Answer:
<h2>3,2 oky na dekh Lena ek bar</h2><h2>2,5</h2>
Answer:
Explanation:
30 N force is pulling total mass of 15 kg , so acceleration in the system of masses
= 30 / 15
= 2 m / s²
Let us now consider forces acting on 9 kg . 30 N is pulling it in forward direction . Tension T in the string attached to it is pulling it in reverse direction
so net force on it
30 - T
Applying Newton's law of motion on it
30 - T = mass x acceleration
30 - T = 9 x 2
30 - 18 = T
T = 12 N
Answer:
F= 4788 N
Explanation:
Because the car moves with uniformly accelerated movement we apply the following formula:
vf²=v₀²+2*a*d Formula (1)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
d=36.9 m
v₀=14.0 m/s m/s
vf= 0
Calculating of the acceleration of the car
We replace dta in the formula (1)
vf²=v₀²+2*a*d
(0)²=(14)²+2*a*(36.9)
-(14)²= (73.8) *a
a= - (196) / (73.8)
a= - 2.66 m/s²
Newton's second law of the car in direction horizontal (x):
∑Fx = m*ax Formula (2)
∑F : algebraic sum of the forces in direction x-axis (N)
m : mass (kg)
a : acceleration (m/s²)
Data
m=1800 Fkg
a= - 2.66 m/s²
Magnitude of the horizontal net force (F) that is required to bring the car to a halt in a distance of 36.9 m :
We replace data in the formula (2)
-F= (1800 kg) * ( -2.66 m/s²
)
F= 4788 N
Answer:
Explanation:
Given that :
The radius of the circular loop = 4.0 m
Maximum Emf 
= 5.0 V
The maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies can be determined via the expression;
= 
= 
5.0 = 
5.0 = 
Explanation:
Initial speed of the incident water stream, u = 16 m/s
Final speed of the exiting water stream, v = -16 m/s
The mass of water per second that strikes the blade is 48.0 kg/s.
We need to find the magnitude of the average force exerted on the water by the blade. The force acting on an object is given by :
Here, 
So, the magnitude of the average force exerted on the water by the blade is 1536 N.
