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madreJ [45]
2 years ago
14

The suspension cable of a 1,000 kg elevator snaps, sending the elevator moving downward through its shaft. The emergency brakes

of the elevator stop the elevator shortly before it reaches the bottom of the shaft. If the elevator fell a distance of 100 m starting from rest, the heat that the brakes must dissipate to bring the elevator safely to rest is (A) 100 J (B) 1,000 J (C) 10,000 J (D) 100,000 J (E) 1,000,000 J
Physics
1 answer:
tester [92]2 years ago
4 0

Answer:

option (E) 1,000,000 J

Explanation:

Given:

Mass of the suspension cable, m = 1,000 kg

Distance, h = 100 m

Now,

from the work energy theorem

Work done by the gravity = Work done by brake

or

mgh = Work done by brake

where, g is the acceleration due to the gravity = 10 m/s²

or

Work done by brake  = 1000 × 10 × 100

or

Work done by brake = 1,000,000 J

this work done is the release of heat in the brakes

Hence, the correct answer is option (E) 1,000,000 J

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A wire of 1mm diameter and 1m long fixed at one end is stretched by 0.01mm when a lend of 10 kg is attached to its free end.calc
Otrada [13]

Answer:

E = 1.25×10¹³ N/m²

Explanation:

Young's modulus is defined as:

E = stress / strain

E = (F / A) / (dL / L)

E = (F L) / (A dL)

Given:

F = 10 kg × 9.8 m/s² = 98 N

L = 1 m

dL = 10⁻⁵ m

A = π/4 (0.001 m)² = 7.85×10⁻⁷ m²

Solve:

E = (98 N × 1 m) / (7.85×10⁻⁷ m² × 10⁻⁵ m)

E = 1.25×10¹³ N/m²

Round as needed.

5 0
2 years ago
A hungry 169169 kg lion running northward at 77.377.3 km/hr attacks and holds onto a 31.731.7 kg Thomson's gazelle running eastw
navik [9.2K]

Answer:  75,242.9 m/s

Explanation:

from the question we are given the following parameters

mass of Lion (ML) = 169,169 kg

velocity of lion (VL) = 777,377.7 m/s

mass of Gazelle (Mg) = 31,731.7 kg

velocity of Gazelle (Vg) = 63,863.8 kg

mass of Lion and Gazelle (M) = 200,900.7 kg

velocity of Lion and Gazelle (V) = ?

The first figure below shows the motion of the Lion and Gazelle with their direction.

The second diagram shows the motion of the Lion and Gazelle with their directions rearranged to form a right angle triangle.

from the triangle formed we can get the velocity of the Lion and Gazelle immediately after collision using their momentum and Phytaghoras theorem

momentum = mass x velocity

momentum of the Lion = 169,169 x 77,377.3 = 13,089,840,463.7 kgm/s

momentum of the Gazelle = 31,731.7 x 63,863.8 = 2,026,506,942.46 kgm/s

momentum of the Lion and Gazelle = 200,900.7  x V

now applying Phytaghoras theorem we have

13,089,840,463.7 + 2,026,506,942.46 =  200,900.7 x V

15,116,347,406.16 = 200,900.7 x V

V = 75,242.9 m/s

7 0
2 years ago
Read 2 more answers
A man walks 30 m to the west, then 5 to the east in 45 seconds
exis [7]
His average speed is  (35m/45s) = 7/9 meters per second.

His average velocity is  (30m W + 5m E) / (45s) = 25 m/s West .
8 0
2 years ago
A rocket in deep space has an exhaust-gas speed of 2000 m/s. When the rocket is fully loaded, the mass of the fuel is five times
notka56 [123]

Answer:

 v_{f} = 1,386 m / s

Explanation:

Rocket propulsion is a moment process that described by the expression

       v_{f} - v₀ =  v_{r} ln (M₀ / Mf)

Where v are the velocities, final, initial and relative and M the masses

The data they give are the relative velocity (see = 2000 m / s) and the initial mass the mass of the loaded rocket (M₀ = 5Mf)

We consider that the rocket starts from rest (v₀ = 0)

At the time of burning half of the fuel the mass ratio is that the current mass is    

       M = 2.5 Mf

       v_{f} - 0 = 2000 ln (5Mf / 2.5 Mf) = 2000 ln 2

       v_{f} = 1,386 m / s

3 0
2 years ago
A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the
Andrej [43]

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15°

velocity = 15m/s

Acceleration = 1.5 m/s²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes  m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ  

F causes the acceleration of the truck: F= m*a

This gives the equation:

T – m1*gsinθ = m1*a  

T = m1(a + gsinθ)

Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:

m2*g*sinθ = (7100 – m2)*a

m2*g*sinθ = 7100a – m2a

m2*gsinθ + m2*a = 7100a

m2* (gsinθ + a) = 7100a

m2 = 7100a/(gsinθ  + a)

m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)

m2 = 2636.8 kg

The load has a mass of 2636.8 kg

6 0
2 years ago
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