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makkiz [27]
3 years ago
7

Thomas Newcomen was the first to produce a working steam engine. Why is the work of James Watt more widely known than the work o

f Newcomen?
Physics
1 answer:
LenaWriter [7]3 years ago
4 0

Answer:

The steam engine of James watt is more efficient than Newcomen ans more suitable for the industrial revolution.

Explanation:

James Watt is more widely know for working steam engine because Watt has created better engine which is suitable for the industrial revolution. The steam engine of James watt is more efficient than Newcomen. Watt developed the condensing arrangement by using piston which lessen the initial pressure leading to effectively worked than Newcomen's

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The Jamaican bobsled team was moving at a velocity of 50 m/s, then they hit the brakes on their sled to decelerate at a uniform
atroni [7]

Answer:

The time it took the bobsled to come to rest is 10 s.

Explanation:

Given;

initial velocity of the bobsled, u = 50 m/s

deceleration of the bobsled, a = - 5 m/s²

distance traveled, s = 250 m

Apply the following kinematic equation to determine the time of motion of the bobsled;

s = ut + ¹/₂at²

250 = 50t + ¹/₂(-5)t²

250 = 50t - ⁵/₂t²

500 = 100t - 5t²

100 = 20t -t²

t² - 20t + 100 = 0

t² -10t - 10t + 100 = 0

t (t - 10) - 10(t - 10) = 0

(t - 10)(t - 10) = 0

t = 10 s

Therefore, the time it took the bobsled to come to rest is 10 s.

3 0
3 years ago
A standing wave of the third overtone is induced in a stopped pipe, 3 m long. The speed of sound is 340 m/s. The number of antin
Leokris [45]

Answer:

overtone- one over the first

n skips by twos

4 antinodes

500 Hz

Explanation: Hope this helps :)

7 0
3 years ago
Explain how and why energy is important to living things. Your answer to this question will become the claim for your scientific
Leto [7]

Answer:

All living organisms need energy to grow and reproduce, maintain their structures, and respond to their environments. Metabolism is the set of life-sustaining chemical processes that enables organisms transform the chemical energy stored in molecules into energy that can be used for cellular processes.

Explanation:

7 0
3 years ago
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
At very close range, which of these forces is stronger than the gravitational force a.) electric only, b.) magnetic only c.) ele
zhenek [66]

Answer: (c) electric and magnetic

Explanation: Gravitational force is the weakest force considering short distances, relative to all other types of forces. However, it is strongest when considering large distances (such as astronomic scales).

7 0
3 years ago
Read 2 more answers
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