Question

2. Чему равно отношение скоростей, приобретенных телом при свободном падении за три и за четыре секунды после начала падения? 3. Кран опускает вниз бадью с раствором массой 400 кг с ускорением 2 м/с2. Определите вес бадьи. 4. Со скалы высотой 45 м в горизонтальном направлении бросают камень со скоростью 10 м/с. Определите время и дальность полета камня. Считать g = 10 м/с2. 5. Найдите силу тяги двигателя тепловоза, который равномерно движется по горизонтальному пути со скоростью 20 м/с. Мощность двигателя равна 4000 кВт. 6. Железнодорожный вагон массой 20 т, движущийся со скоростью 0,56 м/с, сталкивается с неподвижной платформой массой 8 т. Определите их скорость после автосцепки. Трением о рельсы пренебречь. 7. Чему равен период свободных колебаний пружинного маятника, если масса груза равна 0,25 кг, а жесткость пружины 100 Н/м?

Answer 1

Задача 2:

 

$v=g*t$

 

$\frac{v_{4}}{v_{3}}=\frac{t_{4}}{t_{4}}=\frac{4}{3}$

 

Задача 3:

 

$P=m*(g-a)=400*(10-2)=3200 H$

 

Задача 4:

 

$t=\sqrt{\frac{2*h}{g}}=\sqrt{\frac{2*45}{10}}=3 c$

 

$l=v*cos\alpha*t=10*1*3=30$

 

l=30 м

 

Задача 5:

 

$A=F*S=N*t$

 

$\frac{F*S}{t}=N$

 

$F*v=N$

 

$F=\frac{N}{v}=\frac{4000000}{20}=200000 H$

 

Задача 6:

 

$v=\frac{m_{1}*v_{1}}{(m_{1}+m_{2})}=\frac{0,56*20}{20+8}=0,4$

 

Задача 7:

$T=2*\pi*\sqrt{\frac{m}{k}}=3*3,14*\sqrt{\frac{0,25}{100}}=6,28*0,05=0,314c$

Answer 2

Free Fall and Velocity Ratios

The ratio of velocities acquired by a body in free fall after three and four seconds can be calculated using the equations of motion. The velocity of an object in free fall can be calculated using the formula: \( v = u + gt \), where \( v \) is the final velocity, \( u \) is the initial velocity (which is 0 in free fall), \( g \) is the acceleration due to gravity, and \( t \) is the time.

Using this formula, the velocity after three seconds can be calculated as \( v_3 = 0 + (9.8 \, \text{m/s}^2) \times 3 \) and the velocity after four seconds can be calculated as \( v_4 = 0 + (9.8 \, \text{m/s}^2) \times 4 \).

Therefore, the ratio of the velocities acquired by the body in free fall after three and four seconds can be expressed as:

\[ \frac{v_3}{v_4} = \frac{0 + (9.8 \, \text{m/s}^2) \times 3}{0 + (9.8 \, \text{m/s}^2) \times 4} \]

The result of this calculation is the ratio of the velocities after three and four seconds.

Downward Crane and Weight Calculation

When a crane lowers a bucket with a solution of mass 400 kg with an acceleration of 2 m/s\(^2\), the weight of the bucket can be calculated using the formula: \( F = ma \), where \( F \) is the force (weight), \( m \) is the mass, and \( a \) is the acceleration.

Therefore, the weight of the bucket can be calculated as \( F = 400 \, \text{kg} \times 2 \, \text{m/s}^2 \).

Projectile Motion of Thrown Stone

To determine the time and distance of the stone's flight after being thrown horizontally from a 45 m high cliff with a velocity of 10 m/s, the equations of motion for projectile motion can be used. The time of flight can be calculated using the formula: \( t = \frac{2u\sin(\theta)}{g} \), where \( u \) is the initial velocity, \( \theta \) is the angle of projection (which is 90 degrees for horizontal projection), and \( g \) is the acceleration due to gravity.

The horizontal distance traveled can be calculated using the formula: \( d = ut \), where \( d \) is the distance, \( u \) is the initial velocity, and \( t \) is the time of flight.

Tractive Force of a Locomotive

The tractive force of a locomotive moving at a constant speed can be calculated using the formula: \( F = \frac{P}{v} \), where \( F \) is the tractive force, \( P \) is the power of the engine, and \( v \) is the velocity.

Therefore, the tractive force of the locomotive can be calculated as \( F = \frac{4000 \, \text{kW}}{20 \, \text{m/s}} \).

Collision of Railway Wagon and Platform

After the collision of a railway wagon with a mass of 20 tons and a speed of 0.56 m/s with a stationary platform of mass 8 tons, the velocity of the two objects can be calculated using the principle of conservation of momentum.

The total momentum before the collision is equal to the total momentum after the collision, and the velocity of the objects after the collision can be calculated using this principle.

Period of Free Oscillations of a Spring Pendulum

The period of free oscillations of a spring pendulum can be calculated using the formula: \( T = 2\pi\sqrt{\frac{m}{k}} \), where \( T \) is the period, \( m \) is the mass, and \( k \) is the spring constant.

Therefore, the period of free oscillations can be calculated as \( T = 2\pi\sqrt{\frac{0.25 \, \text{kg}}{100 \, \text{N/m}}} \).