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Anestetic
7 days ago
11

In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2. Calculate the amount of force at

which the wrecking ball strikes the building. The wrecking ball strikes the building with a force of N.
Physics
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For anti-ballistic missile system, the time of flight tf is determined by the initial speed v0 of the missile and the maximum ra
Ostrovityanka [3204]
Refer to the diagram below.

This discussion operates under a basic analysis that overlooks air resistance and variations in the terrain the missile traverses.

Let V₀ be the launch velocity, at an angle θ to the horizontal.
The horizontal velocity component equals V₀ cosθ.
If the flight duration is t_{f}, then
r=V_{o} \, t_{f}
where r represents the missile's range.

The time t at which the missile is at ground level is expressed by
0=V_{o} sin\theta \, t- \frac{1}{2}gt^{2}
where g signifies acceleration due to gravity.

t = 0 signifies the missile's launch. Thus
t_{f} = \frac{2V_{o}sin\theta}{g}

Consequently,
r= \frac{2V_{o}^{2} sin\theta cos\theta}{g} = \frac{V_{o}^{2} sin(2\theta)}{g}

Typically, an angle of θ=45° is optimal for achieving maximum range, resulting in
r= \frac{V_{o}^{2}}{g}

This discussion applies more accurately to a scud missile than to a powered, guided missile.

Response:
t_{f} = \frac{r}{V_{o} cos\theta} \\\\ r= \frac{V_{o}^{2} sin(2\theta)}{g}
Usually, θ=45°

6 0
2 months ago
At one instant of time a rocket is traveling in outer space at 2500m/s and is exhausting fuel at a rate of 100 kg/s. If the spee
Keith_Richards [3271]

Thrust is quantified as a reaction force, in accordance with Newton's third law. When a system accelerates or expels mass in one direction, this resulting mass generates a force of equal strength but in the opposite direction on that system. This relationship can be expressed mathematically as:

T = v\frac{dm}{dt}

Where:

v = velocity of the exhaust gases as perceived from the rocket.

\frac{dm}{dt}= Change in mass over time

The provided data is as follows:

v = 1500m/s

\frac{dm}{dt} = 100kg/s

After substitution, we obtain:

T = 1500*100

\therefore T = 1.5*10^5N

7 0
2 months ago
A certain carbon monoxide molecule consists of a carbon atom mc = 12 u and an oxygen atom mo = 17 u that are separated by a dist
Maru [3345]
a) x_{cm} = m₂/ (m₁ + m₂) d, b) x_{cm} = 52.97 pm
7 0
1 month ago
Physics students study a piano being pulled across a room on a rug. They know that when it is at rest, it experiences a gravitat
inna [3103]
The static frictional force exceeds the kinetic frictional force, indicating that the static frictional force is over 1200 N. Explanation: The frictional force opposes the motion of any object on a surface, caused by interactions between the surface molecules and the object. It is known that static friction is typically stronger than kinetic friction (this is the reason initiating motion requires more force than keeping it moving along a surface). Hence, option 3 correctly describes the situation.
3 0
2 months ago
A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Sav [3153]

Response:

A=0.199

Clarification:

We know that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

Frequency of oscillation=

\nu=1.2Hz

Energy for oscillation is 0.51 J

To determine the amplitude of oscillations.

Energy for oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

A=\sqrt{0.0398}=0.199

Therefore, the amplitude of oscillation=A=0.199

4 0
2 months ago
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