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Sergio039 [100]
2 years ago
11

How many calories are equal to one BTU? (One calorie = 4.186 J, one BTU = 1 054 J.)

Physics
1 answer:
I am Lyosha [343]2 years ago
3 0
<h2>Option C is the correct answer.</h2>

Explanation:

We need to find how many calories is 1 BTU.

Given

          1 BTU = 1054 J

          1 calorie = 4.186 J

So we have

          1 BTU = 4.186 x 251.79 J

          1 BTU =251.79 calorie

          1 BTU = 252 calorie.

Option C is the correct answer.

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a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres
shusha [124]

Answer:

Part a) When collision is perfectly inelastic

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b) When collision is perfectly elastic

v_m = \frac{m + M}{2m}\sqrt{5Rg}

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

mv_m = (m + M)v

so we have

v = \frac{mv_m}{m + M}

now we know that in order to complete the circle we will have

v = \sqrt{5Rg}

\frac{mv_m}{m + M} = \sqrt{5Rg}

now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

Now we know that collision is perfectly elastic

so we will have

v = \frac{2mv_m}{m + M}

now we have

\sqrt{5Rg} = \frac{2mv_m}{m + M}

v_m = \frac{m + M}{2m}\sqrt{5Rg}

6 0
2 years ago
Small blocks, each with mass m, are clamped at the ends and at the center of a rod of length L and negligible mass. Compute the
fenix001 [56]

Answer:

Answered

Explanation:

a) Two balls are at a distance of L/2 from the axis of rotation and one block at the center. ( center of rod).

therefore,

I=2\times m\frac{L}{2}^2

I= \frac{1}{2}mL^2

b) two balls at a distance L/4 at the from the axis and 1 ball at a distance 3L/4 from the from the axis.

I= 2\times m\times(L/4)^2 + m(\frac{3L}{4})^2

= \frac{1}{16} mL^2(2+9)= \frac{11}{16}mL^2

5 0
2 years ago
A person on a cruise ship is doing laps on the promenade deck. on one portion of the track the person is moving north with a spe
Dmitry [639]
The resultant motion is given by pithagoras, since the two components (north and east) are perpendicular to each other.
They are asking you about the direction so you have to use trigonometry, finding that the direction is Ф=arctan(3.8/12)=17.57° north of east.
4 0
2 years ago
Read 2 more answers
You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
2 years ago
7. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The oncoming water stream has
NNADVOKAT [17]

Answer:

The magnitude of the average force exerted on the water by the blade is 960 N.

Explanation:

Given that,

The mass of water per second that strikes the blade is, \dfrac{m}{t}=30\ kg/s

Initial speed of the oncoming stream, u = 16 m/s

Final speed of the outgoing water stream, v = -16 m/s

We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :

F=\dfrac{\Delta P}{\Delta t}

F=\dfrac{m(v-u)}{\Delta t}

F=30\ kg/s\times (-16-16)\ m/s

F = -960 N

So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.

6 0
2 years ago
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