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3 years ago

A ball is dropped from rest. how fast is the ball going after 3 seconds

Physics

1 answer:

Airida [17]3 years ago

5 0

Answer:

v = 29.4m/s

Explanation:

Since the ball is dropped at rest,

u = 0m/s

a = 9.81m/s²

Using

v = u + at

After 3 seconds,

v = 0 + (9.81)(3)

v = 29.4m/s

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Two iron bolts of equal Mass one at a hundred see another at 55 Sierra place in the insulated cylinder assuming the heat capacit

malfutka [58]

Answer:

$T_2 = 77.5c$

Explanation:

From the question we are told that

Temp of first bolts $T_1=100$

Temp of 2nd bolt $T_2=55$

Generally the equation showing the relationship between heat & temperature is given by

$q=cm \triangle T$

Generally heat released by the iron bolt = heat gained by the iron bolt

Generally solving mathematically

$-(0.45*m* (T_2-100 \textdegree c)) = 0.45*m*(T_2 -55\textdegree c)$

$-(T_2-100 \textdegree c)) = (T_2 -55 \textdegree c)$

$T_2 +T_2= 100 \textdegree c+55 \textdegree c$

$T_2=\frac{155 \textdegree c}{2}$

$T_2 = 77.5 \textdegree c$

Therefore $T_2 = 77.5 \textdegree c$ is the final temperature inside the container

5 0

2 years ago

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz

gladu [14]

This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle? Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa

Tt is stagnation temperature = ?

T is the static temperature = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa = ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 = Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

6 0

3 years ago

In 2 1/2 hours an airplane travels 1150 km against the wind. It takes 50 min to travel 450 km with the wind. Find the speed of t

Tamiku [17]

Answer:

Velocity of airplane is 500 km/h

Velocity of wind is 40 km/h

Explanation:

$V_a$ = Velocity of airplane in still air

$V_w$ = Velocity of wind

Time taken by plane to travel 1150 km against the wind is 2.5 hours

$V_a-V_w=\frac {1150}{2.5}\\\Rightarrow V_a-V_w=460\quad (1)$

Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours

$V_a+V_w=\frac {450}{50}\times 60\\\Rightarrow V_a-V_w=540\quad (2)$

Subtracting the two equations we get

$V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h$

Applying the value of velocity of wind to the first equation

$V_a-40=460\\\Rightarrow V_a =500\ km/h$

∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h

5 0

3 years ago

How does the force of gravity between two bodies change when the distance between them doubles?

polet [3.4K]

Answer:

If the distance is doubled, the force of gravity between the two bodies is one-fourth as strong as before

Explanation:

The force of gravity between two bodies depends on the mass and distance. But we will focus on distance since that's what the question asks

Therefore, the force of gravity decreases as distance between the bodies increases.

7 0

3 years ago

) a 1.0 kilogram laboratory cart moving with a velocity of 0.50 meter per second due east collides with and sticks to a similar

ZanzabumX [31]

Momentum would be the same before and after the collision
Before the collision:
Momentum of the single cart: 1 * 0.50 = 0.50
After the collision
velocity = 0.25m / s
1 * 0.25 + 1 * 0.25 =
0.25 * (1 + 1) =
0.25 * 2 =
0.50
Now new momentum will be 0.5
answer
the same before and after the collision

4 0

3 years ago

Read 2 more answers

A ball is dropped from rest. how fast is the ball going after 3 seconds​

A ball is dropped from rest. how fast is the ball going after 3 seconds