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2 years ago

7

A boulder rolls off a 100 meter cliff with a horizontal velocity of 2 m/s.

1 answer:

bonufazy [111]2 years ago

7 0

Explanation:

It is given that,

Height of the cliff is 100 m

Its horizontal velocity is 2 m/s

The attached figure shows the sketch of the cliff and the boulder. It will go in the form of parabola. Its trajectory is parabolic in nature. The horizontal distance is called the range of the parabola.

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What is the relationship between force in velocity selector in a bain bridge.

kifflom [539]

Answer:

In a velocity selector, there are two forces namely;

» Electric field Intensity

» Magnetic field density

<u>Relationship</u><u>:</u>

$E _{e}= B$

E is the electric field intensity

B is the magnetic flux density

4 0

2 years ago

What is the mass of an object if a 30 N force makes it accelerate at 6 m/s2

jasenka [17]

Answer:

5 kg

Explanation:

Acceleration = 6 m/s^2

Force = 30 N

Force = mass * acceleration

mass = force / acceleration

mass = 30 / 6

mass = 5 kg

4 0

2 years ago

A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w

trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

$KE = PE$

$\frac{1}{2} mv ^2 = \frac{1}{2} k A^2$

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

$A = \sqrt{\frac{mv^2}{k}}$

Replacing,

$A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}$

$A = 0.0744m$

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

$T = 2\pi \sqrt{\frac{m}{k}}$

Replacing,

$T = 2\pi \sqrt{\frac{0.750}{13}}$

$T= 1.509s$

Now the velocity is described as,

$v = \frac{2\pi}{T} * \sqrt{A^2-x^2}$

$v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}$

We have all the values, then replacing,

$v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}$

$v = 0.2049m/s$

7 0

2 years ago

The property of matter that describes what it is made of

const2013 [10]

Answer:

Miixture

Explanation:

The answer is mixture

4 0

2 years ago

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

Damm [24]

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

6 0

2 years ago