All categories

2 years ago

A stuntman of mass 48 kg is to be launched horizontally out of a spring-

Physics

1 answer:

Damm [24]2 years ago

6 0

The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

The Kinetic energy of the stuntman is equal to the elastic potential energy of the spring.

<h3 /><h3>Velocity: </h3>

This is the ratio of displacement to time. The S.I unit of Velocity is m/s. The velocity of the stuntman can be calculated using the formula below.

⇒ Formula:

mv²/2 = ke²/2
mv² = ke².................. Equation 1

⇒ Where:

m = mass of the stuntman
v = velocity of the stuntman
k = force constant of the spring
e = compression of the spring

⇒ Make v the subject of the equation

v = √(ke²/m)................. Equation 2

From the question,

⇒ Given:

m = 48 kg
k = 75 N/m
e = 4 m

⇒ Substitute these values into equation 2

v = √[(75×4²)/48]
v = √25
v = 5 m/s.

Hence, The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

Learn more about velocity here: brainly.com/question/10962624

You might be interested in

Suppose a 1300 kg car is traveling around a circular curve in a road at a constant

Scrat [10]

Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :

$F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N$

So, the force has a magnitude of 4212 N

4 0

2 years ago

The bright-line spectrum of an element in the gaseous phase is produced as 1. protons move from lower energy states to higher en

lara31 [8.8K]

Answer:

4. Electrons move from higher energy states to lower energy states.

Explanation:

When electrons fall from a higher (excited) energy state to a lower energy state, it loses/gives out energy.

This energy is given out by the emission of photons (quanta of light) by the electron.

7 0

3 years ago

Normalize the equations

tatyana61 [14]

Answer:

Solution is in explanation

Explanation:

part a)

For normalization we have

$\int_{0}^{\infty }f(x)dx=1\\\\\therefore \int_{0}^{\infty }ae^{-kx}dx=1\\\\\Rightarrow a\int_{0}^{\infty }e^{-kx}dx=1\\\\\frac{a}{-k}[\frac{1}{e^{kx}}]_{0}^{\infty }=1\\\\\frac{a}{-k}[0-1]=1\\\\\therefore a=k$

Part b)

$\int_{0}^{L }f(x)dx=1\\\\\therefore Re(\int_{0}^{L }ae^{-ikx}dx)=1\\\\\Rightarrow Re(a\int_{0}^{L }e^{-ikx}dx)=1\\\\\therefore Re(\frac{a}{-ik}[\frac{1}{e^{ikx}}]_{0}^{L})=1\\\\\Rightarrow Re(\frac{a}{-ik}(e^{-ikL}-1))=1\\\\\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1$

$\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1\\\\\frac{a}{k}Re(icos(-kL)+sin(kL)+\frac{1}{i})=1\\\\\frac{a}{k}sin(kL)=1\\\\a=\frac{k}{sin(kL)}$

7 0

3 years ago

A 10kg oil is heated by a source of heat that supplies 800J of heat every minute. Calculate The heat added to oil after 30 minut

goblinko [34]

Answer:

4,200 joules per kilogram per degree Celsius

Explanation:

The specific heat capacity of a material is the energy required to raise one kilogram (kg) of the material by one degree Celsius (°C). The specific heat capacity of water is 4,200 joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.

4 0

2 years ago

A cylindrical storage tank has a radius of 1.35 m. When filled to a height of 3.45 m, it holds 14,014 kg of a liquid industrial

Kruka [31]

Answer:

709.93 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³

From the question above,

D = m/v.............................. Equation 1

Where D = density, m = mass, v = volume.

Note: The volume of the liquid is equal to the volume of the height occupied by the liquid in the container

Since the tank is cylindrical,

v = πr²h........................ Equation 2

Where r = radius of the the tank, h = height of the liquid in the tank

Substitute equation 2 into equation 1

D = m/(πr²h)............... Equation 3

Given: m = 14014 kg, r = 1.35 m, h = 3.45 m, π = 3.14

Substitute into equation 3

D = 14014/(3.14×1.35²×3.45)

D = 14014/19.74

D = 709.93 kg/m³

6 0

3 years ago