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3 years ago

1. height of tower when stone dropped from it reaches ground in 10s is

Physics

1 answer:

Arada [10]3 years ago

7 0

Answer:

490.5 m

Explanation:

we can use the formula s = ut + ½at^2

s= displacement (height of tower)

initial velocity u = 0 m/s

acceleration a = take 9.81 ms-2 (acceleration due to gravity, which is a constant. depending on the given instructions of the question, it may not be exactly 9.81. Some take 10, some take 9.80665)

time t =10s

s = 0t + 1/2 (9.81)(10)^2

s = 490.5 m

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3 years ago

When it reaches an altitude of 3600 m, where the temperature is 5.0∘C and the pressure only 0.68 atm, how will its volume compar

dalvyx [7]

Answer:

The question was incomplete. Here is the complete question.

Explanation:

A helium-filled balloon escapes a child’s hand at sea level and 20.0C. When it reaches an altitude of 3600 m, where the temperature is 5.0∘C and the pressure only 0.68 atm, how will its volume compare to that at sea level?

The ideal gas equation:

$PV = nRT$

P = absolute pressure

V = Volume of a gas

n = no of moles of a gas

R = ideal gas constant

T = Absolute temperature of a gas

For initial and final states:

$P_{i}V_{i} = nRT_{i} \\P_{f}V_{f} = nRT_{f} \\\frac{V_{f} }{V_{i} } = \frac{P_{i}T_{f} }{P_{f}T_{i} }$

= 1.4

7 0

3 years ago

A block of mass 3.6 kg, sliding on a horizontal plane, is released with a velocity of 1.7 m/s. The block slides and stops at a d

pentagon [3]

Answer:

The block would have slid <u>12.6 m</u> if its initial velocity were increased by a factor of 2.8.

Explanation:

Given:

Mass of the block (m) = 3.6 kg

Initial velocity (u) = 1.7 m/s

Final velocity (v) = 0 m/s

Displacement (S) = 1.6 m

First we will find the acceleration of the block.

Using the equation of motion, we have:

$v^2=u^2+2aS\\\\a=\dfrac{v^2-u^2}{2S}$

Now, plug in the given values and solve for 'a'. This gives,

$a=\frac{0-1.7^2}{2\times 1.6}\\\\a=\frac{-2.89}{3.2}=0.903\ m/s^2$

The acceleration is negative as it is resisting the motion.

Now, the initial velocity is increased by a factor of 2.8. So,

New initial velocity = 2.8 × 1.7 = 4.76 m/s

Again using the same equation of motion and expressing the result in terms of 'S'. This gives,

$v^2=u^2+2aS\\\\S=\dfrac{v^2-u^2}{2a}$

Now, plug in the given values and solve for 'S'. This gives,

$S=\frac{0-4.76^2}{2\times -0.903}\\\\S=\frac{-22.6576}{-1.806}=12.6\ m$

Therefore, the block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8

6 0

3 years ago

Need help with physics please

bazaltina [42]

The centripetal force : F = 293.3125 N

<h3>Further explanation</h3>

Given

mass = 65 kg

v = 9.5 m/s

r = 20 m

Required

the centripetal force

Solution

Centripetal force is a force acting on objects that move in a circle in the direction toward the center of the circle

$\large {\boxed {\bold {F = \frac {mv ^ 2} {R}}}$

F = centripetal force, N

m = mass, Kg

v = linear velocity, m / s

r = radius, m

Input the value :

F = 65 x 9.5² / 20

F = 293.3125 N

4 0

2 years ago

In an experiment, a disk is set into motion such that it rotates with a constant angular speed. As the disk spins, a small spher

boyakko [2]

Answer:

L₀ = L_f , K_f < K₀

Explanation:

For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.

This means that the angular momentum before and after the collision changes.

Initial instant. Before the crash

L₀ = I₀ w₀

Final moment. Right after the crash

L_f = (I₀ + mr²) w

we treat the clay sphere as a point particle

how the angular momentum is conserved

L₀ = L_f

I₀ w₀ = (I₀ + mr²) w

w = $\frac{I_o}{I_o + m r^2}$ w₀

having the angular velocities we can calculate the kinetic energy

starting point. Before the crash

K₀ = ½ I₀ w₀²

final point. After the crash

K_f = ½ (I₀ + mr²) w²

sustitute

K_f = ½ (I₀ + mr²) ( $\frac{I_o}{I_o + m r^2}$ w₀)²

Kf = ½ $\frac{I_o^2}{ I_o + m r^2}$ w₀²

we look for the relationship between the kinetic energy

$\frac{K_f}{K_o}$ = $\frac{I_o}{I_o + m r^2}$

$\frac{K_f}{K_o } < 1$

K_f < K₀

we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision

6 0

3 years ago

1. height of tower when stone dropped from it reaches ground in 10s is​

1. height of tower when stone dropped from it reaches ground in 10s is