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astra-53 [7]
3 years ago
5

If an object moves 80 meters North along a straight path for 4 seconds, what is the average velocity?

Physics
1 answer:
scoundrel [369]3 years ago
3 0

Velocity = Distance / Time

Distance is given to be 80 meters

Time is given to be 4 seconds

Velocity = 80 meters / 4 seconds = 20 meters / second

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What is the length of an aluminum rod at 65°C if its length at 15°C is 1.2 meters? A. 0.00180 meter B. 1.201386 meters C. 1.2
Deffense [45]

Answer:

Option B is the correct answer.

Explanation:

Thermal expansion

            \Delta L=L\alpha \Delta T

L = 1.2 meter

ΔT = 65 - 15 = 50°C

Thermal Expansion Coefficient for aluminum, α = 24 x 10⁻⁶/°C

We have change in length

          \Delta L=L\alpha \Delta T=1.2\times 24\times 10^{-6}\times 50=1.44\times 10^{-3}m

New length = 1.2 + 1.44 x 10⁻³ = 1.2014 m

Option B is the correct answer.

3 0
3 years ago
Read 2 more answers
Ma puteti ajuta sa rezolv la fizica o problema ?
xxTIMURxx [149]
Care este problema? Btw why are you speaking Romanian 
7 0
3 years ago
What is the sum of 9260 and 3240?
Ket [755]
The sum is the result of adding 9260 and 3240 together. Each number can be broken down into constituent parts in order to make addition easier. Each place in the number represents its value, so a 2 in the hundreds place represents 200.
You can separate numbers out this way to make it easier to add them. 9260 can be broken down into 9000+200+60 while 3240 is 3000+200+40. You can then add these six numbers together.

60+40 = 100
200+200 = 400
9000+3000 = 12000

Then add your three partial results together to receive the final answer:

12000+400+100 = 12500
4 0
3 years ago
Help!! Please!!
sattari [20]
<span> the answer is 1,500 & 1,700


</span>
6 0
3 years ago
An 56 kg sled is being pulled across the snow, at constant speed,by a horizontal force of 15 N, find the coefficient of kinetic
Stolb23 [73]

Answer:

The coefficient of  kinetic friction = 0.026  

Explanation:

An 56 kg sled is being pulled across the snow, at constant speed,by a horizontal force of 15 N.

Here we have to note that the weight is pulled at a constant speed . This means that the net force acting on the weight is zero.

The external force acting on the body is in the forward direction and the friction acts in the backward direction.

Friction increases as the mass of the body increases.

Friction = u_{k}\times m \times g

We now equate this to the external force of 15 N.

15 = u_{k} \times 56 \times 10

u_{k} = \frac{15}{560}

u_{k} = 0.026

The coefficient of  kinetic friction = 0.026  

8 0
3 years ago
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