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CaHeK987 [17]
3 years ago
15

A car weighing 10.1 kN and traveling at 11.5 m/s without negative lift attempts to round an unbanked curve with a radius of 65.0

m. (a) What magnitude of the frictional force on the tires is required to keep the car on its circular path?
Physics
1 answer:
a_sh-v [17]3 years ago
8 0

Answer: a) 2095N

Explanation:

M = w/g = 10100/9.8 = 1.03 x 10^3 kg

Normal force = Gravitational force

Fn = M x g

Friction is

f = m x v^2 / r = 1.03 x 10^3 x (11.5)^2 / 65

F = 2.10 x 10^3 N

N/b to calculate maximum friction force,

Fmax = Us x mg

Fmax = Us x 10100N

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Read 2 more answers
A force vector points due east and has a magnitude of 140 newtons. A second force is added to . The resultant of the two vectors
ale4655 [162]

Answer:

(a) When the resultant force is pointing along east line, the magnitude and direction of the second force is 280 N East

(b)  When the resultant force is pointing along west line, the magnitude and direction of the second force is 560 N West

Explanation:

Given;

a force vector points due east, F_1 = 140 N

let the second force = F_2

let the resultant of the two vectors = F

(a) When the resultant force is pointing along east line

the second force must be pointing due east

F = F_1 + F_2\\\\F_2 = F - F_1\\\\F_2 = 420 \ N - 140 \ N\\\\F_2 = 280 \ N

F_2 = 280 \ East

(b) When the resultant force is pointing along west line

the second force must be pointing due west and it must have a greater magnitude compared to the first force in order to have a resultant in west line.

F = F_2 - F_1\\\\F_2 = F + F_1\\\\F_2 = 420 \ N + 140 \ N\\\\F_2 = 560 \ N

F_2 = 560 \ West

8 0
2 years ago
projectile motion of a particle of mass M with charge Q is projected with an initial speed V in a driection opposite to a unifor
SIZIF [17.4K]

Answer:

Range, R = MV²/2QE

Explanation:

The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.

Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses  all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.

So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops

Therefore {tex}R = MV²/2QE{/tex}

5 0
3 years ago
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