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Umnica [9.8K]
4 years ago
13

Who invented steam engine​

Physics
2 answers:
mr Goodwill [35]4 years ago
4 0

Thomas Savery

While the Spaniard first patented a steam-operated machine for use in mining, an Englishman is usually credited with inventing the first steam engine. In 1698, Thomas Savery, an engineer and inventor, patented a machine that could effectively draw water from flooded mines using steam pressure.

Neporo4naja [7]4 years ago
3 0

Answer:

It was invented by Thomas Newcomen around 1712.

Explanation:

This would play a major role in cars.

You might be interested in
Name the four fundamental fores at work inside an atom. Tell what each one does.​
STALIN [3.7K]

Answer:

Four fundamental forces are gravitational, electromagnetic, strong, and weak.

Explanation:

The gravitational and electromagnetic interactions, which produce significant long-range forces whose effects can be seen directly in everyday life and the strong and weak interactions, which produce forces at minuscule, subatomic distances and govern nuclear interactions.

6 0
3 years ago
What is the difference between balanced force and action reaction force
Vilka [71]

Answer:

Balanced forces are equal and opposite forces that act on the same object. ... Action-reaction forces are equal and opposite forces that act on different objects, so they don't cancel out. In fact, they often result in motion.

5 0
3 years ago
Read 2 more answers
Light travels at a speed of 2.998*108 m/s. Light takes approximately 3.25 minutes to travel from the Sun to reach a planet. Calc
ANTONII [103]

Answer:

585×10⁸ m

Explanation:

Distance = rate × time

d = (2.998×10⁸ m/s) (3.25 min) (60 s/min)

d = 585×10⁸ m

5 0
3 years ago
Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2
aleksandr82 [10.1K]

Answer:

The curl is 0 \hat x -z^2 \hat y -4xy \hat z

Explanation:

Given the vector function

\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z

We can calculate the curl using the definition

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|

Thus for the exercise we will have

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|

So we will get

\nabla  \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z

Working with the partial derivatives we get the curl

\nabla  \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z

6 0
4 years ago
An open 1-m-diameter tank contains water at a depth of 0.5 m when at rest. As the tank is rotated about its vertical axis the ce
Morgarella [4.7K]

Answer:

Angular velocity (w) = 8.86 rad/s

Explanation:

Angular velocity (w) = \sqrt{} 4ghi/R^{2}

g= 9.81 m/s

R= 0.5

hi (initial depth) = 0.5m

Hence= \sqrt4* 9.81* 0.5/0.5^{2}  = 8.86 rad/s

3 0
3 years ago
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