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allsm [11]
3 years ago
10

Pros and Cons of the Mercator projection as a scientific model.

Physics
1 answer:
Andrew [12]3 years ago
6 0

Answer:

ok i don't know the answer but I send you tomorrow

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A rocket sled accelerates from 10 m/s to 40 m/s in 2 seconds. What is the average acceleration of the sled?
Dafna11 [192]

Answer:

15m/s²

Explanation:

Given parameters:

Initial velocity  = 10m/s

Final velocity  = 40m/s

Time taken  = 2s

Unknown:

Average acceleration  = ?

Solution:

Acceleration is the rate of change of velocity with time;

 Acceleration  = \frac{Final velocity   -  Initial velocity }{time}  

  Acceleration = \frac{40 - 10}{2}    = 15m/s²

6 0
3 years ago
The x-axis of a trajectory represents its _____.
sineoko [7]
The x-acis of a trajectory represents its C
3 0
3 years ago
Read 2 more answers
Machmer Hall is 400 m North and 180 m West of Witless.
yan [13]

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

D = \sqrt{(d_{m})^2+(d_{w})^2}

Where, d_{m} =distance of Machmer Hall

d_{w} =distance of Witless

Put the value into the formula

D = \sqrt{(400)^2+(180)^2}

D=438.63\ m

Hence, The distance from Witless to Machmer is 438.63 m.

5 0
3 years ago
What is the formula to calculate the thickness of a test tube?​
Maslowich

Answer:

Having the inside dimensions (ID) and the outside dimensions (OD) will allow you to figure out the wall thickness on tubing. You would need to subtract the ID from the OD and then divide by two. This number is the wall thickness.

Explanation:

4 0
3 years ago
A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i
Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0
3 years ago
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