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Crank
2 years ago
14

Express each of the following in ms -1 a) 18kmh-1​

Physics
1 answer:
vovangra [49]2 years ago
3 0
1km=1000m; 1hr=3600secs
1km/hr=1000/3600= 5/18m/sec
To convert km/hr into m/sec, multiply the number by 5 and then divide it by 18.

18kmh-1= 18•5=90
90/18=5
5ms-1
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Read 2 more answers
A piece of steel expands 10 cm when ur is heated from 20 to 50 degrees Celsius. How much would it expand if it was heated from 2
faltersainse [42]

The steel would expand by 4. 8 * 10^-3 cm

<h3>How to determine the linear expansion</h3>

The change in length  ΔL is proportional to length  L. It is dependent on the temperature, substance, and length.

Using the formula:

ΔL= α LΔT

where  ΔL  is the change in length  L = 10cm

ΔT  is the change in temperature = 60° - 20° = 40° C

 α  is the coefficient of linear expansion = 1.2 x 10^-5 °C

Substitute into the formula

ΔL = 1.2 * 10^-5 * 10 * 40

ΔL = 4.8 * 10 ^-3 cm

Therefore, the steel would expand by 4. 8 * 10^-3 cm

Learn more about linear expansion here:

brainly.com/question/14325928

#SPJ1

8 0
1 year ago
Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of
Anna [14]

Given data:

* The extension of the steel wire is 0.3 mm.

* The length of the wire is 4 m.

* The area of cross section of wire is,

A=2\times10^{-6}m^2

* The young modulus of the steel is,

Y=2.1\times10^{11}\text{ Pa}

Solution:

The young modulus of the steel in terms of the force and extension is,

Y=\frac{F\times l}{A\times dl}

where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,

Substituting the known values,

\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}

Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.

7 0
10 months ago
A projectile is fired directly upwards at 49.4 m/s. A second projectile is dropped from rest at some higher elevation at the ins
8090 [49]

Answer:

Explanation:

velocity of first projectile after 3 s

v = u - gt

v = 49.4 - 9.8 x 3

= 20 m /s

Velocity of second projectile after 3 s after being dropped from rest

v = u + gt

= 0 + 9.8 x 3

= 29.4 m /s

They will be moving in opposite direction at the time of meeting , so their relative velocity

= 20 + 29.4 = 49.4 m /s

From the frame of reference of the first projectile, the velocity of the second projectile will be 49.4 m /s .

8 0
2 years ago
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