An 1.0 kg mass book, initially at rest, is supported against a vertical wall, being pressed by a horizontal force of F 50 N module, as indicated in the figure. The acceleration of gravity has a modulus of 10.0m / s, the coefficient of static friction between the wall surface and the book is 0.25 and the coefficient of kinetic friction 0.20. Mark the correct statement.
a) The book slides along the wall and the modulus of the friction force is 10.0 N
b) The book remains at rest and the module of the normal force of contact between the book is the wall is 10N
c) The book remains at rest and the modulus of frictional force between the book and the wall is 12.5N.
d) The book remains at rest and the modulus of frictional force between the book and the wall is 10 N
e) The book slides along the wall and the module friction force is 12.5 N.

Answer:

the is answer is Static friction took the test its right.

Explanation:

First write down all the known variables:

d = -1.1m

a = -9.8m/s^2

vi = 0m/s

t = ?

Now use the kinematic equation that relates all the variables:

d = vi*t + (1/2)at^2

(-1.1) = (0)t + (1/2)(-9.8)t^2

-1.1 = -4.9t^2

0.22445 = t^2

0.47s = t

It takes 0.47 seconds for the juice to reach the ground.

I hope this helps you!

Answer:

[tex]a=\frac{F}{m}[/tex]

[tex]a=\frac{v-u}{t}\\a=2\frac{s-ut}{t^2}\\a=\frac{v^2-u^2}{2s}[/tex]

Explanation:

The most common formula from which acceleration of an object can be determined is

F = ma

Where,

F = Force

m = Mass

a = Acceleration

Rearranging the equation we get

[tex]\mathbf{a=\frac{F}{m}}[/tex]

There are equations also from which acceleration can be determined they are

[tex]v=u+at\quad\\s=ut+{\tfrac {1}{2}}at^{2}\\v^{2}=u^{2}+2as[/tex]

Rearranging the equations we get

[tex]\mathbf{a=\frac{v-u}{t}}[/tex]

[tex]\mathbf{a=2\frac{s-ut}{t^2}}[/tex]

[tex]\mathbf{a=\frac{v^2-u^2}{2s}}[/tex]

Where,

a = Acceleration

u = Initial velocity

v = Final velocity

s = Displacement

t = Time taken

when a marble started from rest we can say that

initial velocity of the marble is

[tex]v_i = 0[/tex]

now it took t = 1.4 s to cover total distance of 54 cm by the marble.

so here we will have use kinematics equation in order to find the acceleration of the marble

so we can use

[tex]d = v_i* t + \frac{1}{2}at^2[/tex]

d = 54 cm = 0.54 m

we will plug in all data

[tex]0.54 = 0 * 1.4 + \frac{1}{2}*a * 1.4^2[/tex]

[tex]0.54 = 0 + 0.98*a[/tex]

[tex]a = \frac{0.54}{0.98}[/tex]

[tex]a = 0.55 m/s^2[/tex]

so the acceleration of the marble will be 0.55 m/s^2 on the given ramp

solution is written in attached image.

To find the acceleration of the blocks and the tension in the cable, we can use Newton's second law of motion and set up equations for the hanging mass and the block on the surface. Solving these equations will give us the values we need.

In order to find the acceleration of the blocks and the tension in the cable, we can use Newton's second law of motion which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

In this system, the force of gravity acting on the hanging mass mg pulls it downward, while the tension in the cable T pulls it upward. Since the system is frictionless, the acceleration of both blocks will be the same and can be represented as a.

Applying Newton's second law to each block, we can set up the following equations:

For the hanging mass: mg - T = ma

For the block on the surface: T = mR

where m is the mass of the hanging block, g is the acceleration due to gravity, a is the acceleration of the system, and R is the mass of the block on the surface.

Solving these equations simultaneously will give us the value of the acceleration and the tension in the cable.

Sample A: Iodine

Sample B: Bromine

Sample C: Chlorine.

Answer:

   The minimum length for the slip road = 100 meter.

Explanation:

We have equation of motion, [tex]v^2=u^2+2as[/tex], where u is the initial velocity, u is the final velocity, s is the displacement and a is the acceleration.

 In this case initial velocity = 10 m/s, final velocity = 30 m/s and acceleration = [tex]4m/s^2[/tex], we need to find displacement.

 Substituting

     [tex]30^2=10^2+2*4*s\\ \\ 800=8s\\ \\ s=100meter[/tex]

  The minimum length for the slip road = 100 meter.

If a car is travelling 70 m.p.h that means it is travelling 70 miles PER HOUR and since we want to know half an hour we can do

70mph divided by 2 =35m.p half hour so the car is travelling 35miles in half an hour

Horizontal distance covered is given as

[tex]x = 22 m[/tex]

vertical distance climbed by is given as

[tex]y = 4.8 m[/tex]

now we need top find the angle made by the displacement vector with respect to horizontal

So here we will say that we can use trigonometry to find out the angle with horizontal

The formula is given as

[tex]tan\theta = \frac{opposite \: side}{adjacent \: side}[/tex]

here given that

opposite side = 4.8 m

adjacent side = 22 m

now we can plug in all data in above equation

[tex]tan\theta = \frac{4.8}{22}[/tex]

[tex]tan\theta = 0.22[/tex]

[tex]\theta = tan^{-1}0.22[/tex]

[tex]\theta = 12.3 degree[/tex]

so here the displacement will make 12.3 degree angle with horizontal

When you run in gym class and then climb a rope, the direction angle of your total displacement as measured from the horizontal can be calculated using the inverse tangent function. Given a horizontal displacement of 22m and a vertical displacement of 4.8m, the direction angle is approximately 12.42 degrees.

The direction angle of displacement denoted by Θ in this context, is calculated based on the horizontal and vertical components of your displacement. Since displacement in this scenario forms a right triangle with the horizontal displacement (22m) and vertical displacement (4.8m), we can calculate the angle using the inverse tangent function (also known as arctan or tan^-1).

So, Θ = arctan(vertical displacement/horizontal displacement) = arctan(4.8m / 22m) = 12.42 degrees.

The direction angle of your total displacement, as measured from the horizontal in this context is approximately 12.42 degrees.

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Transformer is used to change the voltage from higher value to lower value

It is done by changing the number of coils in primary and secondary coils

The working equation is given as

[tex]\frac{V_s}{V_p} = \frac{N_s}{N_p}[/tex]

here we have

[tex]V_s[/tex] = Secondary coil voltage = 120 V

[tex]V_p[/tex] = Primary coil voltage = 240,000 V

[tex]N_s[/tex] = Turns in secondary coil

[tex]N_p[/tex] = Turns in primary coil

Now by using the equation above

[tex]\frac{120}{240,000} = \frac{N_s}{N_p}[/tex]

now we have

[tex]\frac{N_p}{N_s} = \frac{240,000}{120}[/tex]

[tex]\frac{N_p}{N_s} = 2000[/tex]

so the ratio must be 2000

2000:1 is the ratio of the number of turns in the primary coil to the secondary coil of a transformer that steps down the voltage from 240,000 V to 120 V.

The ratio of the number of turns in the primary coil ([tex]n_p[/tex]) to the secondary coil ([tex]n_s[/tex]) of a transformer is directly related to the input ([tex]V_p[/tex]) and output voltage ([tex]V_s[/tex]).

The formula to calculate this ratio is:

[tex](V_p / V_s) = (n_p / n_s)[/tex]

Given:

Using the formula:

[tex](240,000 V / 120 V) = (n_p / n_s)[/tex]

This simplifies to:

[tex]2000 = (n_p / n_s)[/tex]

Therefore, the ratio of the number of turns in the primary coil to the secondary coil is 2000:1.

C. About the same amount of time to hit the floor.

I just took a test with this question and got it right :)

Hope this is helpful.

Option C :- about the same amount of time to hit the floor

Because Mass has no effect on how fast things travel  in free fall situations . If you look at your kinematics equations  none of them consider mass as a variable.

however mass does have effect on energy.

Given:

u(initial velocity)=0

v(final velocity)= 10 m/s

t= 4 sec

Now we know that

v= u + at

Where v is the final velocity

u is the initial velocity

a is the acceleration measured in m/s^2

t is the time measured in sec

10=0+ax4

a=10/4

a=2.5 m/s^2

The answer is (C) as

[tex]C (s) + O_2 (g)[/tex] → [tex]CO_2[/tex]

It is commonly known that oxygen has the importance in the combustion reaction and in the absence of this molecule, the no combustion reaction will proceed. SO, for the combustion reaction, it is important that oxygen must be present in the reactant side of the equation than on the product side. Another identity is also present in the combustion reaction as carbon dioxide and water will be present in the product side. In the present options, the oxygen is only present in equation (C) in the reaction side.

So, the right option is

[tex]C (s) + O_2 (g)[/tex] → [tex]CO_2[/tex]

Final answer:

Option C, C (s) + O2 (g) → CO2 (g), is the combustion reaction, which involves the burning of carbon in the presence of oxygen to form carbon dioxide gas.

Explanation:

The student is asking to identify which of the given options is a combustion reaction. A combustion reaction typically involves a substance (usually carbon-based) combusting in the presence of oxygen to form carbon dioxide and water, releasing heat in the process. Among the options given, C. C (s) + O2 (g) → CO2 (g) represents the combustion of carbon (in the form of graphite or diamond), which reacts with oxygen to form carbon dioxide gas. This is a standard example of a combustion reaction.

It's important to note that the other reactions listed, such as combining hydrogen and chlorine to make hydrochloric acid (A), sodium and chlorine to make sodium chloride (B), and the decomposition of calcium carbonate to form calcium oxide and oxygen gas (D), are not combustion reactions.

Given:

v(final velocity)= 7.5 m/s

t(time taken to pedal along)=4.5 s

Displacement (s)=19 m

Now we know that

s=ut+1/2(at^2)

Where s is the displacement measured in m

u is the initial velocity measured in m/sec

a is the acceleration measured in m/s^2.

t is the time taken to cover this distance.

Substituting the given values in the above formula we get

19= 4.5u+1/2(a x 4.5 x 4.5)

20.25 a + 9 u = 38

Now we also know that

v= u + at

Substituting the given values in the above formula we get

7.5= u + 4.5a

Solving for u and a from the above equations we get

u = 0.944m/s

a= 1.45 m/s^2

Thus the initial velocity is 0.944 m/s

Friction in this case will act counter to the direction of the box's movement. The box moves left to right, so the force of friction acts in the opposite direction. (D)

Answer: Hello there!

first, we have two notions of the force of friction:

The force of friction is in the opposite direction of the motion direction.

It acts in the contact point between the objects.

Then if the box is moving from left to right, the friction force should point from right to left, and be in the bottom of the box (the contact point between the box and the floor).

The diagram that shows this is the D.

sound travels through air, not through vacuum = lack of air

A bell can be heard inside a sealed bell jar full of air because sound waves travel through the air, causing vibrations that reach the ear and are interpreted as sound.

The reason a bell can be heard when rung inside a sealed bell jar full of air is due to the ability of sound waves to travel through gases, like air. Sound waves are mechanical waves requiring a medium to propagate, and in the presence of air inside a bell jar, these waves can move by causing the air molecules to vibrate. This sets off a chain reaction of vibrating molecules that eventually reach your ear, translating into the sound you hear. In contrast, if the jar were to be evacuated, leaving a vacuum, there would be no medium for the sound waves to travel through, and the bell wouldn't be heard.

time for which Jill and Scott both traveled is 30 min

as we know that

1 hour = 60 min

now we will convert the time of travel from minutes to hour

[tex]t = \frac{30 min}{60 min}* 1 hour[/tex]

[tex]t  = 0.5 hour[/tex]

now we will have to find the speed of both

For speed we know the formula

[tex]speed = \frac{distance}{time}[/tex]

speed of Jill as he cover 5 km in 30 min

[tex]v_1 = \frac{5}{0.5} = 10 km/h[/tex]

speed of Scott as he cover 10 km in 30 min

[tex]v_2 = \frac{10}{0.5}= 20 km/h[/tex]

so correct answer is

A)Scott had the faster speed since he rode at 20 k/h while Jill only traveled 10 km/h.

Speed of Jill is greater than Scott

As we know that

Speed = Distance / Time

So speed of Jill is = 5 / 0.5     since 30 minutes = 0.5 hours

Speed of Jill = 10 km/hour

And

speed of Scott is = 10 / 0.5    

Speed of Scott = 20 km/min

Here we can see that speed of Scott is greater than Jill

Answer:

B. a ball on a string that stopped at the top of a vertical loop

Explanation:

Motion diagram must represent the direction of velocity of an object in the given figure

here as it is shown in the figure the arrow is vertically downwards which means that the velocity of object at that moment is downwards when it is at the top most point of its circular path

So here

A. a car driving over the top of a hill

Not correct as car is moving tangentially on the top of the hill so its velocity can not be downwards

B. a ball on a string that stopped at the top of a vertical loop

Since ball is stopped at the top most point of its path so here its velocity must be downwards due to free fall so its correct motion diagram for the above condition.

C. a roller coaster car at the top of a vertical loop of a track

Not correct as we know that roller coaster at the top of the loop will move tangentially so its velocity can not be vertically downwards.

D. a bug on the top of a rolling hula hoop

Bug is at the top of rolling hula hoop so its velocity is again tangentially and not vertical downwards.

Now the vertical velocity of the ball thrown at an angle 10° is given as

Voy(initial vertical velocity)= 30m/s x sin 10

Voy(initial vertical velocity)= 5.2m/s

Now the ball is decelerating with an acceleration due to gravity equivalent to 9.8m/s^2.

Let Vy be the final velocity and that is equal to zero in this case.

Now

Vy= Voy- tx9.8

Where t is the time at which the vertical velocity becomes 0.

Substituting the values we get

0= 5.2-tx9.8

9.8t=5.2

t=0.53 secs

Answer:

at the top of the ball’s trajectory

Explanation:

weight = mg

W on earth = 800N

mg = 800 N        

m(9.8 m/s²)= 800N

dividing 9.8 both sides

m=800/9.8          ( as g=9.8m/s² on earth)

m=81.63 kg

Now weight on Mars

w=mg

 =(81.63kg) ×(3.76m/s²)

w=306N

The astronaut's weight on Mars is approximately 306.53 N.

Weight is the measure of the force of gravity acting on an object. It is often expressed as newtons (N) or pounds (lbs). Although the words "weight" and "mass" are sometimes used interchangeably in everyday English, they are not the same thing.

We can use the following formula for calculating the weight of Astronuat:

Weight on Mars = Weight on Earth * (Gravity on Mars / Gravity on Earth)

Given:

Weight on Earth = 800 N

Gravity on Mars = 3.76 m/s²

Gravity on Earth = 9.8 m/s²

Weight on Mars = 800 N * (3.76 m/s² / 9.8 m/s²)

Weight on Mars ≈ 306.53 N

Therefore, the astronaut's weight on Mars is approximately 306.53 N.

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Answer: Option A

Explanation: In an elastic collision, the two objects only will have the same velocity after the collision if the had it before.

Now, in an inelastic collision the kinetic energy is not conserved, so there is a chance that the two objects that started with a different velocity end with the same, and that case is called "Perfectly inelastic" it occurs when the maximum kinetic energy is absorbed, and both objects "stick" to each other and move together as one, so the correct option is option A.

Time before projectile hits wall

= 88.2 m / 29.4 m/s = 3 seconds

Vertical velocity of projectile after three seconds

= 3*9.8 = 29.4 m/s

Horizontal velocity of projectile after three seconds, assuming no air resistance

= 29.4 m/s  (given)

Conclusion:

velocity of projectile when it hits the wall

= < 29.4, -29.4> m/s

= sqrt(29.4^2+29.4^2) m/s east-bound at 45 degrees below horizontal

= 41.58 m/s east-bound at 45 degrees below horizontal.

Final answer:

The projectile's velocity when it strikes the wall is a combination of its initial horizontal velocity and the vertical velocity it gains due to gravity. The horizontal velocity remains constant at 29.4 m/s east, but the final velocity also includes a downward vertical component due to gravity. The exact final velocity would require calculating the vertical component at impact but is not simply 29.4 m/s east.

Explanation:

The question is a classic example of a projectile motion problem in physics, specifically dealing with a projectile launched horizontally. The key concept here is that horizontal motion and vertical motion are independent of each other. Since the projectile is launched horizontally at a speed of 29.4 m/s without any initial vertical component, this speed will not change throughout its flight because horizontal velocity remains constant when neglecting air resistance. The vertical motion of the projectile, however, is influenced by gravity, which causes it to accelerate downwards at 9.8 m/s2.

To find the velocity at the moment it strikes the wall, we need to separately consider the horizontal and vertical components of velocity. The horizontal component remains 29.4 m/s east, as stated. To find the vertical component, we need to calculate the time it takes for the projectile to reach the wall. This is done by dividing the horizontal distance by the horizontal velocity (time = distance / velocity = 88.2 m / 29.4 m/s).

Without calculating the exact time here, it's important to understand that by the time it hits the wall, the projectile has gained a vertical velocity component due to gravity. The final velocity is a vector combination of the horizontal and vertical components. Therefore, the answer is not simply 29.4 m/s east, as that disregards the vertical component. The exact answer would involve calculating the vertical velocity at the time of impact and then combining it with the horizontal velocity to find the magnitude and direction of the final velocity vector. Options A, C, and D involve assumptions or calculations not fully provided here, but the question's premise indicates that the projectile's velocity upon impact is a combination of both horizontal and vertical components.

Here when an object is placed on the level floor then in that case there are two forces on the object

1). Weight of object downwards (mg)

2). Normal force due to floor which will counterbalance the weight (N)

so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

2). Normal force due to ground

So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

[tex]F_g = F_n + F_a[/tex]

given that

[tex]F_g = mg[/tex]

here

m = 30 kg and

g = acceleration due to gravity = 10 m/s^2

[tex]F_n = 150 N[/tex]

now from above equation

[tex]30*10 = 150 + F_a[/tex]

[tex]F_a = 300 - 150 = 150 N[/tex]

So force applied by the person must be 150 N

the answer is

144 N

i got it right

speed of the car at t = 10 s is given as 10 m/s

now at t = 30 s car comes to rest so v = 0

now in this velocity time graph we can say slope will show the acceleration of the car as we know that acceleration is rate of change in velocity

[tex]a = \frac{v_f - v_i}{\Delta t}[/tex]

as we know that

[tex]v_f = 0[/tex]

[tex]v_i = 10 m/s[/tex]

[tex]\Delta t = (30 - 10) = 20 s[/tex]

now from above formula

[tex]a = \frac{0 - 10}{20} = -0.5 m/s^2[/tex]

So slope of the graph will represent the acceleration of car which is 0.5 m/s^2

Actually the correct answer is going to be Linear, sloping downward.

Please mark Brainliest if this answer helped you.

i think its 200 i hope that helps

Answer:

78.1 m/s

Just took it.

Explanation:

here rocket starts from rest

It turns the booster and continue with constant acceleration

[tex]a = 4 m/s^2[/tex]

now in order to find the final velocity we can use kinematics equation

[tex]v_f = v_i + at[/tex]

here we know that

[tex]v_i = 0[/tex]

t = 3 s

now plug in all data in above equation

[tex]v_f = 0 + 4*3[/tex]

[tex]v_f = 12 m/s[/tex]

so the rocket will move rightwards with speed 12 m/s after 3 s of boosters

The periodic table on the left separates elements into three groups: the metals (green in the table), nonmetals (orange), and metalloids (blue). Most elements are metals. They are usually shiny, very dense, and only melt at high temperatures.

On the periodic table, representative metals are on the left side, transition metals and inner transition metals are in the centre. Transition metals are found in groups 3-11, while inner transition metals are in the bottom two rows. The reactivity of these metals varies, affecting how they are isolated from ores.

In the periodic table, the metals are generally located on the left side and in the centre. Specifically, representative metals are located on the left side of the table, while transition metals and inner transition metals can be found in the middle. The elements boron, silicon, germanium, arsenic, antimony, and tellurium, known as metalloids or semimetals, act as a divider between metals and nonmetals.

Transition metals are further differentiated as they are located in groups 3-11 of the periodic table and are characterized by their partially filled d orbitals. The inner transition metals are located in the two rows at the bottom of the periodic table, below group 3.

The reactivity of these metal groups varies, from the highly active metals such as scandium and iron, to almost inert elements, like platinum. This variety in reactivity determines the methods used to isolate the elements from their ores, with more active metals being more difficult to reduce.

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A scientist treats a cell with a chemical that destroys the chloroplast. As a result, photosynthesis will be stopped.

It is defined as a process by which plants and other photosynthetic organisms convert the light energy in to chemical energy  through the process of cellular respiration.

Some of the energy which is converted  is stored in molecules of carbohydrates like sugar and starches  which are made up of from carbon dioxide and water . Photosynthetic organisms which can perform photosynthesis  are algae and cyanobacteria. Photosynthesis is largely responsible for producing and maintaining the content of oxygen in earth's atmosphere.

The process begins with proteins absorbing light energy  which are called reaction centers and contain a green pigment which is called chlorophyll . In plants ,these pigments are present inside organelles called chloroplasts while in bacteria they are present in plasma membrane.

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The answer to this would be plastic. The other options are conductors. Remember, metal is a conductor as well as the salt ions in salt water.