An aircraft travels a distance of 50km in a straight line

Take a close look at the energy transfers and transformations shown in the above diagram. Which type of energy is transformed in

labwork [276]

Kinetic Energy

How to calculate

Kinetic energy is given by K.E.

$\\ \sf\longmapsto K.E=\dfrac{1}{2}mv^2$

Where

m is denoted to mass
v is denoted to velocity

4 0

3 years ago

Help help hep help???????????

Sergeu [11.5K]

Momentum

mava + mbvb = mava '+ mbvb'

(300 x 10)+(150 x 0) = (300 x 4.12)+(150 x vb')

3000=1236+150vb'

1764 = 150vb'

vb'=+11.76 m/s ≈ +11.8 m/s (positive sign, to the right)

3 0

2 years ago

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero. (a)

Marysya12 [62]

This question is incomplete, the complete question is;

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.

(a) Determine the forces and and the couple

(b) Determine the sum of the moments about the right end of the beam.

(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?

Answer:

a)

the x-component of the force at A is $A_{x}$ = 0

the y-component of the force at A is $A_{y}$ = 400 N

the couple acting at A is; $M_{A}$ = 146 N-m

b)

the sum of the momentum about the right end of the beam is; ∑ $M_{R}$ = 0

c)

the equivalent force acting at the left end is; F = -400J ( N)

the couple acting at the left end is; M = - 146 N-m

Explanation:

Given that;

The sum of the forces acting on the beam is zero ∑f = 0

Sum of the moments about the left end of the beam is also zero ∑ $M_{L}$ = 0

Vector force acting at A, $F_{A}$ = $A_{x}i$ + $A_{y}j$

Now, From the image, we have;

a)

∑f = 0

$F_{A}$ - 600j + 200j = 0i + 0j

$A_{x}i$ + $A_{y}j$ - 600j + 200j = 0i + 0j

$A_{x}i$ + ( $A_{y}$ - 400)j = 0i + 0j

now by equating i- coefficients'

$A_{x}$ = 0

so, the x-component of the force at A is $A_{x}$ = 0

also by equating j-coefficient

$A_{y}$ - 400 = 0

$A_{y}$ = 400 N

hence, the y-component of the force at A is $A_{y}$ = 400 N

we also have;

∑ $M_{L}$ = 0

$M_{A}$ - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0

$M_{A}$ - 30 N-m - 228 N-m + 112 Nm = 0

$M_{A}$ - 146 N-m = 0

$M_{A}$ = 146 N-m

Therefore, the couple acting at A is; $M_{A}$ = 146 N-m

b)

The sum of the moments about right end of the beam is;

∑ $M_{R}$ = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)( $A_{y}$ ) + $M_{A}$

∑ $M_{R}$ = (108 N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m

∑ $M_{R}$ = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m

∑ $M_{R}$ = 0

Therefore, the sum of the momentum about the right end of the beam is; ∑ $M_{R}$ = 0

c)

The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.

The equivalent force at the left end will be;

F = -600j + 200j (N)

F = -400J ( N)

Therefore, the equivalent force acting at the left end is; F = -400J ( N)

Also couple acting at the left end

M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)

M = -(30 N-m) + (112 N-m) - ( 228 N-m))

M = 112 N-m - 258 N-m

M = - 146 N-m

Therefore, the couple acting at the left end is; M = - 146 N-m

7 0

2 years ago

1. How much heat must be absorbed by 375 grams of water to raise its

antiseptic1488 [7]

Answer:

39225J

Explanation:

Given parameters:

Mass of water = 375grams of water

Change in temperature = 25°C

Specific heat capacity of water = 4.184J/g°C

Unknown:

Amount of heat absorbed = ?

Solution:

To solve this problem, we use the expression below:

H = m c Ф

H is the heat absorbed

m is the mass

c is the specific heat capacity

Ф is the change in temperature

Insert the parameters and solve;

H = 375 x 4.184 x (25) = 39225J

5 0

2 years ago

A train travels due south at 25 m/s (relative to the ground) in a rain that is blown toward the south by the wind. The path of e

olasank [31]

Answer:

Explanation:

Given

Train travels towards south with a velocity if $v_t=25\ m/s$

Rain makes an angle of $\theta =66^{o}$ with vertical

If an observer sees the drop fall perfectly vertical i.e. horizontal component of rain velocity is equal to train velocity

suppose $v_r$ is the velocity of rain with respect to ground then

$v_r\sin\theta =v_t$

$v_r\times \sin (66)=25$

$v_r=27.36\ m/s$

Therefore velocity of rain drops is 27.36 m/s

8 0

3 years ago