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IRINA_888 [86]
3 years ago
7

What was the name of the first steam-powered locomotive that was developed by george stephenson?

Physics
1 answer:
Natalija [7]3 years ago
4 0
It was called the Blucher, and was made in 1814.
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Salsk061 [2.6K]

Answer: about 100 km[kilometers] thick

Explanation:

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Two students, Jenny and Cho, are investigating motion.
IgorLugansk [536]

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1: a measuring instruments the students should use for time is a stopwatch

2: a measuring instruments the students should use for distance is a measuring tape

Explanation:

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1 year ago
PLEASE HELP ASAP!!!
musickatia [10]

Answer:

22.2 W

Explanation:

First of all, we calculate the work done by moving the wagon, using the formula:

W=Fdcos \theta

where

F = 20 N is the magnitude of the force

d = 1000 m is the displacement of the wagon

\theta=0^{\circ} is the angle between the direction of the force and of the displacement (assuming the force is applied in the direction of motion)

Substituting, we find

W=(20)(1000)=20,000 J

Now we can find the power generated, which is equal to the ratio between the work done and the time taken:

P=\frac{W}{t}

where

W = 20,000 J

t = 15 min = 900 s

Substituting,

P=\frac{20000}{900}=22.2 W

And the same value in Joules/second (remember that 1 Watt = 1 Joule/second)

5 0
3 years ago
a pillow , a textbook and a paper airplane are dropped from the top of a tall building at the same time. consider what you have
MAVERICK [17]

A textbook would hit the ground first


Factors:

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-Paper airplane will glide to the ground do to its wings and will hit the ground last

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3 years ago
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A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0
Dafna11 [192]

Answer:

(a). The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). The time is 0.493\times10^{-7}\ sec.

Explanation:

Given that,

Electric field E=-6.00\times10^{5}i\ N/C

Time = 5.0 sec

Distance 7.00 cm

(a). We need to calculate the acceleration

Using formula of force

F=F_{e}

ma=qE

a=\dfrac{Eq}{m}

Where, E = electric field

m = mass of proton

q = charge of proton

Put the value into the formula

a=\dfrac{-6.00\times10^{5}\times1.6\times10^{-19}}{1.67\times10^{-27}}

a=-5.74\times10^{13}\ m/s62

The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). We need to calculate the initial peed

Using equation of motion

v^2-u^2=2as

Where, s = distance

Put the value into the formula

0-u^2=2\times(-5.74\times10^{13})\times7.00\times10^{-2}

u=\sqrt{2\times5.74\times10^{13}\times7.00\times10^{-2}}

u=2834783.94

u=2.83\times10^{6}\ m/s

The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). We need to calculate the time interval over which the proton comes to rest

Using formula

t=\dfrac{u}{a}

Where, u = initial velocity

a = acceleration

Put the value into the formula

t=\dfrac{2.83\times10^{6}}{5.74\times10^{13}}

t=0.493\times10^{-7}\ sec

Hence, (a). The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). The time is 0.493\times10^{-7}\ sec.

5 0
3 years ago
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