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3 years ago

What type of pipe wrenches are designed for turning and holding where marrying is not objectionable?

1 answer:

3 0

The type of pipe wrenches that are designed for turning and holding where marrying is not objectionable is called the chain wrench. This tool has two serrated that are adjusted to grip the pipe. This is very useful for those objects where marrying is not objectionable since this can be adjusted for that purpose.

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Suppose a car travels 106 km at a speed of 28 m/s and uses 1.9 gals of gasoline in the process. Only 30% of the gasoline goes in

USPshnik [31]

Answer:

a) The magnitude of the force is 968 N

b) For a constant speed of 30 m/s, the magnitude of the force is 1,037 N

Explanation:

NOTE: The question b) will be changed in other to give a meaningful answer, because it is the same speed as the original (the gallons would be 1.9, as in the original).

Information given:

d = 106 km = 106,000 m

v1 = 28 m/s

G = 1.9 gal

η = 0.3

Eff = 1.2 x 10^8 J/gal

a) We can express the energy used as the work done. This work has the following expression:

$W=F\cdot d$

Then, we can derive the magnitude of the force as:

$F=\frac{W}{d}=\frac{\eta\cdot (G\cdot Eff)}{d}=\frac{0.3*1.9*(1.8*10^8)}{106*10^3} =968\,N$

b) We will calculate the force for a speed of 30 m/s.

If the force is proportional to the speed, we have:

$F_2=F_1(\frac{v_2}{v_1} )=968(\frac{30}{28} )=968*1.0714=1,037\,N$

6 0

3 years ago

What is the ideal banking angle for a gentle turn of 1.20-km radius on a highway with a 105 km/h speed limit (about 65 mi/h), as

Mnenie [13.5K]

Answer:

4.14°

Explanation:

given:

r = 1.2 km

v = 105 km/h

1) convert your given

a) r = 1.2 km to m = 1200m

b) v = 105 km/h to m/s = 29.2 m/s

2) plug into your ideal banking angle equation

$tan^-1$ ( $\frac{v^2}{rg}$ ) = $\frac{29.2^2}{(1200)(9.8)}$ = 4.14°

8 0

2 years ago

What is weight?

OverLord2011 [107]

Answer:

Explanation:

The answer is C) the mass of an object

8 0

3 years ago

Read 2 more answers

A 10.0 g bullet moving at 300m/s is fired into a 1.00 kg block at rest. The bullet emerges (the bullet does not get embedded in

chubhunter [2.5K]

Answer:

v' = 1.5 m/s

Explanation:

given,

mass of the bullet, m = 10 g

initial speed of the bullet, v = 300 m/s

final speed of the bullet after collision, v' = 300/2 = 150 m/s

Mass of the block, M = 1 Kg

initial speed of the block, u = 0 m/s

velocity of the block after collision, u' = ?

using conservation of momentum

m v + Mu = m v' + M u'

0.01 x 300 + 0 = 0.01 x 150 + 1 x v'

v' = 0.01 x 150

v' = 1.5 m/s

Speed of the block after collision is equal to v' = 1.5 m/s

5 0

3 years ago

A pendulum has a mass of 1.5 kg and starts at a height of 0.4 m. If it is released from rest, how fast is it going when it reach

ella [17]

Answer:

A. 2.8 m/s

Explanation:

Suppose that at the height of 0 m, the path of the pendulum is lowest.

If we use law of conservation of energy, the pendulum will have zero kinetic energy or K.E when it is at highest point, because K.E happens during movement of object and at the highest point all the energy will be P.E

P.E= mgh

Similarly, when the pendulum reaches at the lowest point, the height becomes zero and the P.E also becomes zero. Now all the energy will be K.E

K.E= 1/2 m v^2

In question, we are asked about the speed as the pendulum it reaches the lowest point of its path. Like we mentioned P.E will be zero at lowest point because of zero height. And also we will use law of conservation of energy because no energy has been lost from system.

K.E= P.E

1/2 m v^2 = mgh

Taking sq.root at both sides

v= Under root 2 gh

v=Under root 2x 9.8 m/s x0.4 m

v=Under root 7.84

v=2.8 m/sec

Hope it helps!

3 0

3 years ago