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olga2289 [7]
1 year ago
14

Mr John slides from rest a height of 12.2m down a smooth plane inclined at an angle 30° to the horizontal,.. calculate the dista

nce covered...?​
Physics
1 answer:
pentagon [3]1 year ago
5 0

Answer:

24.4 m

Explanation:

sin 30 =  12.2 / x

x = 12.2 / sin 30    =  24.4 m

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The position of a particle s given by x=3-2t+3t^2. What is its instantaneous velocity and instantaneous acceleration as t=3.
Leno4ka [110]

Explanation:

velocity = dx/dt

\frac{dx}{dt}  = 6t - 2 = 16

acceleration =

\frac{ {d}^{2}x }{d {t}^{2} }  = 6 \ \\

5 0
2 years ago
Is it true or false
tester [92]

Answer:

false

Explanation:

8 0
2 years ago
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1,
mash [69]

Answer:

a) t₁ = 4.76 s, t₂ = 85.2 s

b) v = 209 ft/s

Explanation:

Constant acceleration equations:

x = x₀ + v₀ t + ½ at²

v = at + v₀

where x is final position,

x₀ is initial position,

v₀ is initial velocity,

a is acceleration,

and t is time.

When the engine is on and the sled is accelerating:

x₀ = 0 ft

v₀ = 0 ft/s

a = 44 ft/s²

t = t₁

So:

x = 22 t₁²

v = 44 t₁

When the engine is off and the sled is coasting:

x = 18350 ft

x₀ = 22 t₁²

v₀ = 44 t₁

a = 0 ft/s²

t = t₂

So:

18350 = 22 t₁² + (44 t₁) t₂

Given that t₁ + t₂ = 90:

18350 = 22 t₁² + (44 t₁) (90 − t₁)

Now we can solve for t₁:

18350 = 22 t₁² + 3960 t₁ − 44 t₁²

18350 = 3960 t₁ − 22 t₁²

9175 = 1980 t₁ − 11 t₁²

11 t₁² − 1980 t₁ + 9175 = 0

Using quadratic formula:

t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22

t₁ = 4.76, 175

Since t₁ can't be greater than 90, t₁ = 4.76 s.

Therefore, t₂ = 85.2 s.

And v = 44 t₁ = 209 ft/s.

3 0
2 years ago
A piece of wood has a mass of 25g and a volume of <br> 10cm3 What is its density?
Tasya [4]

Answer:

<h3>The answer is 2.5 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question we have

density =  \frac{25}{10}  =  \frac{5}{2}  \\

We have the final answer as

<h3>2.5 g/cm³</h3>

Hope this helps you

3 0
3 years ago
The driver then tests the brakes on the car and safely comes to a complete stop with constant acceleration from 26.8 meters per
Temka [501]

Answer:

62.78

Explanation:

5 0
2 years ago
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