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

What is the thickness of the Lithosphere

Salsk061 [2.6K]

Answer: about 100 km[kilometers] thick

Explanation:

7 0

3 years ago

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Two students, Jenny and Cho, are investigating motion.

IgorLugansk [536]

Answer:

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:

pls mark brainliest

8 0

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:

-Textbook weighs most

-Pillow is flat and fluffy not very aerodynamic) also is very light

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

3 0

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