = a₁ = - 3 m/s²
Net acceleration = a = -(g + a₁ ) = - ( 9.8 + 3 ) = - 12.8 m/s²
( Acceleration is consider negative if it is in opposite direction of velocity )
Now
2as = Vf² - Vi²
⇒ 2 * (-12.8) *s = 0 - 10²
⇒-25.6 *s = -100
⇒ s = 100/ 25.6
⇒ s = 3.9 m
Downward motion:
Vi= 0 m/s
s = 3.9 m
Gravity acting downward = g = 9.8 m/s²
Air resistance acting upward = a₁ = - 3 m/s²
Net acceleration = a = g - a₁ = 9.8 - 3 = 6.8 m/s²
Now
2as = Vf² - Vi²
⇒ 2 * 6.8 * 3.9 = Vf² - 0
⇒ Vf² = 53. 125
⇒ Vf= 7.29 m/s
Answer:
P = 5.22 Kg.m/s
Explanation:
given,
mass of the projectile = 1.8 Kg
speed of the target = 4.8 m/s
angle of deflection = 60°
Speed after collision = 2.9 m/s
magnitude of momentum after collision = ?
initial momentum of the body = m x v
= 1.8 x 4.8 = 8.64 kg.m/s
final momentum after collision
momentum along x-direction
P_x = m v cos θ
P_x = 1.8 x 2.9 x cos 60°
P_x = 2.61 kg.m/s
momentum along y-direction
P_y = m v sin θ
P_y = 1.8 x 2.9 x sin 60°
P_y = 4.52 kg.m/s
net momentum of the body
P = 5.22 Kg.m/s
momentum magnitude after collision is equal to P = 5.22 Kg.m/s
Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg
