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kifflom [539]
2 years ago
9

Write down an example scenario of an object that has acceleration

Physics
1 answer:
Grace [21]2 years ago
5 0

Answer:

An object which experiences either a change in the magnitude or the direction of the velocity vector can be said to be accelerating. This explains why an object moving in a circle at constant speed can be said to accelerate - the direction of the velocity changes.

if a car turns a corner at constant speed, it is accelerating because its direction is changing. The quicker you turn, the greater the acceleration. So there is an acceleration when velocity changes either in magnitude (an increase or decrease in speed) or in direction, or both.

Explanation:

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The magnitude of the force associated with the gravitational field is constant and has a value FF . A particle is launched from
uysha [10]

Answer:

The kinetic energy of the particle will be 12U₀

Explanation:

Given that,

A particle is launched from point B with an initial velocity and reaches point A having gained U₀ joules of kinetic energy.

Constant force = 12F

According to question,

The kinetic energy is

U_{0}=Fx....(I)

Constant force = 12F

A resistive force field is now set up ,

Resistive force is given by,

F_{r}=12F

When the particle moves from point B to point A then,

We need to calculate the kinetic energy

Using formula for kinetic energy

U=F_{r}x

Put the value of F_{r}

U=12Fx

Now, from equation (I)

U=12U_{0}

Hence, The kinetic energy of the particle will be 12U₀.

7 0
3 years ago
you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k
alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}

→ v_{p}=150 km/h ,  v_{w}=50 km/h

→ v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11 km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is tan^{-1}\frac{50}{150}=18.4°

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0
3 years ago
Suppose scientists discover two new moons.The average surface temperature of one of the moons is –180°C, but the temperature can
s344n2d4d5 [400]
The answer is a in the center of active volcanos
8 0
3 years ago
Read 2 more answers
A projectile is fired with an initial velocity of 120.0 meters per second at an angle, θ above the horizontal. If the projectile
k0ka [10]

Answer:

θ = 62.72°

Explanation:

The projectile describes a parabolic path:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = x₀+ vx*t   Formula (1)

vx = v₀x

Where:  

x: horizontal position in meters (m)

x₀: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s

v₀x: Initial speed in x  in m/s

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

y= y₀+(v₀y)*t - (1/2)*g*t² Formula (2)

vfy= v₀y -gt Formula (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 120 m/s  , at an angle  θ above the horizontal

v₀x= 55 m/s

x-y components of the initial  velocity ( v₀)

v₀x = v₀*cosθ Equation (1)

v₀y = v₀*sinθ   Equation (2)

Calculating of the angle θ

We replace data in the  Equation (1)

55 =  120*cosθ

cosθ = 55 / 120

\theta = cos^{-1}(  \frac{55}{120} )

θ = 62.72°

3 0
3 years ago
An object accelerates in a direction that is always perpendicular to its motion.
anzhelika [568]
The object will not be able to accelerate perpendicular to direction of motion
7 0
3 years ago
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