You hike two thirds of the way to the top of a hill at a speed
of 3.5 mi/h and run the final third at a speed of 6.5 mi/h.

The acceleration of the crate  after the frictional force is overcame is determined as 1 m/s².

Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity with time.

The  acceleration of the crate is calculated from the net force acting on the crate as shown below;

F(net) = applied force - frictional force

F(net) = 10 N - 5N

F(net) = 5 N

Apply Newton's second law of motion to calculate the acceleration of the crate.

F(net) = ma

where;

a = F(net) / m

a = 5/5

a = 1 m/s²

Thus, the acceleration of the crate  after the frictional force is overcame is determined as 1 m/s².

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Memoir is a type of narrative nonfiction: Monique went through many trials and tribulations and decided to write a book about this particular period in her life.

Because it combines elements of creative writing with the relation of facts required to produce a work of nonfiction, narrative nonfiction is a type of writing that is occasionally referred to as creative nonfiction. This genre comprises of writing that is factually accurate yet written with a creative flair that contributes to improving the work's flow and appeal. The genre of narrative nonfiction can include journalism and some genres of biography, essay, memoir, and personal essay. When compared to plain nonfiction, this genre is very new, and as the genre has developed, so too has the scrutiny that is applied to this kind of writing.

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Answer:

Final speed of car = 9.834 m/s

Displacement = 0.035 km

Explanation:

Part 1

Final velocity, v of an object moving with an acceleration a m/s² for t seconds and an initial velocity u m/s is given by the formula
v = u + at m/s

[tex]\textrm{Initial velocity } u = 22.1 km/h\\\\\textrm{Convert this to m/s:}\\\\\textrm{1 km/h = 1000m/3600 s}\\\\ 22.1 km/h = 22.1 x 1000/3600 = 6.139 m/s\\\\\boxed{\textrm{Final velocity } v = u + at = 6.138 + 0.84 x 4.4 = 9.834 m/s}[/tex]

Displacement s in m for a constant acceleration of a m/s² for a time t seconds is given by the formula:
[tex]x = ut+\dfrac{1}{2}at^2[/tex]
where u is initial velocity

Plugging in the values we get
[tex]s = 6.139\cdot 4.4 + \dfrac{1}{2}\cdot0.84\cdot (4.4)^2 = 35.143 \;\mathrm {meters}[/tex]
35.143 meters = 35.143/1000 =  0.035 km

Answer:

1)  9.83 m/s (2 d.p.)

2)  0.04 km (2 d.p.)

Explanation:

Constant Acceleration Equations (SUVAT)

[tex]\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}[/tex]  

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Given:

Convert km/h into m/s by dividing by 3.6:

[tex]\implies \sf 22.1\;km/h=\dfrac{22.1}{3.6}\;m/s[/tex]

Substitute the given values into the formula and solve for v:

[tex]\begin{aligned}v&=u+at\\\implies v & = \dfrac{22.1}{3.6}+0.84(4.4)\\v & = 6.1388...+3.696\\v & = 9.834888...\\v & = 9.83 \sf \;m/s\;(2\:d.p.)\end{aligned}[/tex]

Therefore, the final speed of the car was 9.83 m/s (2 d.p.).

Substitute the given values into the formula and solve for s, remembering to use the initial velocity in m/s:

[tex]\begin{aligned}s & = ut+\dfrac{1}{2}at^2\\\implies s &=\dfrac{22.1}{3.6}(4.4)+\dfrac{1}{2}(0.84)(4.4)^2\\s &=27.0111...+8.1312\\s &=35.1423111...\sf m\end{aligned}[/tex]

Convert meters into kilometers by dividing by 1000:

[tex]\implies \dfrac{35.1423111...}{1000}=0.0351423111...=0.04 \sf \; km\;(2\:d.p.)[/tex]

Therefore, the displacement of the car after 4.4 s was 0.04 km (2 d.p.).

Answer:

VPA

Explanation:

Answer:

Bannister = 3:59.4 sec = 239.4 sec

Guerruoj = 3:43.13 = 223.13 sec

1 mi = 5280 ft

G = 5280 / 223.13 = 23.66 ft/sec

B = 5280 / 239.4 = 22.06 ft/sec

In 223.13 sec B runs 223.13 * 22.06 = 4922 ft

G Wins by 358 ft = 119.3 yds = 109.1 meters

The straight line in the velocity versus time graph has a physical significance as it depicts constant acceleration of the motion.

Acceleration is a physical quantity which is defined as the change of velocity per unit time it is a vector quantity and therefore can be both positive and negative.

Velocity versus time graph is the depiction of emotion in which the y-axis represents the velocity of the motion and the x-axis represents the time the graph can either be a positive state line or a curve depending on the acceleration of the motion.

According to the question the velocity versus time graph in our case is a straight line.

A straight line in a velocity versus time graph means that motion has a constant acceleration.

The following conclusion can also be reached from the use of differential calculus as we differentiate velocity with respect to time we get a new physical quantity that is known as acceleration.

On differentiating velocity with respect to time we will get a new value that will represent the change in velocity per unit time and since from the definition of acceleration we know that acceleration is change in velocity per unit time therefore we can conclude that the straight line in the velocity time graph depicts constant acceleration.

Therefore, The straight line in the velocity versus time graph has a physical significance as it depicts constant acceleration of the motion.

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Explanation:

I solved it in the picture but I'm not a hundred percent sure

The horizontal distance is 82.455m and vertical distance is 19.734m

Let us first find the x and y components of velocity

The velocity is given to be 41m/s

the angle of projectile is given to be 29 degree

The x component of velocity = Vx =v cosα =41 x 0.8746 =35.85 m/s

The y component of velocity= Vy= v sinα=41 x 0.484 =19.85 m/s

The horizontal displacement x= Vxt = 35.85 x 2.3 =82.455 m

To solve for vertical distance y = Vyt+1.2at^2

y = 19.85 x 2.3 + 1/2 x -9.8 m/s^2 x 2.3

  = 19.734m

Hence, the horizontal distance is 82.455m and vertical distance is 19.734m

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Yes, the stars going supernova will have impact on allowing life to exist in nearby planets.

When a star goes supernova, a large amount of radiation is emitted. So, if a nearby planet inhabiting life has a ozone layer, it will be depleted. This results in increased harmful radiation levels on the ground. There will be mutations in organisms that can survive the radiations. Those that are not strong will die.

A supernova is the explosion of a star. This phenomenon happens when a star's nuclear fuel in its core runs out. As a result the pressure created by the burning in its core becomes completely null. Thus the gravity is only force acting, so the star collapses on itself emitting huge quantities of light and harmful radiation.  

Therefore, the stars going supernova will have impact on allowing life to exist in nearby planets.

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Answer:

1281

Explanation:

what is the magnitude of the resultant of a 61m north and 21m south vector

Mass on the planet would be 10/3 times the mass on Earth weigh on this new planet.

A quantitative measure of the fundamental characteristic of all matter, inertia. In essence, it's the resistance of a mass of matter to changing its direction or speed in reaction to the force applied to it. The much more weight a body has, the less change an applied force produces.

Let the mass of Earth = Me

Let the radius of Earth = Re

Mass of the planet, Mp = 10Me

Radius of the planet, Rp = 3Re

Weight on other planet, W = ?

Weight on planets is determined by the gravitational force

Gravitational force, F = GMm/r²

where, G = gravitational constant

M = Mass of the planet

m = our mass

r = radius of the planet

Gravitational force on Earth:

F = G X Me X ma/ Re².....eq (1) (ma = mass on earth)

Gravitational force on planet :

F = G X 10Me X mₓ/ (3Re)²....eq (2) (mx = mass on other planet)

Dividing both the equations, we get

mx = 10/3ma

Therefore, mass on the planet would be 10/3 times the mass on Earth.

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The total forces acting on the rope is 125 N towards the right side.

125 N of total forces are being applied to the rope's right side.

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We can see that the speed of the car is obtained as 12 miles per hour.

The term speed is the ratio of the distance covered to the time taken. We can see that the distance that have been covered by the boy is  24 miles. The total time that have been taken to cover this distance is 30 mins or 0.5 hours.

We now come up with the following;

Distance = 24 miles

time =  0.5 hours

We now want to compute the speed of the boy in miles per hours so we have;

Speed = 24 miles /  0.5 hours = 12 miles per hour

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1. The weight of a 12.5 kg mass is 122.625 N

2. The mass of a 2.5 N weight is 0.255 kg

3. The mass of a 200 N weight is 20.4 kg

4. The scale read for a 4 kg mass is 39.2 N

The mass of a substance is the quantity of matter present in that substance.

The quantity of matter or mass of a substance is constant and is measured using a weighing balance.

The weight of a substance is the product of the mass of a substance and the acceleration due to gravity.

The weight of a substance varies and is measured using a spring balance and the unit is N.

Weight = m * g

where;

m is mass in kg

g = 9.81 m/s²

Considering the given substances:

1. The weight of a 12.5 kg mass = 12.5 * 9.81

The weight of a 12.5 kg mass = 122.625 N

2. The mass of a 2.5 N weight = 2.5/9.81

The mass of a 2.5 N weight = 0.255 kg

3. The mass of a 200 N weight = 200/9.81

The mass of a 200 N weight = 20.4 kg

4. The scale read for a 4 kg mass = 4 * 9.81

The scale read for a 4 kg mass = 39.2 N

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A Vector is a quantity which has both magnitude as well as direction.

vectors follow parallelogram law of addition and uses cosine of the angle between the vectors.

The addition of two vectors, i.e. c, is given by the formula

c=[tex]\sqrt{a^2+b^2+2abcos\alpha }[/tex]

where [tex]\alpha[/tex] is the angle between the vectors

now we know that maximum value of cos[tex]\alpha[/tex] can be is 1

and minimum value of cos [tex]\alpha[/tex] can be-1

now the maximum value cos[tex]\alpha[/tex] is 1 at the angle of π/2

and the minimum value of cosα is -1 is at the angle of -π

Therefore, A and B should have 0 degree between them and in the same direction to have the maximum value

there fore they should be in same direction and 0 degree meaning they should be on same line

A and B will have minimum value when the angle is 180 degree which means they should be on same line and be inn opposite direction to each other.

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The kinetic energy of of given data is K.E. = 1,047.62 J

To calculate kinetic energy using the formula,

K.E. = 1/2 m v²

where,

K.E is kinetic energy

m is mass

v is velocity

So, Putting values,

K.E. = 1/2 x 29 x (8.5m/s)²

K.E. = 1,047.62 J

The energy an object has as a consequence of motion is known as kinetic energy. A force must be applied to an object in order to accelerate it. We must put in effort in order to apply a force. After the work is finished, energy is transferred to the object, which then moves at a new, constant speed.

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Answer:

A. Parallel connection

Answer:

Parallel circuit to be exact

When a flowerpot falls from a window sill 36.5 m

above the sidewalk, then the velocity of the flowerpot is 26.7 m/s.

From Newton's third equation of motion,

v^2 = u^2 + 2gh

where,

h is the height of the object or body from ground

u is the initial velocity of the body or object

v is the final velocity of the body or object

g is the acceleration due to gravity

Now, as we know that

Flowerpot is at rest. So, u = 0

g = 9.81m/s^2

h = 36.5m

By substituting all the values, we get

v^2 = 2 × 9.81 × 36.5

= 716.13

v = 26.7m/s

Thus, we concluded that when a flowerpot falls from a window sill 36.5 m

above the sidewalk, then the velocity of the flowerpot is 26.7 m/s.

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The apple takes 2.05 seconds to fall due to gravity.

Newton's Second Law of Motion and Newton's Law of Universal Gravitation serve as the foundation for the formula for the acceleration caused by gravity. The most practical formula for estimating the acceleration caused by gravity is given by these two laws: g = G*M/R2, where g is the acceleration caused by gravity, G is the universal gravitational constant, M is mass, and R is distance.

Everything has a gravitational pull on every other object, according to Newton's Law of Universal Gravitation. The force is inversely proportional to the square of the distance between the centers of both objects and directly proportional to the masses of both objects. This is determined mathematically by the formula F = G*(m1*m2/d2), where F is the force and G is the gravitational constant of the universe.

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The package strike the ground relative to the point directly below it was released at 213.3m.

Given values,

Velocity of plane 45 m/s

Acceleration due to gravity 9.8 m/s²

horizontal component of plane velocity is

ux = 45 m/s

Vertical component is zero

In the same way

ay = - 9.8  m/s²

ax = .0 m/s^2

Motion in y direction = -110 m

From equation of motion time of flight is

d = u x t + 1/2 at²

where,

d is distance

u is velocity initial

a is acceleration

t is time

So, putting values

-110 = 0 + 1/2 (-9.8)t²

-110 = -4.9t²

t² = 110 / -4.9

t² = 22.45

t = √22.45

t = 4.74

Motion in x direction,

d = u x t + 1/2 at²

d = ( 45 x 4.74) + 1/2 (0) 4.74²

d = 213.3 + 0

d = 213.3m

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Answer: C

Explanation:

The bigger the satellite, the longer it will take to get around the satellite. Hence, the velocity would get smaller.

r*v=4r*?

r*v/4r=?

r/4r * v=?

1/4 * v=?

?=v/4 ==> C

Answer:

It's the last one- Finley > Xander > Max

Explanation:

You just divided the velocity by the time

The acceleration, in m/s² , that riders experience is 50.3 m/s²

Since the riders move in a circle, they undergo circular motion and will have a centripetal acceleration.

So, the centripetal acceleration, is given by a = v²/r where

Given that a rider in one swing is moving at 34 m/s with respect to the ground in a 46- m -diameter circle. We have that

So, substituting the values of the variables into the equation for the acceleration, we have

a = v²/r

a = (34 m/s)²/23 m

a = 1156 m²/s²/23 m

a = 50.26 m/s²

a ≅ 50.3 m/s²

So, the acceleration, in m/s² , that riders experience is 50.3 m/s²

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(a) The volume of asphalt needed is 6400 m³.

(b) The number of dump truck loads required is 320.

(c) The gross mass of each truck is 34400 kg.

The dimension of the two-lane highway 20 km long, 8 m wide, and 4 cm thick.

The mass of a dump truck is 20 metric tons, and carrying capacity is 20 m³ and the density is 0.72 g/cm³.

The two-lane highway:

length, l = 20 km = 20 × 10³ m

breadth, b = 8 m

Thickness, h = 4 cm = 4 × 10⁻² m

(a) The volume of the asphalt will be:

V = l × b × h

V = 20 × 10³ × 8 × 4 × 10⁻² m³

V = 64 × 10² m³

V = 6400 m³

V = 6400 × 10⁶ cm³

(b) The carrying capacity of 20 m³ asphalt.

Therefore, the number of truckloads required will be:

= 6400 m³ / 20 m³

= 320

(c) Now,

Density, d = 0.72 g/cm³ = 720 kg/m³

The empty mass is 20 metric tons which is equal to 20 ×10³ kg.

Therefore, the gross mass of each truck will be

m = 20 ×10³ kg + 720 kg/m³ × 20 m³

m = 20000 kg + 14400 kg

m = 34400 kg

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The frequency of the simple harmonic motion is 2 Hertz , the maximum acceleration is 3 u and the maximum velocity is equal to 2.48 m/s.

Oscillation is the repetitive or periodic variation, typically in time, of some measure about a central value or between two or more different states.

Time period in periodic oscillation refers to the time taken by the system to complete one oscillation.

Given:

Time period of the oscillation = 0.5 seconds

Amplitude of the oscillation = 2 cm

For determining frequency we use the formula of frequency.

f = 2π/T

Where,

f is the frequency of the oscillation

T is the time period of the oscillation

On substituting values we get,

Frequency (f) = 1/0.5

Frequency (f) = 2 Hertz

For determining maximum acceleration we use,

a = A × w × w

Where,

a is the acceleration

A is amplitude of the SHM

w is the angular frequency

Now,

a = A × w × w

a = 0.02 × 12.4× 12.4

a = 3 u

For maximum velocity we use,

v = A × w

Where v is maximum velocity

v = 0.02 × 12.4

v = 2.48 m/s

Therefore, the frequency of the simple harmonic motion is 2 Hertz , the maximum acceleration is 3 u and the maximum velocity is equal to 2.48 m/s.

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The acceleration of driver with given values is - 1.771 m/s².

To know acceleration used formula is the rate of change of the velocity of an object with respect to time that is,

a = dv / dt

Where,

dv is the small change in velocity

dt  is change in time of an object

Given,

The distance traveled by the driver is +2170 m

The time taken by the driver is 35. 0 s

The initial velocity of driver is + 62.0 m/s

Therefore , to get the acceleration of the driver with the equation of motion

v = u + at

Here v is the final velocity which is taken zero

Then by putting the values in the above equation,

0 = (+62. o m/s) + (a) 35.0s

a = -  (+62. o m/s) / ( 35.0s)

a = - 1.771 m/s²

The negative sign of acceleration shows that, the driver is slowing then acceleration is in the opposite as the velocity.

So, the acceleration

a = -1.771 m/s².

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According to the give value of the mass of lemonade and change in the temperature, the specific heat capacity of the lemonade is 3.81 J/kg/k.

The formula for the specific heat capacity (C) = Q/ m × ΔT

Q is energy added and the value, which is 334J

m is the mass of the lemonade, which is 300g = 0.3 kg

Δ T is the change in temperature, which is 19°C =(273+19)= 292 K

So, C = 334/ 0.3 × 292

         = 3.81 J/kg/k

So the specific heat capacity of the lemonade is 3.81 J/kg/k.

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The precession is the motion of Earth along its orbital path.

Earth's rotation or Earth's spin is the rotation of planet Earth around its own axis.

Precession of the Earth's axis consists of a cyclic wobbling in the orientation of Earth's axis of rotation with a period of 25,772 years.

The precession of the Earth's axis  wobble is due to tidal forces caused by the gravitational influences of the Sun and Moon.

Thus, precession is the motion of Earth along its orbital path.

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