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4 years ago

How far below an initial straight-line path will a projectile fall in one second

Physics

1 answer:

Anni [7]4 years ago

3 0

Answer:

9.8 m

Explanation:

The rate of free fall of a body does not depend on its mass or shape in the absence of air resistance. This property of gravitational force is proved by the famous leaning tower of Pisa experiment by Galileo.

The gravitational force is responsible for the gravitational acceleration of a body.

When considering the gravitational acceleration near the surface of Earth it has the constant value g = 9.8 m/s irrespective of the mass of the body.

This phenomenon states that every projectile falls at the same rate such that its velocity keeps on increasing to maintain the constancy of g.

So, when a projectile fall initially from a certain height for one second, it would have traveled only 9.8 m.

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In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 2.15 kg. The

Maksim231197 [3]

Answer:

0.339 kgm²

Explanation:

We know the period of this pendulum, T = 2π√(I/mgh) where I = moment of inertia of the object about the pivot axis, m = mass of object = 2.15 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.163 m.

Since T = 2π√(I/mgh), making I subject of the formula, we have

I = mghT²/4π²

Now since it takes 241 s to complete 113 cycles, then it takes 241 s/113 cycles to complete one cycle.

So, T = 241 s/113 = 2.133 s

So, Substituting the values of the variables into I, we have

I = mghT²/4π²

I = 2.15 kg × 9.8 m/s² × 0.163 m × (2.133 s)²/4π²

I = 15.63/4π² kgm²

I = 0.396 kgm²

Now from the parallel axis theorem, I = I' + mh² where I' = moment of inertia of object with respect to its center of mass about an axis parallel to the pivot axis

I' = I - mh²

I' = 0.396 kgm² - 2.15 kg × (0.163 m)²

I' = 0.396 kgm² - 0.057 kgm²

I' = 0.339 kgm²

7 0

3 years ago

Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of m

OverLord2011 [107]

To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

$E_m = \frac{1}{2} m\omega_1^2A_1^2$

The energy of the system having mass 2m is,

$E_{2m} = \frac{1}{2} (2m)\omega_1^2A_1^2$

For the two expressions mentioned above remember that the variables mean

m = mass

$\omega =$ Angular velocity

A = Amplitude

The energies of the two system are same then,

$E_m = E_{2m}$

$\frac{1}{2} m\omega_1^2A_1^2=\frac{1}{2} (2m)\omega_1^2A_1^2$

$\frac{A_1^2}{A_2^2} = \frac{2\omega_2^2}{\omega_1^2}$

Remember that

$k = m\omega^2 \rightarrow \omega^2 = k/m$

Replacing this value we have then

$\frac{A_1}{A_2} = \sqrt{\frac{2(k/m_2)}{(k/m_1)^2}}$

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m_1}{m_1}}$

But the value of the mass was previously given, then

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m}{2m}}$

$\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}$

$\frac{A_1}{A_2} = 1$

Therefore the ratio of the oscillation amplitudes it is the same.

5 0

3 years ago

A jet airliner moving initially at 503 mph

Varvara68 [4.7K]

Answer:

The new speed is 1230.28 m/h

Explanation:

The jet airliner moving initially at 503 mph to the east

The wind is blowing at 855 mph in a direction 52° north of east

At first let us distribute the velocity of the wind into east component

and north component

→ The east component is 855 cos(52) m/h

→ The north component is 855 sin(52) m/h

Now we have two components of velocity in the east direction

and one component of velocity in the north direction

The new speed is the resultant of the east and north components

→ The east components are 503 m/h and 855 cos(52) m/h

→ The north component is 855 sin(52) m/h

Add the components of the speeds in direction of east

→ The east component = 503 + 855 cos(52) = 1029.39 m/h

→ The north component = 855 sin(52) = 673.75 m/h

Now we can find the new speed as a resultant speed of the east and

north components

→ The new speed = $\sqrt{(1029.39)^{2}+(673.75)^{2}}$

→ The new speed = 1230.28 m/h

The new speed is 1230.28 m/h

3 0

4 years ago

Which consequence can occur if a mother uses drugs during pregnancy?

saveliy_v [14]

The answer is b. birth defects hope this helps

5 0

4 years ago

Read 2 more answers

The front of a 1300 kg car is designed to absorb the shock of a collision by having a "crumple zone" in which the front 1.20 m o

Novay_Z [31]

Answer:

The collision last 0.83 s

Explanation:

Using the equation of motion,

S = (v+u)t/2 ........................ Equation 1

making t the subject of the equation,

t = 2S/(v+u)..................... Equation 2

Where S = distance, v = final velocity, u = initial velocity, t = time

Given: v = 0 m/s, u = 29.0 m/s, S = 1.20 m,

Substituting these values into equation 2

t = 2(1.2)/(29+0)

t = 2.4/29

t = 0.83 sec.

Thus the collision last 0.83 s

7 0

3 years ago