. Emily pushes a 38.8 kg grocery cart of groceries by exerting a 76.0 N force on the handle inclined at 40.0 degrees below the horizontal. What are:
a. The horizontal and vertical components of Emily’s force






b. The acceleration of the cart?

We consider an elementary mass located at this point to be in a stable equilibrium.

An equilibrium is said to be stable if small, externally induced displacements from that state produce forces that tend to oppose the displacement and return the body or particle to the equilibrium state.

Also a body is said to be in a stable equilibrium, if the opposing forces acting on a body are equal.

Thus, along the line through the centers of the Moon and the Earth, there is a point where the gravitational forces exerted by the Moon and the Earth, respectively, are equal. We consider an elementary mass located at this point to be in a stable equilibrium.

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Words are too small for me -_-

The Earth's axis' orientation in space gradually shifts due to the precession.

Precession is the name given to this motion, which is characterized by a cyclic wobble in the direction of the Earth's axis of rotation over a period of 25,772 years. The third motion of the Earth to be discovered, following the much more evident daily rotation and yearly revolution, is precession. The axial tilt of the Earth gradually changes over time as a result of precession. The points on the Earth's orbit where the equinoxes and solstices fall will shift over time. As a result, 13,000 years from now, the location of the summer solstice will change to that of the winter solstice. The Earth's axis changes its direction due to precession. In relation to the stars, the celestial poles steadily shift.

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The direction of vector A which is parallel but opposite to the vector I is is 305°

Direction of vector I = 125°

Vector I makes a parallel line with vector A. Consider vector A and I to be collinear. Collinear lines makes up an angle of 180°

[tex]Angle_{A}[/tex] + [tex]Angle_{I}[/tex] = 180°

[tex]Angle_{A}[/tex] + 125° = 180°

[tex]Angle_{A}[/tex] = 55°

Direction of vector A = 360° - 55° = 305°

The direction of a vector is the angle, the vector creates with positive x-axis in counter clockwise direction.

Therefore, the direction of vector A is 305°

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The distance of the ball that would land which was launched horizontally at a speed (s) of 3.4 m/s from a table that is 7.8 m tall (d) will land is 4.29m

Consider the ball moving horizontally off the table. When it moves, gravitational force acts on it pulling the ball downwards. In addition to that objects that are falling down would accelerate quickly i.e., let’s take an object falling at the speed of x, after time t it would travel at 2x, after 2t it would be 4x.

We need to find time to calculate distance by using speed equation.

For calculating time, let’s use the freefall formula,

d = 1/2 gt²

t = √(2d) / g

t = √2 (7.8)/ 9.8

t = 1.262 s

To calculate distance, we could use speed equation as we have both Time and Velocity

s = d / t

d = s t

d = 1.262 * 3.4

d = 4.29 m

Thus, the distance of the ball that would land is 4.29 m

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The magnitude of its Vfy i.e Vertical componet is mathematically given as

V=705.6 m/s

This is further explained below.

Generally, At this point, we need to do an analysis of the vertical motion of the item, which may be thought of as a free fall motion with a constant acceleration.

g = 9.8

As a result, we are able to employ the suvat equation that is as follows:

v = u +at

As a result, the following expression may be used to determine the magnitude of the object's vertical velocity after t = 72 seconds:

v = 0 + (9.8) (72)

V=705.6 m/s

In conclusion, the magnitude

V=705.6 m/s

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The impulse of the car is 365kgm/s and average force experienced is 437.5N.

acceleration is defined as the change in velocity per unit time.

average force is defined as the force experienced by a body while the acceleration is constant.

given:

initial velocity (u)=0m/s

final velocity(v)=5.20m/s

time(t)=0.832s

mass (m)=70 kg

a)

first we will find acceleration

a=5.2/0.832

=6.25m/s²

now average force experienced (F)= mass*acceleration

F=70×6.25F

=437.5N

now impulse (I) is given by F×t=437.5×0832I

=365kg×m/s

therefore the impulse is 365kg×m/s

b)

Now we will find the average force applied.

Average force experienced is 437.5N

therefore,

The impulse of the car is 365kgm/s and average force experienced is 437.5N.

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The magnitude of current flowing through the conductor is 0.5A.

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The airplane starts form rest, then accelerates by rate of [tex]3m/s^2[/tex] until it reaches 102m/s at the end of the runway. so we have:

[tex]Vo=0m/s\\Vf=102m/s\\a=3m/s^2\\\\[/tex]

From the equations of uniformly accelerated motion (since the airplane's acceleration is uniform) [tex]'Vf^2=Vo^2+2aS' \\[/tex] is compatible to find the minimum runway length (S).

[tex]Vf^2=Vo^2+2aS\\Vf^2-Vo^2=2aS\\S=Vf^2-Vo^2/2a\\[/tex]

[tex]S=((102m/s)^2-(0m/s)^2)/2m/s^2\\S=(10404m^2/s^2-0m^2/s^2)/2m/s^2\\S=(10404m^2/s^2)/2m/s^2\\S=5202m[/tex]

The height of the 2nd spot is 3 times the height of the 1st spot (Option A)

We'll begin by obtaining the height of the first spot. This can be obtained as follow:

v = 2x / t

343 = 2x / 4

Cross multiply

2x = 343 × 4

2x = 1372

Divide by 2

x = 1372 / 2

Height of first spot = 686 m

v = 2x / t

343 = 2x / 12

Cross multiply

2x = 343 × 12

2x = 4116

Divide by 2

x = 4116 / 2

Height of second spot = 2058 m

Comparison = Height of 2nd spot / Height of 1st spot

Height of 2nd spot / Height of 1st spot = 2058 / 686

Height of 2nd spot / Height of 1st spot = 3

Cross multiply

Height of 2nd spot = 3 × Height of 1st spot

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The impulse is equal to the rate of change of momentum.

The term impulse refers to the product of force and time. Let us recall that from Newton's law, impulse is proportional to the change in the momentum of the ball.

We can write from Newtons law;

F. t = Mv - Mu

F = force

t = time

M = mass

v = final velocity

u = initial velocity

Thus, the impulse is equal to the rate of change of momentum.

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Answer:

A)  4.95 m

B)  1.48225 m

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.

Consider the horizontal and vertical motion of the projectile separately.

The horizontal component of velocity is constant, as there is no acceleration horizontally.

Resolving horizontally, taking → as positive:

[tex]u=9.00\quad v=9.00 \quad a=0\quad t=0.550[/tex]

[tex]\begin{aligned}\textsf{Using} \quad s & = \left(\dfrac{u+v}{2}\right)t:\\\\s&= \left(\dfrac{9+9}{2}\right)(0.550)\\s&= (9)(0.550)\\ \implies s&= 4.95\:\sf m\\\end{aligned}[/tex]

As the projectile is fired horizontally, the vertical component of its initial velocity is zero.

Acceleration due to gravity = 9.8 ms⁻²

Resolving vertically, taking ↓ as positive:

[tex]u=0\quad a=9.8\quad t=0.550[/tex]

[tex]\begin{aligned}\textsf{Using} \quad s & = ut+\dfrac{1}{2}at^2:\\\\s&= (0)(0.550)+\dfrac{1}{2}(9.8)(0.550)^2\\s&= 0+(4.9)(0.3025)\\\implies s&= 1.48225\:\sf m\\\end{aligned}[/tex]

The density of 100 ml of liquid mercury in the graduated cylinder that weighs 25 grams is 13.6 g / ml

ρ = m / V

ρ = Density

m = Mass

V = Volume

V = 100 ml

Mass of cylinder = 25 g

Combined mass of cylinder and mercury = 1385 g

Mass of mercury = 1385 - 25 = 1360 g

ρ = 1360 / 100

ρ = 13.6 g / ml

Density is the amount of molecules present in a given volume of the substance. It is denoted by ρ.

Therefore, the density of 100 ml of liquid mercury in the graduated cylinder that weighs 25 grams is 13.6 g / ml

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Answer:

Explanation:

The correct option will be the option (d) that is 1.09 A.

According to the equation given in the question:

That is; P=i^2R                  where I= current, R= resistance

           P=30 W

           R=25 ohm

            30=i^2*25

              I^2=30/25

             I^2=1.2

             I=1.09A

So, the correct answer of the given question will be 1.09A

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(a) The current drawn is 5A.

(b) The resistance of the filament is 48Ω.

(a) The formula for power is:

P = VI

therefore, current drawn is;

[tex]I = \frac{P}{V}[/tex]

[tex]I = \frac{1200}{240}[/tex]

I = 5A

Therefore, the current drawn is 5A.

(b) Using, ohm's law:

V = IR

therefore, resistance of the filament is:

[tex]R = \frac{V}{I}[/tex]

[tex]R = \frac{240}{5}[/tex]

R = 48Ω

Therefore, the resistance of the filament is 48Ω.

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Answer: Vx is changing  and  Vx is greatest near the beginning of the launch.

Explanation: Think of Vx as a bullet fired from a gun. As soon as it is fired it starts to slow down so Vx is changing and at the beginning Vx has its greatest velocity because it can only go slower from there.

answers are B and C. horizontal velocity never changes so A cannot be the answer. that makes B one of the answers. I would also say that the horizontal velocity is greatest near the beginning of the launch as it just starts, since it stays at a constant pace after, so D is not an answer. hope this helps ^_^

It is an example of acceleration due to the wind.

Acceleration: the rate at which the speed and direction of a moving object vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting. Both effects contribute to the acceleration of all other motions.

Acceleration is a vector quantity since it has both a magnitude and a direction. A vector quantity is also velocity. The velocity vector change during a time interval divided by the time interval is the definition of acceleration.

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Proper vector notations are V = 42i  km/hr, r = 6.5j m, and a = -1.9j m/s².

In physics, a vector is a quantity with both magnitude and direction. It is typically represented by an arrow whose length is proportional to the magnitude of the quantity and whose direction is the same as that of the quantity.

A) Let us assume that the west direction is the positive X-axis.

In vector notation, the X-axis is denoted by "i". So, the westward velocity of 42 km/hr  will be written as

V = 42i  km/hr

B) Let us assume that the North direction is the positive Y- axis.

In vector notation, the  Y-axis is denoted by "j". Also, the position vector is represented by the letter 'r'. So position 6.5 measured in m, that is North of the reference point will be written as

r = 6.5j m

C) Let us assmue thaat the downward direction is negative  Y-axis.

In vector notation, the  Y- axis is denoted by "j". so, the downward acceleration measured in m/s2 that has a magnitude of 1.9 will be written as

a = 1.9 × (-j) m/s²

a = -1.9j m/s²

So, proper vector notations are V = 42i  km/hr, r = 6.5j m, and a = -1.9j m/s².

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If a bat hits a ball with 85 N of force the bat will also exert the same force

85N.

You are probably already aware of the force the ball applies to the wall when it is thrown against it. Similar to how the wall exerts a force on the ball, the ball bounces off the wall. Similar to how the earth drags you down with gravity. You might not be aware of it, but you are also pushing the planet with the same amount of power. The third law of Newton is a result of this amazing event.

Newton's Third Law states that if object A exerts a force on object B, then object B must apply a force on object A that is equal in magnitude and directed in the opposite direction.

An object interacts with another object when a push or pull is applied to it. A product of interaction is force. Frictional force, for example, falls into the category of contact force. Non-contact force, on the other hand, includes gravitational force. Newton's third rule of motion explains that when two bodies contact, they exert force on one another. This force is referred to as an action and reaction pair.

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When Two Objects Are Charged By Rubbing Or Peeling, both objects have both types of charge, but there are different amounts of each type on each object. (One object has more negative charges than positive chrages and the other object has more positive charges than negative charges); option C.

Charges are the inherent properties of the particles of matter which determines how the atoms will react when placed in an electric filed.

There are two types of charges in an atom:

The electrons in an atom can be easily transferred from one atom to another, hence objects can be charged by rubbing or peeling one against the other.

When when two objects are rubbed or peeled against each other, electrons move from one object to the other.

The object that acquires the electrons become negatively charged whereas the object that loses the electrons become positively charged. However, both objects have both electrons and protons present in them.

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The gravitational force between the two objects whose mass is reduced to one-third of its original mass is 400 N

We know that,

F = G[tex]m_{1}[/tex][tex]m_{2}[/tex] / r²

F = Gravitational force

G = Gravitational constant

[tex]m_{1}[/tex], [tex]m_{2}[/tex] = Mass of two objects

r = Distance between two objects

G = 3600 N

[tex]m_{1}'[/tex] = [tex]m_{1}[/tex] / 3

[tex]m_{2}'[/tex] = [tex]m_{2}[/tex] / 3

F' = G [tex]m_{1}'[/tex] [tex]m_{2}'[/tex] / r² = G[tex]m_{1}[/tex][tex]m_{2}[/tex] / 9 r²

F' = F / 9

F' = 3600 / 9

F' = 400 N

Gravitational force is the force of attraction between two objects due to the presence of gravity.

Therefore, the gravitational force between the two objects whose mass is reduced to one-third of its original mass is 400 N

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The vertical speed of the ball increases and its horizontal speed remains constant as usual because air friction is being neglected.

By definition, air resistance refers to the forces that work against an object's relative motion as it travels through the atmosphere. These drag forces cause the object to move more slowly by acting in opposition to the speed of the incoming flow. Since drag is the part of the net aerodynamic force operating in the opposite direction to the direction of the movement, it differs from other resistance forces in that it directly depends on velocity.

Fluid friction includes friction in the air. These friction forces differ from the conventional model of surface friction in that they rely on velocity. Only rare examples can be dealt with analytically since the velocity dependence may be extremely intricate. Air resistance is roughly proportional to velocity and can be represented in the form of very low speeds for microscopic particles.

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Answer:

because

Explanation:

ok just because

The distance to the destination is 18,000 meters or 18 kilometers. the formula for calculating the distance is speed × Time.

Using the formula:

Distance = Speed × Time

Speed = 15 m/s (meters per second)

Time = 20 minutes

First, we need to convert the time from minutes to seconds, as the speed is given in meters per second.

1 minute = 60 seconds

Time in seconds = 20 minutes × 60 seconds/minute = 1200 seconds

Now, we can calculate the distance:

Distance = 15 m/s × 1200 seconds

Distance = 18,000 meters

So, the distance to the destination is 18,000 meters, or 18 kilometers (since 1 kilometer = 1000 meters).

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On the day of the summer solstice, at noon, Eratosthenes was aware of a deep well on the Nile close to Aswan where the sun shone all the way to the bottom. Because the sun was not overhead in Alexandria, the length of that shadow was not zero.

We believe that the light rays from the sun that reach us are all traveling in the same direction; they are parallel, as the sun is thought to be extremely far away. No shadow would be formed by a vertical stick at point A, a "moderate" shadow by a stick at point B, and a very long shadow by a stick at point C. A radius of the earth leading to point A is shown on the diagram, and another leading to point B. A little geometry reveals that the angle between those two radii and the difference between the shadow angles are the same. In our example, the shadows indicate that during our journey from Florida to Pennsylvania, we had rotated the earth by about 10 degrees.

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Answer:

See below

Explanation:

On Earth , turtle weight  = 5 * 9.81 =49 N

  Moon =  490.5 * 1/6 = 8.18 N

  Jupiter    490.5 * 2.64 = 129.4 N

129.4 - 8.18 N = 121.2 N more on Jupiter than moon

Work done on the book is 10.595 Joules .

Work done on the book would be equal to the Potential Energy

Work Done = Potential Energy

W = m g h

where ,  ' m ' represents mass of the book

' g ' stands for  Acceleration due to gravity

' h ' represents the height to which the book is raised

So, according to the given question ,

m = 5.1 kg

g = 9.8 m / s ²

h = 21.2 cm

  =  0.212 m

W = m g h

    = 5.1 kg x 9.8 m / s ² x 0.212 m

    = 10 . 595 Joules

Hence, Work done on the book is 10.595 Joules .

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The correct answer is 2.26 m.

Initial velocity of elevator u =0

Constant upward acceleration of the elevator a= 2.90 m/s²

Height of the bolt from the floor of the elevator h=3.00 m

Initial velocity of the bolt in the elevator u = 0 m/s

Elevator is moves upward with a constant acceleration 2.90 m/s²

so the net acceleration a' = (a+g) m/s²

=( 2.90 + 9.8) m/s²

a' = 12.7 m/s²

Initial velocity of the bolt u = 0 m/s

Distance traveled by the bolt h = 3.00 m

Substitute these values in the equation

h=ut+ 1/2 ( a )t², we get

3.00 = (0)x+1/2 (12.7)(t^2)

3.00 = 6.35 t²

t²= 3.00/6.35

t = 0.68 s

i.e., after 0.68 s the bolt reaches the floor of the elevator.

(b)For the observer in the elevator acceleration

a'=(a+g) m/s²

=( 2.90 + 9.8) m/s²

a'= 12.7 m/s²

Time taken bolt to reach the floor of the elevator t = 0.68 s From the equation v=u+a't we get

v=(0)+(12.7) (0.68)

v = 8.636 m/s

Velocity of the bolt when it reaches the floor of the elevator according to an observer into elevator 8.636 m/s

ii) An observer standing on the floor landings of the buildings: for observer standing on the floor landing, the acceleration of the

bolt

appears a = g m/s²

= 9.8m/s²

Time taken the bolt to reach the floor t = 0.68 s

Initial velocity of the bolt u = 0 m/s

Substitute these values in the equation

v=u+at we get

v=0+(9.8) (0.68 s)

v= 8.636 m/s

Velocity of the bolt when it reaches the floor of the elevator according to

an observer standing on the floor landing = 8.636  m/s

c) i) According to an observer in the elevator in this situation acceleration a= 12.7 m/s²

time t= 0.68 s initial velocity u = 0

substitute these values in the equation 1

s = ut +1/2 at²

We get

s =(0) (0.659)+1/2(12.7 ) (0.68)²

= 2.93 m

Distance traveled by the bolt according to the observer on the elevator 2.93 m

iii)According to an observer standing on the floor landing in this situation

acceleration a=g m/s²

= 9.8m/s²

Time t= 0.68 s

Initial velocity u = 0 m/s Substitute these values in the equation

1 ut + 1/2 at ²,

s=ut+1/2at²

we get

s=(0)+1/2(9.8) (0.68)²

= 2.26 m

Distance traveled by the bolt according to the observer standing on the floor landing 2.26 m.

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