What is the mass of a person running at 2.5 m/s with momentum of<br> 200 kg m/S

A printed circuit board (PCB) is supported by a chassis that is attached to a vibrating motor. The board is 1.6 mm thick, 200 mm

Vedmedyk [2.9K]

jtejyrkekdkeludmydmumud

5 0

3 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

4 0

3 years ago

A force off 700 newtons is applied to a 600 kg bowling ball. What is the acceleration of the bowling ball once the force is appl

Readme [11.4K]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{700}{600} = \frac{7}{6} \\ = 1.1666666...$

We have the final answer as

Hope this helps you

6 0

3 years ago

What is an essential characteristic of an object in equilibrium?

bulgar [2K]

Answer:

2) zero acceleration

Explanation:

Motion can be defined as a change in the location (position) of a physical object or body with respect to a reference point.

This ultimately implies that, motion would occur as a result of a change in location (position) of an object with respect to a reference point or frame of reference i.e where it was standing before the effect of an external force.

Mathematically, the motion of an object is described in terms of time, distance, speed, velocity, position, displacement, acceleration, etc.

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

Generally, an object is said to be in equilibrium when neither the energy possessed by the object not state of motion changes with respect to time. Thus, the vector sum of all the forces acting upon an object that's in equilibrium is zero.

In conclusion, an essential characteristic of an object in equilibrium is zero (0) acceleration because there's no change in its velocity with respect to time.

5 0

3 years ago

Earth's subsystems are the geosphere, atmosphere, hydrosphere, and biosphere. Energy and matter are transferred when Earth's sub

Leona [35]

Answer:

ok

Explanation:

The Earth is made of several subsystems or "spheres" that interact to form a complex and continuously changing whole called the Earth system. Scale

Processes operating in the Earth system take place on spatial scales varying from fractions of millimeters to thousands of kilometers, and on time scales that range from milliseconds to billions of years.

Examples of instantaneous - breathing; rotation of the Earth; earthquake

Examples of long term - making coal; plate tectonics

Cycles

The Earth system is characterized by numerous overlapping cycles in which matter is recycled over and over again. Cycles involve multiple spheres and systems interactions.

Examples of cycles: day and night; rock cycle; seasons

Energy

The Earth system is powered by energy from two major sources: the Sun and the planet's internal heat.

Humans and the Earth System

People are part of the Earth system and they impact and are impacted by its materials and processes.

6 0

3 years ago

What is the mass of a person running at 2.5 m/s with momentum of 200 kg m/S