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Alex787 [66]
3 years ago
8

How much time is needed for a 15,000 w engine to do 1,800,000 j of work?

Physics
2 answers:
Hatshy [7]3 years ago
8 0
We know that;
power=W/t
15000Watts=1800000J/t
t=1800000J/15000Watts
t=120s
OlgaM077 [116]3 years ago
6 0

Answer:

<h3>The Right Answer is <u>120 s </u></h3><h3><u>C</u></h3><h3><u></u></h3>

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A current is flowing in a wire in direction 3i + 4j where the direction of the magnetic field is 5j + 12k. The force on the wire
nasty-shy [4]

Answer:

F = 0.768 i ^ - 0.576 j ^ + 0.24 k ^

the correct answer is "b"

Explanation:

The magnetic force is

          F = i l x B

The bold are vectors, in this case they give us the direction of the current and the magnetic field, for which we can solve as a determinant

         

F = i \left[\begin{array}{ccc}x&y&z\\3&4&0\\0&5&12\end{array}\right]

resolver

     F = i ^ (4 12 - 0) + j ^ (0- 3 12) + k ^ (3  5 - 0)

     F = i (48 i ^ - 36 j ^ + 15 k⁾

in this case i is the value of the current flowing through the cable

     i = 16 mA = 0.016 A

     F = 0.768 i ^ - 0.576 j ^ + 0.24 k ^

When reviewing the different answers, the correct answer is "b"

6 0
3 years ago
Oil (SAE 30) at 15.6 oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the c
solong [7]

Answer:

q = 0.0003649123 m²/s = (3.65 × 10⁻⁴) m²/s

Explanation:

For laminar flow between two parallel horizontal plates, the volumetric flow per metre of width is given as

q = (2h³/3μ) (ΔP/L)

h = hydraulic depth = 4mm/2 = 2mm = 0.002 m

μ = viscosity of oil (SAE 30) at 15.6°C = 0.38 Pa.s

(ΔP/L) = 26 KPa/m = 26000 Pa/m

q = (2h³/3μ) (ΔP/L)

q = (26000) × (2(0.002³)/(3×0.38))

q = 0.0003649123 m²/s = (3.65 × 10⁻⁴) m²/s

5 0
3 years ago
Read 2 more answers
A warm hockey puck has a coefficient of restitution of 0.50, while a frozen hockey puck has a coefficient of restitution of only
AlexFokin [52]

Answer:

the required height is 0.2449 m only

Explanation:

Given the data in the question;

Initial height = 2m

so speed of the puck before hitting the ground will be;

u² = 2gh

Initial speed u_ball = √2gh

u_ball = √( 2 × 9.8 × 2 )

u_ball =  √39.2

u_ball = 6.26 m/s

given that; FOR THE FROZEN PUCK, coefficient of restitution = 0.35 only

R = - (v_ball - v_ground / u_ball - u_ ground)

so

0.35 = - (v_ball - 0 / 6.26 - 0)

0.35 = -v_ball / - 6.26

-v_ball = 0.35 × (- 6.26)

-v_ball = -2.191 m/s

v_ball = 2.191 m/s

to get the height;

v² = 2gh

h = v² / 2g

we substitute

h = (2.191)² / 2×9.8

h = 4.800481 / 19.6

h = 0.2449 m

Therefore, the required height is 0.2449 m only

5 0
3 years ago
A 3.00 kg crate is pulled along a frictionless surface with a force of 55.0 N at 25.00 to the
Xelga [282]
0.161 is the answer by the way
7 0
3 years ago
Read 2 more answers
A spring is compressed so that it has 7.2 J of elastic potential energy. A 0.3 kg ball is placed on top of the spring. When the
VLD [36.1K]

The height of the ball is 2.45 m.

To calculate the height of the ball, we use the formula below.

<h3>Formula</h3>
  • E = mgh................ Equation 1

Where:

  • E = Elastic energy of the spring
  • m = mass of the ball
  • h = height of the ball
  • g = acceleration due to gravity.

make h the subject of the equation

  • h = E/mg...............Equation 2

From the question,

Given:

  • E = 7.2 J
  • m = 0.3 kg
  • g = 9.8 m/s²

Substitute these  values into equation 2

  • h = 7.2/(0.3×9.8)
  • h = 7.2/2.94
  • h = 2.45 m

Hence, The height of the ball is 2.45 m.

Learn more about height here: brainly.com/question/1739912


5 0
3 years ago
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