You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after
1 answer:
Answer: d = 33 cm or 0.33 m
Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:
W = F.d.cosθ
F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.
For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:
W = F.d.cos(0)
W = F.d
To determine d:
d =
d =
d = 0.33 m
The distance d the block ice moved is 33 cm.
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Answer:
The average recoil force on the gun during that 0.40 s burst is 45 N.
Explanation:
Mass of each bullet, m = 7.5 g = 0.0075 kg
Speed of the bullet, v = 300 m/s
Time, t = 0.4 s
The change in momentum of an object is equal to impulse delivered. So,
For 8 shot burst, average recoil force on the gun is :
So, the average recoil force on the gun during that 0.40 s burst is 45 N.
The gravity force between Jupiter and Io will be 6.343 × 10²² N.
<h3>What is Newton's law of gravitation?</h3>
Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.
Given data;
Mass of Jupiter,
Mass of moon of Jupiter,]
The gravitational constant is,
Distance between Jupiter and Io, R = 421,700 km = 4,217,00,000 m
The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.
The gravitational force is found as;
Hence, the gravity force between Jupiter and Io will be 6.343 × 10²² N.
The complete question is
"Jupiter has a mass of 1.9 × 1027 kg, and its moon Io has a mass of 8.9 × 1022 kg. Their centers are separated by a distance of 421,700 km what is the force of gravity acting on Io? "
To learn more about Newton's law of gravitation, refer to the link.
#SPJ1
Answer:
(a) Ff = 0.128 N
(b μk = 0.1135
Explanation:
kinematic analysis
Because the hockey puck moves with uniformly accelerated movement we apply the following formulas:
vf=v₀+a*t Formula (1)
d= v₀t+ (1/2)*a*t² Formula (2)
Where:
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s
Calculation of the acceleration of the hockey puck
We apply the Formula (1)
vf=v₀+a*t v₀=5.8 m/s , vf=0
0=5.8+a*t
-5.8 = a*t
a= -5.8/t Equation (1)
We replace a= -5.8/t in the Formula (2)
d= v₀*t+ (1/2)*a*t² , d=15.1 m , v₀=5.8 m/s
15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²
15.1= 5.8*t-2.9*t
15.1= 2.9*t
t = 15.1 / 2.9
t= 5.2 s
We replace t= 5.2 s in the equation (1)
a= -5.8/5.2
a= -1.115 m/s²
(a) Calculation of the frictional force (Ff)
We apply Newton's second law
∑F = m*a Formula (3)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Look at the free body diagram of the hockey puck in the attached graphic
∑Fx = m*a m= 115g * 10⁻³ Kg/g = 0.115g , a= -1.12 m/s²
-Ff = 0.115*(-1.115) We multiply by (-1 ) on both sides of the equation
Ff = 0.128 N
(b) Calculation of the coefficient of friction (μk)
N: Normal Force (N)
W=m*g= 0.115*9.8= 1.127 N : hockey puck Weight
g: acceleration due to gravity =9.8 m/s²
∑Fy = 0
N-W=0
N = W
N = 1.127 N
μk = Ff/N
μk = 0.128/1.127
μk = 0.1135
