Imagine that person B is more massive than person A in the picture above.

An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

ololo11 [35]

Answer

given,

mass of jogger = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed = 1.3 m/s in 61 ° north of east.

jogger one

$P_1 = m_1 v_1 \hat{i}$

$P_1 = 67 \times 2.3\hat{i}$

$P_1 = 154.1 \hat{i}$

$P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}$

$P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}$

$P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}$

$P_2 = 44.12\hat{i} +79.59\hat{j}$

now

P = P₁ + P₂

$P = 198.22 \hat{i} +79.59 \hat{j}$

magnitude

$P = \sqrt{198.22^2 + 79.59^2}$

$P =213.60 kg.m/s$

$\theta = tan^{-1}\dfrac{79.59}{198.22}$

$\theta = 21.87$

the angle is $\theta = 21.87$ north of east

7 0

3 years ago

A 1500 kg elevator, suspended by a single cable with tension 16.0 kN, is measured to be moving upward at 1.2 m/s. Air resistance

kirill115 [55]

Tension in the Cable is 0.87 m/s6^2, Elevator's speed after it has moved 10m is 1.6*10^5 N, Work done by gravity is 1.47 * 10^3 N, Elevator Kinetic Energy is 13897.5J and Elevator Speed after rising to 10m is 4.312 m/s.

Tension is a pulling force that operates in one dimension along the cables' axes in the opposite direction from the direction of the applied force. The combined weight of the elevator box and the passenger riding inside it, in the case of an elevator, provides the pulling force in the cables is called Tension.

A moving object or particle's kinetic energy, which depends on both mass and speed, is one of its characteristics. The type of motion can be vibration, translation, rotation around an axis, or any combination of these. Kinetic energy is a type of energy that an item or particle possesses as a result of motion.

We know that,

Tension in the Cable

T = m(g+a) = g+a = T/m = 16 * 103 / 1500 = 10.67 m/s2

a = 10.67 - 9.8 = 0.87 m/s6^2

Elevator's speed after it has moved 10m.

U^2 = u^2 +2as

= 1.22 +2*0.87*10

=1.6*10^5 N

Work done by gravity = mg * 10 = 14700 * 10 = 1.47 * 10^3 N

Elevator Kinetic Energy = 1/2 mv^2 = 1/2*1500*18.53 = 13897.5J

Elevator Speed after rising to 10m ,

U^2 = u^2 +2as = 1.2 +2*0.87*10 = 18.6

U =(18.6)^1/2=4.312 m/s

Learn more about Kinetic Energy here

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6 0

1 year ago

If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t

solong [7]

Refer to the diagram shown below.

Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.

The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m

The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²

The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s

Answer: 1.375 s