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d1i1m1o1n [39]
3 years ago
13

Consider a small car of mass 1200 kg and a large sport utility vehicle (SUV) of mass 4000 kg. The SUV is traveling at the speed

limit (v ???? 35 m/s). The driver of the small car travels so as to have the same kinetic energy as the SUV. Find the speed of the small car.
Physics
1 answer:
Karolina [17]3 years ago
6 0

Answer:

63.9 m/s

Explanation:

Parameters given:

Mass of small car, m = 1200 kg

Mass of SUV, M = 4000 kg

Speed of SUV, V = 35 m/s

Their kinetic energy of the small car is equal to the kinetic energy of the SUV, hence:

0.5 * m * v² = 0.5 * M * V²

=> 0.5 * 1200 * v² = 0.5 * 4000 * 35²

600 * v² = 2450000

v² = 2450000/600

v² = 4083.3

=> v = 63.9 m/s

The speed of the small car is 63.9 m/s.

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A 65-kg bungee jumper, who is attached to one end of an 85-m long bungee cord that has its other end tied to a bridge, jumps off
irinina [24]

Answer:

The impulse delivered to the bungee jumper is 1.32 kN.s

Explanation:

The situation can be shown graphically as shown in the figure.

Impulse delivered to the bungee jumper = Area under the curve.

The curve represents a triangle and the area of traiangle = (1/2)base×height

The base of the triangle from the graph = 1.2 seconds.

The height of the triangle from the graph = 2.2 kN

Thus,

<u>Impulse = (1/2)×(1.2 seconds)×(2.2 kN) = 1.32 kN.s</u>

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3 years ago
A european car manufacturer reports that the fuel efficiency of the new microcar is 28.5 km/l highway and 22.0 km/l city. what a
Studentka2010 [4]

We know that:

1 mile = 1.61 km

1 gal = 3.8 L

 

Therefore converting the fuel efficiency rates:

highway = (28.5 km/L) * (1 mile / 1.61 km) * (3.8 L / 1 gal) = 67.27 mile / gal

<span>city = (22.0 km/L) * (1 mile / 1.61 km) * (3.8 L / 1 gal) = 51.93 mile / gal</span>

4 0
3 years ago
A cord of negligible mass runs around two massless, frictionless pulleys. A canister with mass m = 20 kg hangs from one pulley.
photoshop1234 [79]

(a) 196 N

The equation of the forces on the side of the cord where the force F is applied is:

F-T=0 (1)

where T is the tension in the cord.

On the other side of the cord, the equation of the forces on the canister is

T-mg = ma (2)

where

m = 20 kg is the mass of the canister

g=9.8 m/s^2 is the acceleration of gravity

a is the acceleration

From (1),

T=F

Substituting into (2),

F-mg = ma\\F=m(g+a)

We want the canister to move at constant speed, so

a = 0

And therefore:

F=mg=(20)(9.8)=196 N

b) 2.0 cm

The cord is inextensible, this means that the acceleration of its parts are the same. Therefore, the acceleration of the free end must be the same as the acceleration of the canister: and this means that the two parts also cover the same distance in the same time.

Therefore, the free end of the cord must be moved exactly the same as the canister, by 2.0 cm.

c) 3.92 J, the same

The work done by the tension in the cord is

W_T = T d

where

T is the tension

d = 2.0 cm = 0.02 m is the displacement

As we said in part (a), the tension in the cord is equal to the force applied to the free end:

T = F

So

T = 196 N

Therefore, the work done by the tension is

W=(196)(0.02)=3.92 J

And since the force applied (F) is the same, then the work done by you when pulling the cord is exactly the same.

(d) -3.92 J

The weight of the canister is

F_g = mg =(20 kg)(9.8 m/s^2)=196 N

However, the direction of the force of gravity is opposite to the displacement. Therefore, the work done by gravity is negative:

W_g = - F_g d

And substituting,

W_g=-(196)(0.02)=-3.92 J

(e) Zero

The net work done on the canister can be simply calculated by adding the work done by the tension in the cord and the weight of the canister:

W=W_T+W_g = 3.92 + (-3.92 ) = 0

This is in agreement with the work-energy theorem, which states that the work done on an object is equal to its change in kinetic energy. In this situation, the canister is moving at constant speed, so its kinetic energy is not change: therefore,

\Delta K = 0 (change in kinetic energy = 0)

and so, the work done on it is also zero.

(f) The pulley system changes the direction of the force applied

This is a simple pulley system, which means  that the system does not multiply the force applied in input. In fact, the mechanical advantage of the system is

MA=\frac{F_{out}}{F_{in}}

where:

F_{out} is the output force, which is the weight of the canister

F_{in} is the force in input, which is F

So, the mechanical advantage is 1:

MA=\frac{196 N}{196 N}=1

From a point of view of energy, therefore, there is no advantage in this system.

However, the advantage offered by the pulley system concerns the direction of the force: in fact, it changes the direction of the applied force (which is F, downward) into the tension of the cord (which is upward on the canister).

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Answer:

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Explanation:

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3 years ago
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(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
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