A rock dropped from a 5 m height accelerates at 10 m/s and strikes the ground 1 s later. If the rock is dropped

from a height of 2,5 m, its acceleration of fall is

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.

The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2

Fd = (1/2)mv^2

F = (1/2)mv^2/d.

Plug in m = 20 kg, v = 3 m/sec, d = 40 m.

83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.

Explanation:Hope I helped :)

Answer:

The maximum speed of the car is 35 m/s

The total distance traveled by the car is 658.33 m

Explanation:

Given;

initial velocity of the car, u = 15 m/s

acceleration of the car, a = 2 m/s²

time of car motion, t = 10 s

(i)

Initial distance traveled by the car is given by;

d₁ = ut + ¹/₂at²

d₁ = (15 x 10) + ¹/₂(2)(10)²

d₁ = 150 + 100

d₁ = 250 m

The maximum speed of the car during this is given by;

v² = u² + 2ad₁

v² = (15)² + (2 x 2 x 250)

v² = 1225

v = √1225

v = 35 m/s

(ii)

The final distance cover by the car during the deceleration of 1.5 m/s².

Note: the final or maximum speed of the car becomes the initial velocity during deceleration.

v² = u² + 2ad₂

where;

v is the final speed of the car when it stops = 0

0 = u² + 2ad₂

0 = (35²) + (2 x - 1.5 x d₂)

0 = 1225 - 3d₂

3d₂ = 1225

d₂ = 1225 / 3

d₂ = 408.33 m

The total distance traveled by the car is given by;

d = d₁ + d₂

d = 250 m + 408.33 m

d = 658.33 m

Answer:

p₂ / p₁ = 2 (v₁ / v₂)

Explanation:

The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,

         p = m v / √[1+ (v/c)² ]

for the case of speeds much lower than the speed of light this expression is close to

         p = m v

In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small

        p₂ = 2 p₁

       p₂/p₁ = 2

in consecuense

       m v₂ = 2 m v₁

       v₂ = 2 v₁

consider particles of equal mass.

By the time their speeds increase they enter the relativistic regime

        p₂ = mv₂ /√(1 + v₂² /c²)

        p₁ = m v₁ /√(1 + v₁² / c²)

let's look for the relationship between these two moments

       p₂ / p₁ = mv₂ / mv₁   [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)

from the initial statement

      p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)

we take c from the root

      p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]

this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1

      p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]

      p₂ / p₁ = 2 (v₁ / v₂)

we see the value of the moment depends on the speed of the particles

Answer:

2

Explanation:

there are 2 significant figures in there

Answer:

metals like iron or nickel

Explanation:

Answer:

Explanation:

m1v1=m2v2

m1=70 kg

m2=10 g=0.01 kg

v2=500 m/s

m1v1=m2v2

v1=m2v2/m1

v1=0.01*500/70

v1=0.07

Answer:

[tex]P=2.57\times 10^{-7}\ N/m^2[/tex]

Explanation:

Given that,

A radio wave transmits 38.5 W/m² of power per unit area.

A flat surface of area A is perpendicular to the direction of propagation of the wave.

We need to find the radiation pressure on it. It is given by the formula as follows :

[tex]P=\dfrac{2I}{c}[/tex]

Where

c is speed of light

Putting all the values, we get :

[tex]P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2[/tex]

So, the radiation pressure is [tex]2.57\times 10^{-7}\ N/m^2[/tex].

Answer:

the velocity of the acorn

Explanation:

just do in in real life and see

Answer:

it is probably the velocity of the acorn

Answer:

Approximately [tex]35.2\; \rm m[/tex].

Explanation:

Given:

Initial velocity: [tex]u = 13\; \rm m \cdot s^{-1}[/tex].

Acceleration: [tex]a = -2.40\; \rm m \cdot s^{-2}[/tex] (negative because the car is slowing down.)

Implied:

Final velocity: [tex]v = 0\; \rm m \cdot s^{-1}[/tex] (because the car would come to a stop.)

Required:

Displacement, [tex]x[/tex].

Not required:

Time taken, [tex]t[/tex].

Because the time taken for this car to come to a full stop is not required, apply the SUVAT equation that does not involve time:

[tex]\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{{\left(0\; \rm m \cdot s^{-1}\right)}^2 - {\left(13\; \rm m \cdot s^{-1}\right)}^2}{2\times \left(-2.40\; \rm m\cdot s^{-2}\right)} \approx 35.2\; \rm m \end{aligned}[/tex].

In other words, this car would travel approximately [tex]35.2\; \rm m[/tex] before coming to a stop.

The 400-foot altitude limit was put in place for the sake of airspace safety, and there is a risk to country security as well as the privacy of citizens.

Unmanned aerial vehicles (UAVs), sometimes known as drones, are used for a variety of jobs, from routine to extremely dangerous. These robotic-looking planes can be seen practically everywhere, from delivering groceries to your home to rescuing avalanche victims.

The 400-foot altitude restriction was ultimately implemented for airspace safety. Given the breadth of the airspace above 400 feet, the likelihood of a drone colliding with a human aircraft is exceedingly remote, but the consequences might be disastrous.

Any aerial vehicle that uses software to fly autonomously or that may be controlled remotely by a pilot is referred to as a drone. Numerous drones come equipped with cameras to gather visual data and propellers to stabilize flying paths. Drone technology has been incorporated into industries like videography, search and rescue, agriculture, and transportation.

When in uncontrolled (Class G) airspace, your drone must be flown 400 feet above the ground or less.

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Answer:

A plane mirror is placed to the right of an object. The image formed by the mirror will be a virtual image that appears to be on the left of the mirror.

Explanation:

Complete Question

A rolling ball moves from [tex]x_1 = 8.0 \ cm[/tex] to [tex]x_2 = - 4.1 \ cm[/tex] during the time from [tex]t_1 = 2.9 s[/tex]  to  [tex]t_2 = 6.0s[/tex]

What is its average velocity over this time interval?

Answer:

The velocity is  [tex]v = 3.903 \ m/s[/tex]

Explanation:

From the question we are told that

    The first position of the ball is  [tex]x_1 = 8.0 \ cm[/tex]

    The second position of the ball is  [tex]x_2 = - 4.1 \ cm[/tex]

Generally the average velocity is mathematically represented as

       [tex]v = \frac{ x_1 - x_2}{t_2 - t_1}[/tex]

=>    [tex]v = \frac{ 8 - -4.1 }{ 6 - 2.9 }[/tex]

=>    [tex]v = 3.903 \ m/s[/tex]

Answer:

Work = Mass * Gravity * Height and is measured in Joules. Imagine you find a 2 -Kg book on the floor and lift it 0.75 meters and put it on a table. Remember, that “force” is simply a push or a pull. If you lift 100 kg of mass 1-meter, you will have done 980 Joules of work.

Explanation:

Answer:

The final velocity of the car is 16.893 m/s

The final velocity of the truck is 4.393 m/s

Explanation:

Given;

mass of the car, m₁ = 715 kg

mass of the truck, m₂ = 1490 kg

initial velocity of the car, u₁ = 0

initial velocity of the truck, u₂ = 12.5 m/s

let the final velocity of the car, = v₁

let the final velocity of the truck, = v₂

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(715 x 0) + (1490 x 12.5) = 715v₁ + 1490v₂

18625 = 715v₁ + 1490v₂ -----equation (1)

Apply one-directional velocity formula;

u₁ + v₁ = u₂ + v₂

0 + v₁ = 12.5 + v₂

v₁ = 12.5 + v₂

Substitute v₁ into equation (1)

18625 = 715(12.5 + v₂) + 1490v₂

18625 =8937.5 + 715v₂ + 1490v₂

18625 - 8937.5 = 715v₂ + 1490v₂

9687.5 = 2205v₂

v₂ = 9687.5 / 2205

v₂ = 4.393 m/s

solve for v₁

v₁ = 12.5 + v₂

v₁ =  12.5 + 4.393

v₁ = 16.893 m/s

Answer:

false

Explanation:

Answer:

θ = 122°

Explanation:

Components of a Vector

A vector in the plane can be defined by its rectangular components:

[tex]\vec A =<x,y>[/tex]

Or also can be given by its polar components:

[tex]\vec A =<r,\theta>[/tex]

Where r is the magnitude of the vector and θ is the angle it forms with the positive direction of x.

The relation between them is:

[tex]r=\sqrt{x^2+y^2}[/tex]

[tex]\displaystyle \theta=\arctan\frac{y}{x}[/tex]

It's given the x-component of vector A is x=-25 m and the y-component is y=40 m

(a)

The magnitude of the vector is:

[tex]r=\sqrt{(-25)^2+40^2}[/tex]

[tex]r=\sqrt{625+1600}[/tex]

[tex]r=\sqrt{2225}[/tex]

[tex]r\approx 47.2\ m[/tex]

(b)

[tex]\displaystyle \theta=\arctan\frac{40}{-25}[/tex]

[tex]\displaystyle \theta=\arctan (-1.6)[/tex]

The calculator gives us the value

θ = -58°

But the real angle lies on the second quadrant since x is negative and y is positive, thus:

θ = -58° + 180° = 122°

θ = 122°

Answer: True

Explanation:

Answer:

Gordon Allport’s

Explanation:

edge2o2o

The cardinal, central, and secondary traits are all part of Gordon Allport’s categorized traits. The Correct option is A

Who was Gordon Allport ?

Gordon Willard Allport was born on 11 November 1897 and died 9 October 1967. He was an American psychologist. Allport was first psychologists who studied on personality. he has developed theory of personality. which was one of the greatest finding in the study of personality psychology. He was Appointed  as a social science instructor at Harvard University in 1924,

Gordon Allport was  a great trait theorist who categorized personality traits into three categories cardinal, central, and secondary.

Hence option A is Correct.

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Answer:

Magnetic field lines can be drawn by moving a small compass from point to point around a magnet. At each point, draw a short line in the direction of the compass needle.When opposite poles of two magnets are brought together, the magnetic field lines join together and become denser between the poles.

Explanation:

Answer:

sorry I wish I could it help you

Answer:

480.2 m

Explanation:

The following data were obtained from the question:

Speed of sound (v) = 343 m/s.

Time (t) = 2.8 s

Distance (x) of the cliff =?

The distance of the cliff from the woman can be obtained as follow:

v = 2x /t

343 = 2x /2.8

Cross multiply

2x = 343 × 2.8

2x = 960.4

Divide both side by the coefficient of x i.e 2

x = 960.4/2

x = 480.2 m

Therefore, the cliff is 480.2 m away from the woman.

The distance should be 480.2 m

Since A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. And, there is the velocity of 343 m/s

[tex]v = 2x \div t\\\\343 = 2x \div 2.8\\\\2x = 343 \times 2.8[/tex]

2x = 960.4

x = 480.2 m

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Answer:

option 3

Explanation:

can i get brainliest

As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.

Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.

The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:

n + (-w) = 0

n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N

The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be

f = 0.67 (51.94 N) ≈ 35 N

so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.

The block starts moving as soon as x = 35 N, at which point f = 35 N.

At any point after the block starts moving, we have

f = 0.48 (51.94 N) ≈ 25 N

so that x = 25 N is the required force to keep the block moving at a constant speed.

As x  is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force

Given data :

mass of block at rest ( m ) = 5.3 kg

Coefficient of static friction ( μ_s ) =0.67

Coefficient of kinetic friction is ( μ_k ) = 0.48

Horizontal force applied to block = x  

First step : magnitude of normal force ( n ) when object is at rest

n = w            where w = m*g

n - w = 0

n - ( 5.3 * 9.81 ) = 0     ∴  n = 51.94 N

Second step : Required magnitude of x before the movement of object

F =  μ_s * n

F = 0.67 * 51.94  = 34.79 N  ≈ 35 N

∴ The object will start moving once F and x = 35 N

Final step : Magnitude of x  after object start moving

F = μ_k  * n

  = 0.48 * 51.94 = 24.93 N  ≈ 25 N

∴ object will continue to accelerate at a constant speed once F and x = 25N

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Answer:

High specific heat -> takes more energy to raise/lower object's temperature

Low specific heat -> takes less energy to raise/lower object's temperature

Explanation:

The specific heat capacity is the amount of heat required to raise the temperature of something per unit of mass.

A high specific heat value for an object means it takes more energy to raise or lower that object's temperature. A low specific heat value for an object means it does not take very much energy to heat or cool that object.

Answer:

Acceleration = 18g

Explanation:

Given the following data;

Initial velocity, u = 26m/s

Final velocity, v = 0

Time = 0.15 secs

To find the acceleration;

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

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]

Substituting into the equation, we have;

[tex]a = \frac{0 - 26}{0.15}[/tex]

[tex]a = \frac{26}{0.15}[/tex]

Acceleration = 173.33m/s2

To express it in magnitude of g;

Acceleration = 173.33/9.8

Acceleration = 17.7 ≈ 18g

Acceleration = 18g

Answer:

Option 1

Explanation:

I always see animals do that

Answer:

= 720000 [k]

Explanation:

The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.

[tex]5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ][/tex]

= 720000 [k]