An attempt to explain observations of the natural world is

(a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/ s?

Aleksandr-060686 [28]

Answer:

1.4719 m per sec

Explanation:

Hello

Kinetic energy is the energy associated with the movement of objects. Although there are many forms of kinetic energy

the formula to use is

$E=\frac{mv^{2} }{2}$

where m is the mass of the object and v the velocity

lets see the kinetic energy of the sprinter running

$E=\frac{65 Kg*10(\frac{m}{s} ^)){2} }{2} \\\\E=\frac{65 *100 }{2} \\E=3250 Joules\\\\$

Now, the elephant must have the same kinetic energy

$E=\frac{m*v_{2} ^{2} }{2} \\\\E*2=m*v_{2} ^{2}\\ \frac{2E}{m} =v_{2} ^{2} \\\sqrt{\frac{2E}{m} } =v_{2} \\\\\\v_{2} =\sqrt{\frac{2*3250}{3000} }\\ \\v_{2} =1.4719 \frac{m}{s} \\\\$

it works only the positive root, so the elephant must to walk to 1.4719 m/s to have the same kinetic energy.

Have a great day

8 0

3 years ago

Ok plz answer and tell me how to do it

kirza4 [7]

Answer: 25N

method: total force in the right hand direction is 100N and total force in the left hand direction is 125N. To get the net force, we add forces if they are in the same direction and substract if they are in opposite directions. since 100N and 125N are in opposite directions, we substract the larger value from the smaller value. Then we get 25N in the left hand direction as the final answer.

4 0

3 years ago

Which device may be used to throttle and vaporize liquid refrigerant before it enters the compressor?

elena55 [62]

Answer: An expansion valve.

Explanation:
The expansion valve is a copper tubing that contains a small orifice. The orifice restricts the flow of the refrigerant and makes it undergo a severe drop in pressure and temperature when it exits the orifice.
The temperature drop provides the cooling required in the cooling coils of the refrigerator.

3 0

3 years ago

Bumper cars A and B undergo a collision during which the momentum of the combined system is conserved.

dexar [7]

Answer:

1. P⃗ A,i+P⃗ B,i=P⃗ A,f+P⃗ B,f

Explanation:

Collision occurs when two bodies moving at different velocities exert force on each other through colliding. Collision can either be elastic or inelastic.

For elastic collision, both energy and momentum of the bodies is conserved i.e they separates after collision.

For inelastic collision, only momentum is conserved but not energy. The body sticks together after colliding and moves with a common velocity.

According law of conservation of momentum which states that the sum or momentum of two bodies before collision is equal to the momentum of the bodies after collision.

Given two bumper cars A and B that undergoes collision during which momentum is conserved, the type of collision that exists between the bumpers is inelastic collision

.

If initial momentum of A is PiA and initial momentum of B is PiB, the sum of their momentum before collision is expressed as;

PiA + PiB ... (1)

If final momentum of A is PfA and initial momentum of B is PfB, the sum of their momentum after collision is expressed as;

PfA+PfB ... (2)

According to the law, equation 1 is equal to equation 2

The fore the formula that states the law is expressed as;

PiA + PiB = PfA+PfB

5 0

3 years ago

A 0.160kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.820m/s . It has a head-on coll

IRISSAK [1]

Answer:

A) v_{f1} = -3.2 m / s, B) LEFT , C) v_{f2} = -0.12 m / s, D) LEFT

Explanation:

This is a collision exercise that can be solved using momentum conservation, for this we define a system formed by gliders, so that the forces during the collision are internal and the moment is conserved.

Let's use the subscript 1 for the lightest glider m1 = 0.160 kg and vo1 = 0.820 m / s

subscript 2 for the heaviest glider me² = 0.820 kg and vo2 = -2.27 m / s

Initial instant. Before the crash

p₀ = m₁ v₀₁ + m₂ v₀₂

Final moment. After the crash

p_f = m₁ v_{f1} + m₂ v_{f2}

p₀ = p_f

m₁ v₀₁ + m₂ v₀₂ = m₁ v_{f1} + m₂ v_{f2}

as the shock is elastic, energy is conserved

K₀ = K_f

½ m₁ v₀₁² + ½ m₂ v₀₂² = ½ m₁ $v_{f1}^2$ + ½ m₂ $v_{f2}^2$

m₁ (v₀₁² - v_{f1}²) = m₂ (v_{f2}² -v₀₂²)

let's make the relationship

(a + b) (a-b) = a² -b²

m₁ (v₀₁ + v_{f1}) (v₀₁-v+{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

let's write our two equations

m₁ (v₀₁ -v_{f1}) = m₂ (v_(f2) - v₀₂) (1)

m₁ (v₀₁ + v_{f1}) (v₀₁-v_{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

we solve

v₀₁ + v_{f2} = v_{f2} + v₀₂

we substitute in equation 1 and obtain

M = m₁ + m₂

$v_{f1} = \frac{m_1-m_2}{M} v_o_1 + 2 \frac{m_2}{M} v_f_2$

$v_f_2 = \frac{2m_1}{M} v_o_1 + \frac{m_2-m_1}{M} v_o_2$ vf2 = 2m1 / mm vo1 + m2-m1 / mm vo2

we calculate the values

m₁ + m₂ = 0.160 +0.3000 = 0.46 kg

v_{f1} = $\frac{ 0.160 -0.300} {0.460} \ 0.820 + \frac{2 \ 0300}{0.460} \ (-2.27)$

v_{f1} = -0,250 - 2,961

v_{f1} = - 3,211 m / s

v_{f2} = $\frac{2 \ 0.160}{0.460} \ 0.820 + \frac{0.300 - 0.160}{0.460 } \ (-2.27)$

v_{f2} = 0.570 - 0.6909

v_{f2} = -0.12 m / s

now we can answer the different questions

A) v_{f1} = -3.2 m / s

B) the negative sign indicates that it moves to the left

C) v_{f2} = -0.12 m / s

D) the negative sign indicates that it moves to the LEFT

3 0

3 years ago