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REY [17]
3 years ago
14

A body decelerates uniformly to a constant speed and after some time it accelerates uniformly Draw the shape of speed time graph

for such a motion . Label the three sections of this graph. What is the quantity which is measured by the area occupied under speed time graph ?

Physics
1 answer:
Alex73 [517]3 years ago
7 0

Answer:

here I am just giving an idea of how the graph will be like ...

In the pic..

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

:)

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2. The word used to represent a chemical reaction is a/an a. Atom b. Formula O c. Equation d. Symbol​
Slav-nsk [51]

Answer:

A formula I am pretty sure

5 0
3 years ago
A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The
IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s

Time taken by the ball to reach the highest point is 0.14 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

v=u+at\\\Rightarrow v=0+32.1\times 0.518\\\Rightarrow v=16.62\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-16.62^2}{2\times -100\times 3.28}\\\Rightarrow s=0.42\ ft

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

8 0
3 years ago
What is the pressure at 5000 km below the surface of the earth
attashe74 [19]

Answer:

The centre of the earth is harder to study than the centre of the sun." Temperatures in the lower mantle the reach around 3,000-3,500 degrees Celsius and the barometer reads about 125 gigapascals, about one and a quarter million times atmospheric pressure.

Explanation:

7 0
2 years ago
a ball is thrown inclined in to the air with a initial velocity of u. if it reaches the maximum height in 6 seconds , find the r
olga55 [171]

Answer:

11:1

Explanation:

At constant acceleration, an object's position is:

y = y₀ + v₀ t + ½ at²

Given y₀ = 0, v₀ = u, and a = -g:

y = u t − ½g t²

After 6 seconds, the ball reaches the maximum height (v = 0).

v = at + v₀

0 = (-g)(6) + u

u = 6g

Substituting:

y = 6g t − ½g t²

The displacement between t=0 and t=1 is:

Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]

Δy = 6g − ½g

Δy = 5½g

The displacement between t=6 and t=7 is:

Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]

Δy = (42g − 24½g) − (36g − 18g)

Δy = 17½g − 18g

Δy = -½g

So the ratio of the distances traveled is:

(5½g) / (½g)

11 / 1

The ratio is 11:1.

8 0
3 years ago
A projectile is fired vertically from Earth's surface with an initial speed v0. Neglecting air drag, how far above the surface o
lidiya [134]
This may helpv^2=u^2+2as. v=0 at top of flight. a=acceleration of gravity(vo^2)/2a=s.
3 0
3 years ago
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