Jenny is doing her homework when Jenny writes, the pencil exerts a force of 5N on the paper. The area of the pencil in contact with the paper is 0.5mm2 calculate the pressure of the pencil in the paper give the unit

Answer:

Cause you are their family and they need you. Do you wanna be someone who abandoned them or be the one they look to and dont dislike into adulthood. Also you signed onto this job no one said it was going to be easy but you making your life easier but harder for someone who is just a kid who still needs you.

Answer:

Explanation:

We will use the equation F - f = ma, which is a fancy way of stating Newton's 2nd Law.

F = +50.0,

f = -30.0,

m = 7.60 kg. Therefore:

50.0 - 30.0 = 7.60a and

20.0 = 7.60a and

a = 20.0/7.60 so

a = 2.63 m/s/s to the right

Answer

Explanation:

Simple arithmetic cannot be used if two or more vectors are not along the same direction. Vectors can be added when two vectors are in the same direction but for the subtraction of vectors, if the vectors are not in the same direction you then subtract.

Answer:

where is the question? I can only see the diagram

d) is an atom of Sodium and an atom of Chlorine.

e) is a molecule of water.

Answer:

The tension in the thread is approximately 3.2103225 Newtons

Explanation:

The given parameters are;

The side length of the aluminum cube, s = 5 cm

The R.D. of aluminum = 2.7

The relative density of alcohol into which the aluminum cube is placed = 0.8

The volume of the aluminum cube, V = s³ = (5 cm)³ = 125 cm³

The mass of the aluminum cube, V = The volume of the cube

The mass of the aluminum cube, m = V × R.D. × Density of water

m₁ = 0.000125 × 2.7 × 970 = 0.327375 kg

The mass of the alcohol displaced, m₂ = 0.000125 × 0.8 × 970 = 0.097

The tension in the string, T = The apparent weight of the aluminum cube, Wₐ

∴ The tension in the thread, T = Wₐ = (m₁ - m₂) × g

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

∴ T ≈ (0.327375 - 0.000125) × 9.81 = 3.2103225

The tension in the thread, T ≈ 3.2103225 N.

Answer:

Push with force of 1N

Explanation:

I have explained in the paper.

Goodluck

You must push with 1 newton  on the pencil to make the balloon pop.

The definition of force in physics is: The push or pull on a massed object changes its velocity.

An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force.

A spring balance can be used to calculate the Force. Newton is the SI unit of force.

Applied pressure is = 100  N/cm².

Cross sectional area = 0.01 cm².

The required force = pressure × area

= 100 N/cm² × 0.01 cm²

= 1 Newton.

Hence, you must push with 1 newton  on the pencil to make the balloon pop.

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Answer:

t = 180 / 1.4 = 129 sec   (time to swim horizontally across river)

S = 129 sec * V     where V is speed of current and S is the distance he will be carried downstream

The problem does not specify V the speed of the river

Answer:

Explanation:

From the given information:

a) the distance(D) showing how dar downstream he will land can be computed as follows:

Assuming the current of the river = 0.2 m/s

[tex]D = \dfrac{180 \ m \times 0.2 \ m/s}{1.4 \ m/s}[/tex]

D = 36 m ÷ 1.4

D = 25.71 m

The required time (t) to reach the other side is:

time (t) = 180 m/ 1.4 m/s

time (t) = 128.57 seconds

Answer:

B) 2.7 g of aluminium has a volume of 1 cm^3

Explanation:

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

[tex]Density = \frac{mass}{volume}[/tex]

If the density of aluminum is 2.7 g/cm³, it simply means that 2.7 g of aluminium has a volume of 1 cm³

Check:

Given the following data;

Mass = 2.7 grams

Volume = 1 cm³

Substituting into the formula, we have;

[tex]Density = \frac{2.7}{1}[/tex]

Density = 2.7 g/cm³

Answer:

a

Explanation:

Answer:

(a) 12 m/s

(b) At t = 0, x'(t) = 0

At t = 5.0 s, x'(t) = 15 m/s

At t = 10.0 s, x'(t) = 12 m/s

(c) i. The initial velocity = 2.0 cm/s

The initial position = 50 cm

The initial acceleration = 0.125 cm/s²

ii. 16 seconds

2. (a) Approximately 0.85 m/s

(b) 2.094 J

(c) i. Approximately 49.74 m/s

ii. Approximately 86.38°

3. (i) 4·i + 5·j

(ii) -2·i - j - 2·k

(iii) 5·i + 4·j - 3·z

(iv) 8

(v) (a) The magnitude is 2.8 cm, the direction is East

(b) The magnitude is ((14·√3)/5) cm, the direction is North

(c)  The magnitude is ((14·√3)/5), the direction is South

Explanation:

x(t) is given as follows;

x(t) = b·t² - c·t³

Where;

b = 2.4 m/s(²) (we note that the unit of b for the term to be distance in m/s²)

c = 0.120 m/s³

(a) At t = 0, we have;

x(0) = b × 0² - c × 0³ = 0

At t = 10.0 s, we have;

x(10.0) = 2.4 m/s² × (10.0 s)² - 0.120 m/s³ × (10.0 s)³ = 120 m

The average velocity = (Total distance)/(Total time)

∴ The average velocity of the car for the time interval t = 0 to t = 10.0 s, [tex]v_{ave}[/tex], is given as follows;

[tex]v_{ave}[/tex] = (120 m - 0 m)/(10.0 s - 0 s) = 12 m/s

The average velocity of the car for the time interval t = 0 to t = 10.0 s, [tex]v_{ave}[/tex]  = 12 m/s

(b) The instantaneous velocity, [tex]v_{inst}[/tex], is given as follows;

[tex]v_{inst} = \lim \limits_{t \to 0} \left( v_{ave}\right) = \lim \limits_{t \to 0} \left( \dfrac{\Delta x}{\Delta t} \right) = \dfrac{dx}{dt} = x'(t)[/tex]

[tex]x'(t) = \dfrac{d\left (b \cdot t^2 - c \cdot t^3\right)}{dt}[/tex]

x'(t) = 2·b·t - 3·c·t²

At t = 0, x'(t) = 2 × b × 0 - 3 × c × 0² = 0

At t = 5.0 s, x'(t) = 2 × 2.4 m/s² × 5.0 s - 3 × 0.120 m/s³ × (5.0 s)² = 15 m/s

At t = 10.0 s, x'(t) = 2 × 2.4 m/s² × 10.0 s - 3 × 0.120 m/s³ × (10.0 s)² = 12 m/s

(c) x(t) = 50 cm + (2.0 cm/s)·t - (0.0625 cm/s²)·t²

i. The initial velocity is the instantaneous velocity, x'(t), at time, t = 0

x'(t) = 2.0 cm/s - 2 × 0.0625 cm/s² × t

At t = 0, x'(0) = 2.0 cm/s - 2 × 0.0625 cm/s² × 0 = 2.0 cm/s

The initial velocity, x'(0) = 2.0 cm/s

The initial position = The position at time t = 0 = x(0)

x(0) = 50 cm + (2.0 cm/s) × 0 - (0.0625 cm/s²) × 0² = 50 cm

The initial position, x(0) = 50 cm

The initial acceleration, x''(0) = 2 × 0.0625 cm/s² = 0.125 cm/s²

ii. x'(t) = 2.0 cm/s - 2 × 0.0625 cm/s² × t

When the velocity of the turtle, x'(t) = 0 we have;

0 = 2.0 cm/s - 2 × 0.0625 cm/s² × t

∴ t = (2.0 cm/s)/(2 × 0.0625 cm/s²) = 16 seconds

The velocity of the turtle is zero after 16 seconds

2. The mass of the large fish, m₁ = 15.0-kg

The speed of the large fish, v₁ = 1.1 m/s

The mass of the smaller fish, m₂ = 4.50 kg

The speed of the small (stationary) fish, v₂ = 0

The initial momentum = 15.0 kg × 1.1 m/s + 4.50 kg × 0 = 16.5 kg·m/s

The initial momentum = 16.5 kg·m/s

The final momentum = (15.0 kg + 4.50 kg) × v₃ = 19.50 kg × v₃

The final momentum = 19.50 kg × v₃

Where;

The total initial momentum = The total final momentum

We get;

16.5 kg·m/s = 19.50 kg × v₃

∴ v₃ = (16.5 kg·m/s)/(19.50 kg)

v₃ = (16.5/19.50) m/s = (11/13) m/s ≈ 0.85 m/s

∴ The speed of the large fish just after it eats the small, v₃ ≈ 0.85 m/s

(b) The initial kinetic energy, K.E.₁ = (1/2) × 15 kg × (1.1 m/s)² = 9.075 J

The final kinetic energy, K.E.₂ = (1/2) × 19.5 kg × (11/13 m/s)² = 363/52 J

The mechanical energy dissipated, ΔE = K.E.₁ - K.E.₂

ΔE = (9.075 - 363/42) J = 1089/520 J ≈ 2.094 J

The mechanical energy dissipated, ΔE = 2.094 J

(c) i. We have the total momentum = 110 × 8.8· j + 85 × 7.2· i = 9.680·i + 612·i

The velocity after collision, v = (9.680·i + 612·i)/(110 + 85) = 49.64·j + 3.14·i

The magnitude of the velocity, v = √(49.64² + 3.14²) ≈ 49.74 m/s

ii. The direction, θ = arctan(49.64/3.14) ≈ 86.38°

3. (i) [tex]\underset{A}{\rightarrow} + \underset{B}{\rightarrow}[/tex] = (i + 2·j - k) + 3·i + 3·j + k = 4·i + 5·j

(ii) [tex]\underset{A}{\rightarrow} - \underset{B}{\rightarrow}[/tex] = (i + 2·j - k) - (3·i + 3·j + k) = -2·i - j - 2·k

(iii) [tex]\underset{A}{\rightarrow} \times \underset{B}{\rightarrow}[/tex] = (2 + 3)·i - (1 + 3)·j + (3 - 6)·z = 5·i + 4·j - 3·z

(iv) [tex]\underset{A}{\rightarrow} \cdot \underset{B}{\rightarrow}[/tex] = 1×3 + 2 × 3 + (-1)×1 = 8

(v) (a) [tex]\underset{A}{\rightarrow} + \underset{B}{\rightarrow}[/tex] = 2.8·cos(60°)·i + 2.8·sin(60°)·j + (2.8·cos(60°)·i - 2.8·sin(60°)·j

[tex]\underset{A}{\rightarrow} + \underset{B}{\rightarrow}[/tex] = 5.6·cos 60°·i = 2.8·i

The magnitude = 2.8, the direction is east

(b) [tex]\underset{A}{\rightarrow} - \underset{B}{\rightarrow}[/tex] = 2.8·cos(60°)·i + 2.8·sin(60°)·j - (2.8·cos(60°)·i - 2.8·sin(60°)·j

[tex]\underset{A}{\rightarrow} - \underset{B}{\rightarrow}[/tex] = 5.6·sin(60°)·j= ((14·√3)/5)·j

The magnitude = ((14·√3)/5), the direction is North

(c)  [tex]\underset{B}{\rightarrow} - \underset{A}{\rightarrow}[/tex] = (2.8·cos(60°)·i - 2.8·sin(60°)·j - (2.8·cos(60°)·i + 2.8·sin(60°)·j)

[tex]\underset{B}{\rightarrow} - \underset{A}{\rightarrow}[/tex] = -5.6·sin(60°)·j= (-(14·√3)/5)·j

The magnitude = ((14·√3)/5), the direction is South

Answer:

The Most Famous Astronomers of All Time. Karl Tate, SPACE.com. ...

Claudius Ptolemy. Bartolomeu Velho, Public Domain. ...

Nicolaus Copernicus. Public Domain. ...

Johannes Kepler. NASA Goddard Space Flight Center Sun-Earth Day. ...

Galileo Galilei. NASA

Answer:

D bcz the slope rose the fastest

Explanation:

I DON'T WANNA BE QUESTIONED, THIS IS BECAUSE y POSITION ROSE FASTER THAN x POSITION

Answer:

The untrue option is C: "More arrowheads on lines mean a brighter light"

Explanation:

Ray diagrams are used to show how the light behaves with things like mirrors or lenses. Where we only study how the direction of the light changes when it interacts with these objects.

The "light" is represented with arrows, where again, the only thing we care is the direction of the light, so the first statement is true, the arrowheads show the direction of the light, and only that.

The intensity of the light, in this context, has no effect on how light behaves, so there is not a necessity of representing the intensity of the light, thus, more arrowheads on lines do not mean a brighter light. It may only be used to represent changes in direction of the light.

Finally, we know that light travels in straight pats (the pats can be curved in some cases, like with large gravitational fields, but this is not the case of a ray diagram) so the lines that represent the light should always be straight, thus option B is also true.

The untrue option is C: "More arrowheads on lines mean a brighter light"

Answer: i dont know

Explanation:

Answer:

[tex]{ \bf \: time \: of \: flight : s = 0} \\ { \tt \: s \: is \: displacement} \\ \\ s = ut + \frac{1}{2} a {t}^{2} \\ 0 = u \sin( \theta) t - \frac{1}{2} g {t}^{2} \\ u \sin( \theta)t = \frac{1}{2} g {t}^{2} \\2 u \sin( \theta) = gt \\ { \boxed{ \tt{formular : time \: of \: flight = \frac{2u \sin( \theta) }{g} }}} \\ t = \frac{2 \times 10 \times \sin(30 \degree) }{10} \\ { \green{ \boxed{ \tt{ \bf{answer : { \tt{time \: of \: flight =1 \: second }}}}}}} \\ \\ { \tt{becker \: jnr}}[/tex]

Answer:

Thermal capacity of the object = 720 j/°C

Explanation:

Given:

Specific heat capacity of metal = 360J/(kg°C)

Mass of given object = 2 kg

Find:

Thermal capacity of the object

Computation:

Thermal capacity = Specific heat x Mass

Thermal capacity of the object = Specific heat capacity of metal x Mass of given object

Thermal capacity of the object = 360 x 2

Thermal capacity of the object = 720 j/°C

Answer:

A. You would weigh the same on both planets because their masses and the distance to their centers of gravity are the same.

Explanation:

Given that Planets A and B have the same size, mass.

Let the masses of the planets A and B are and respectively.

As masses are equal, so .

Similarly, let the radii of the planets A and B are and respectively.

As radii are equal, so .

Let my mass is m.

As the weight of any object on the planet is equal to the gravitational force exerted by the planet on the object.

So, my weight on planet A,

my weight of planet B,

By using equations (i) and (ii),

.

So, the weight on both planets is the same because their masses and the distance to their centers of gravity are the same.

Hence, option (A) is correct.

Answer:

B...................

Answer:

Option A is correct: "A concave mirror forming a larger, virtual image"

Explanation:

There are two types of mirror:

Concave (converging) mirrors: a concave mirror is a mirror that reflects the light inward

Convex (diverging) mirrors: a convex mirror is a mirror that reflects the light outward

The image formed by a mirror can also be of two types:

Real image: it is formed on the same side of the object, with respect to the mirror

Virtual image: it is formed on the opposite side of the object, with respect to the mirror

In the figure of this problem, we see that:

< The mirror reflects the light from the object inward --> so it is a concave mirror

< The image is formed on the other side of the mirror --> it is a virtual image

So the correct option is

A) A concave mirror forming a larger, virtual image

Answer:

The Reactants                                                       Products                          

Copper sulfate + zinc             →                        Zinc sulfate + copper

aluminum + copper chloride     →                    Aluminum chloride + Copper

Copper + Oxygen                     →                      Copper oxide

Explanation:

The Reactants                                                       Products                          

Copper sulfate + zinc                                        Zinc sulfate + copper

aluminum + copper chloride                            Aluminum chloride + Copper

Copper + Oxygen                                             Copper oxide

First reaction can be described as a single replacement reaction

Second reaction is also a single replacement reaction just like the first reaction

The third is a synthesis reaction ( i.e. reaction in which elements combine to form single compound )

Answer:

There is a lot of instruments used for measurement, may I ask which one are you referring to?

Answer:

The statements that best compares radiation and conduction are the following: both transfer thermal energy from warm objects to cooler objects and both can transfer energy through matter. The difference is that conduction is through direct heat transfer; whereas, radiation transfers heat through waves.

Answer:

you have gay

Explanation:

699999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999ooooooooooooooooooooooooooo

Answer:

Answer: C. -12 newtons

Explanation: Since, the block is in static equilibrium,

the net force on the block is zero i.e. balance forces are acting on the block.

In the horizontal direction,

force applied+force due to friction = 0

20 N + (-20N) = 0

In the vertical direction as well, net force is zero:

Force due to gravity + force exerted by the ground = 0

12 N + force exerted by the ground = 0

⇒force exerted by the ground = 0 -12 N = -12 N

Thus, the correct option is C. -12 N where negative sign means that the direction of the force due to ground is opposite to the force of gravity.

Answer:

4.71m/s

Explanation:

Average speed = Total distance travelled ÷ Total time taken.

80/17=4.71

4.71m/s

Answer:

In this question,

                                                = 80m.

                                     = 17s

So, Average speed would be 80 ÷ 17

= [tex]\frac{80}{17}[/tex]

 = 4.71 m/s. or 4.71 meter per second.

Answer:

Explanation:Gravitational force always acts on the ball in the downward direction, thus an acceleration due to gravity always point in the downward direction with a constant magnitude  10.6m/s  

Answer:

Explanation:

Virtual images are always right side up while real images are always upside down. Therefore, the is a virtual image that is smaller than the original.

Answer:

Theres 2 meanings first is:

the action or process of registering or of being registered.

The other one is:

a combination of stops used when playing the organ.

The meaning of registration is the action of process of registering to some applications or websites which includes some terms and conditions also.

Registration is the process or action of registering into a website or application or to any institute through online or offline mode. By registering we are giving some personal and professional details as per the requirements of the source which need us to be registered.

For example, if we are applying for a central examination, we have to register our name and details in the official website of the examination board.

The registration process bring us some terms and conditions which we can apply or accept to continue the registration process. After successful registration we can have an acknowledgement for the action.

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Answer:

Initial velocity (u) = 40 m/s

Distance travel in last 5 seconds = 100 m

Explanation:

Given:

Acceleration (a) = 8 m/s²

Final velocity (v) = 0 m/s

Find;

1] Initial velocity before 5 sec

2] Distance travel in last 5 seconds

Computation:

1] Initial velocity before 5 sec

v = u + at

0 = u + (-8)(5)

u - 40 = 0

Initial velocity (u) = 40 m/s

2] Distance travel in last 5 seconds

s = ut + (1/2)(a)(t²)

s = (40)(5) + (1/2)(-8)(5²)

s = 200 - 100

Distance travel in last 5 seconds = 100 m

Answer:

Explanation:

The equation for frequency is

[tex]f=\frac{v}{\lambda}[/tex] where

f = frequency, v is velocity, and lambda is wavelength. Filling in:

[tex]262=\frac{343}{\lambda}[/tex] and

[tex]\lambda=\frac{343}{262}[/tex] so

[tex]\lambda=1.31m[/tex]

The wavelength of the note played on the piano is 1.31m.

Given the data in the question;

Wavelength of the note; [tex]\lambda = \ ?[/tex]

Wavelength is simply referred to as the spatial period of a periodic wave i.e when the shapes of waves are repeated, the distance over which they are repeated is called wavelength. It is expressed as;

[tex]\lambda = \frac{v}{f}[/tex]

Where v is velocity and f is frequency

We substitute our given values into the expression above;

[tex]\lambda = \frac{v}{f} \\\\\lambda = \frac{343m/s}{262s^{-1}} \\\\\lambda = 1.309m\\\\\lambda = 1.31m[/tex]

Therefore, the wavelength of the note played on the piano is 1.31m.

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