Good conductors are the materials which allow electricity to pass through them easily whereas, bad conductors are the materials which do not allow electricity to pass through them. Good conductors offer less resistance to the flow of current, whereas the bad conductors do not transmit current through them.
C. The forces on an object traveling at terminal velocity are balanced.
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
The new velocity after 4 s is 40 m/s
The height of the spaceship above the ground after 5 seconds is 1,127.5 m
The given parameters for the first question;
- initial velocity of the car, u = 76 m/s
- acceleration of the car, a = - 9 m/s²
The new velocity after 4 s is calculated as;
v = u + at
v = 76 + (-9)(4)
v = 76 - 36
v = 40 m/s
(5)
The given parameters;
- height above the ground, h = 500 m
- velocity of spaceship, u = 150 m/s
The height of the spaceship above the ground after 5 seconds is calculated as;
Learn more here: brainly.com/question/24527971
Responder:
Explicación:
Para que podamos calcular el tiempo que le tomará al esquiador permanecer en el aire, encontraremos el tiempo de vuelo como se muestra;
T = 2Usin theta / 2
theta = 90 grados
U = 25 m / s
T = 25sin90 / 2 (9,8)
T = 25 / 19,62
T = 1,27 segundos
Por lo tanto, los cielos usarán 1.27 segundos en el aire.
La distancia horizontal es el rango;
Rango R = U√2H / g
R = 25√2 (80) /9,8
R = 25√160 / 9,8
R = 25 * √16,326
R = 25 * 4.04
R = 101,02 m
Por tanto, la distancia horizontal recorrida por el esquiador es 101,02 m
