4. Compara. La composición de la gasolina para los coches cambia del invierno al verano. La mezcla de los componentes de la gaso
1 answer:
You might be interested in
Answer:
1) a = 6.14 km
2) b = 4.69 km
Explanation:
Let the first building be A, second building be B and third building be C.
Now, bearing of A = 4.76 km in a direction 37° north of east
Bearing of B = 69° west of north
Bearing of C = 28° east of south
Thus if this 3 points form a triangle, we will have the following angles;
Angle at point A = 28 + (90 - 37) = 81°
Angle at point B = 28 + (90 - 69) = 49°
Angle at point C = 180 - (81 + 49) = 50°
Now, the distance between second and third building is "a" which is represented by BC in the triangle attached while the given distance of 4.76 represents side AB. Thus;
Using sine rule, we can find "a".
a/sin 81 = 4.76/sin 50
a = 6.14 km
B) distance between first and third building is AB in the triangle depicted by "b".
Similar to the first problem, we will use sine rule again.
b/sin49 = 4.76/sin 50
b = 4.69 km
Answer:
a)-2m/s^2
b)27.2m/s
Explanation:
Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)
W=455N=weight
W=mg
W=455N=weight
The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration
for point b we use the equations of motion with constant acceleration to find the velocity
Where
Vf = final speed
Vo = Initial speed =0
A = acceleration =2m/s
X = displacement =6.8m
Solving
Answer:
correct option is d) 7.0 x 10^-7 N
Explanation:
given data
distance = 175 picometers = 1.75 × m
to find out
electrical force
solution
we know atomic no of uranium is 92
and charge on electron is = 1.6 × C
and electrical force is express as
electrical force = .............1
put here value we get
electrical force =
electrical force = 6.921 × N
so correct option is d) 7.0 x 10^-7 N
Answer:
Explanation:
We apply Newton's second law at the crate :
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
m=90kg : crate mass
F= 282 N
μk =0.351 :coefficient of kinetic friction
g = 9.8 m/s² : acceleration due to gravity
Crate weight (W)
W= m*g
W= 90kg*9.8 m/s²
W= 882 N
Friction force : Ff
Ff= μk*N Formula (2)
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W = 0
N = W
N = 882 N
We replace the data in the formula (2)
Ff= μk*N = 0.351* 882 N
Ff= 309.58 N
We apply the formula (1) in x direction:
∑Fx = m*ax , ax=0
282 N - 309.58 N = 90*a
a= (282 N - 309.58 N ) / (90)
a= - 0.306 m/s²
Kinematics of the crate
Because the crate moves with uniformly accelerated movement we apply the following formula :
vf²=v₀²+2*a*d Formula (3)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
v₀ = 0.850 m/s
d = 0.75 m
a= - 0.306 m/s²
We replace the data in the formula (3)
vf²=(0.850)²+(2)( - 0.306 )(0.75 )