Anyone know this question?

The turnbuckle is tightened until the tension in the cable AB equals 2.3 kN. Determine the vector expression for the tension T a

brilliants [131]

Answer:

a) 0.83984 i + 0.41992 j - 2.0996 k KN

b) T_ac = 1.972888 KN

Explanation:

Given:

- The tension in cable AB = 2.3 KN

Find:

a) Determine the vector expression for the tension T as a force acting on member AD.

b) Also find the magnitude of the projection of T along the line AC.

Solution:

part a)

- Find unit vector AB:

vector (AB) = 2 i + j - 5 k

mag (AB) = sqrt (2^2 + 1^2 + 5^2)

mag (AB) = sqrt(30)

unit (AB) = ( 1 / sqrt(30) )* ( 2 i + j - 5 k )

- Find Tension vector:

vector (T) = unit(AB)* 2.3 KN

= 0.83984 i + 0.41992 j - 2.0996 k

- The projection of T onto AC can be found from the dot product of vector T to unit vector (AC)

- For unit vector (AC)

vector (AC) = 2 i - 2 j - 5 k

mag (AC) = sqrt (2^2 + 2^2 + 5^2)

mag (AC) = sqrt(33)

unit (AC) = ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

- Compute the projection:

T_ac = vector T . unit (AC)

T_ac = (0.83984 i + 0.41992 j - 2.0996 k) . ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

T_ac = 0.2923947572 - 0.146973786 - 1.827467232

T_ac = 1.972888 KN

4 0

3 years ago

A train of mass 3.3 × 10^6 kg is moving at a constant speed up a slope inclined at an angle of 0.64°

AnnyKZ [126]

Answer:

C. 39 m/s

Explanation:

First we need to calculate the total force required to move the train along the inclined plane. So, it is clear that the work done will be equal to the component of the weight that is parallel to the inclined plane, because there is no frictional force present:

Force = F = mg Sin θ

where,

m = mass of train = 3.3 x 10⁶ kg

g = 9.8 m/s²

θ = Angle of Inclination = 0.64°

Therefore,

F = (3.3 x 10⁶ kg)(9.8 m/s²)Sin 0.64°

F = 3.612 x 10⁵ N

Now, the formula for power is:

P = FV

V = P/F

where,

V = Velocity of Train = ?

P = Power of Engine = 14 MW = 1.4 x 10⁷ W

Therefore,

V = 1.4 x 10⁷ W/3.612 x 10⁵ N

V = 38.75 m/s

which is approximately equal to:

4 0

3 years ago

Why do the pvc plastic wells used in this weeks elisa need to be sticky on the inner walls and how may your results change if a

Semmy [17]

Answer:

The stickiness in the inner walls allows them to be easily coated with the desired antigens, this translates in the use of a smaller amount of antigen. If the walls weren't sticky there's a possibility the antigen won't stick to them and therefore the result of the ELISA can be a false negative.

I hope you find this information useful and interesting! Good luck!

5 0

3 years ago

A series ac circuit contains a resistor, an inductor of 150 mh, a capacitor of 5.00 m f , and a source with vmax = 240 v operati

antoniya [11.8K]

Ω=2*pi*f = 2*pi*50 = 314.16

Inductive reactance, ZL = jωL = j*314.16*0.15 = j42.12
Capacitance reactance, ZC = 1/(jωC) =1/(j*314.16*0.005) = -j0.64
Impendance, Z = V/I = 240/0.1 = 2400

Now,
Z=R+j(ZL+ZC) => 2400 = R+ j(45.12-0.64) => 2400 = R + j44.48
Additionally,
2400^2 = R^2+44.48^2 => R = Sqrt (2400^2-44.48^2) = 2399.58 ohms.

Phase angle = arctan (44.48/2399.58) = 1.06°

4 0

3 years ago

A 2 kg toy cart and a 6 kg toy cart have a spring compressed between them. When the spring expands, it sends the 2 kg toy cart o

harkovskaia [24]

Answer:

The speed of second toy cart is 4 m/s.

(c) is correct option

Explanation:

Given that,

Mass of first toy cart = 2 kg

Mass of second toy cart = 6 kg

Speed of first toy cart = 12 m/s

We need to calculate the speed of second toy cart

Using formula of momentum

$m_{1}v_{1}=m_{2}v_{2}$

Where, m₁ = mass of first toy cart

m₂ = mass of second toy cart

v₁ = velocity of first toy cart

v₂ = velocity of second toy cart

Put the value into th formula

$2\times12=6\times v_{2}$

$v_{2}=\dfrac{2\times12}{6}$

$v_{2}=4\ m/s$

Hence, The speed of second toy cart is 4 m/s.

(c) is correct option

4 0

4 years ago