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Stolb23 [73]
2 years ago
7

30. The length of mercury thread when it is at 0°C, 100°C and at an unknown temperature 0 is 25mm, 225mm and 175mm respectively.

What is the value of 02 A. 85.0°C B. 80.0°C C. 75.0°C D. 70.0°C ​
Physics
1 answer:
yulyashka [42]2 years ago
3 0

Answer:

75°C

Explanation:

175−25/225−25=x−0/100−0

150/200=x/100

x=150×100/200

= 75°C

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Two paths lead to the top of a big hill. One is steep and direct, while the other is twice as long but less steep. How much more
umka2103 [35]

Answer:

None.

Explanation:

  • Gravitational potential energy, depends only of the mass on which the gravity is doing work, and the displacement produced by this force.
  • As displacement depends only on the final and initial positions (in this case, the height of the hill), if we choose as our zero reference level the bottom of the hill, the change in gravitational potential energy will be as follows:

       \Delta U = U_{f} -U_{0} = m*g*h - 0  = m*g*h

  • As we can see, the only value of distance involved is the height of the hill , so it is independent of the distance travelled.
6 0
3 years ago
Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4
kati45 [8]

Answer:

velocity  = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = \frac{\pi }{4} × (4.50×10^{-2})²  × velocity

solve it we get

velocity  = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = \frac{\pi }{4} × 3 ×  (4.50×10^{-2})²  × velocity

velocity = 52.4 m/s

5 0
3 years ago
A uniform rod of length L is pivoted at L/4 from one end. It is pulled to one side through a very small angle and allowed to osc
ludmilkaskok [199]

Answer:

T= 4.24sec

Explanation:

We are going to use the formula below to calculate.

T=2\pi \sqrt{\frac{L}{g} }

Where T is period

           L is length of rod

       g is acceleration due to gravity =     9.8m/s^{2}

From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this L_{O}

L_{O} = 3/4 * 5.95m

        = 4.4625m

thus   T=2\pi \sqrt{\frac{L_{O} }{g} }

          T=2\pi \sqrt{\frac{4.4625 }{9.8} }

          T= 4.24sec

8 0
3 years ago
Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea
Kipish [7]

Answer:

6.5 m/s

Explanation:

We are given that

Distance, s=100 m

Initial speed, u=1.4 m/s

Acceleration, a=0.20 m/s^2

We have to find the final velocity at the end of the 100.0 m.

We know that

v^2-u^2=2as

Using the formula

v^2-(1.4)^2=2\times 0.20\times 100

v^2-1.96=40

v^2=40+1.96

v^2=41.96

v=\sqrt{41.96}

v=6.5 m/s

Hence, her final velocity at the end of the 100.0 m=6.5 m/s

5 0
2 years ago
Technician A says the manual lever position (MLP) switch is sometimes called a transmission range switch or a neutral safety swi
Phantasy [73]

Answer:

i think it is Technician a

3 0
3 years ago
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