17. A ball rolls off a lab table at 5 m/s horizontally. The table is 1.2 meters tall.

a. How long is it in the air?

b. How far from the table does it hit the ground?

c. What are the x and y speeds just before it hits the ground?

The distance that separates the charges is 0.7 m.

The distance between the charges is determined by applying Coulomb's law of electrostatic force as shown below.

F = kq₁q₂/r²

where;

r² = kq₁q₂/F

r² = (9 x 10⁹ x 8 x 10⁻⁴ x 3 x 10⁻⁴) / (4.4 x 10³)

r² = 0.49

r = √0.49

r = 0.7 m

Thus, the distance that separates the charges is 0.7 m.

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Answer: x(t2)−x(t1) over the time interval [t1,t2]

Explanation: hope this helps

Answer:

Assume that the air resistance on the supplies is negligible, and that [tex]g = 9.8 \; {\rm m\cdot s^{-2}}[/tex].

The plane need to drop the supplies when it is horizontally approximately [tex]700\; {\rm m}[/tex] away from the hill.

The supplies will hit the tree.

Explanation:

Let [tex]u_{y}[/tex] and [tex]v_{y}[/tex] denote the initial and final vertical velocity of the supply; [tex]u_{y} = 0\; {\rm m\cdot s^{-1}}[/tex] since the plane was flying horizontally.

Let [tex]x_{y}[/tex] denote the vertical displacement of the supply; [tex]x_{y} = 350\; {\rm m} - 810\; {\rm m} = (-460)\; {\rm m}[/tex].

Let [tex]a_{y}[/tex] denote the vertical acceleration of the supply; [tex]a = (-g) = (-9.8)\; {\rm m\cdot s^{-2}}[/tex].

Make use of the SUVAT equation [tex]{v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}[/tex] to find [tex]v_{y}[/tex], the final vertical velocity of the supply:

[tex]\begin{aligned} {v_{y}}^{2} &= {u_{y}}^{2} + 2\, a_{y}\, x_{y} \end{aligned}[/tex].

[tex]\begin{aligned} v_{y} &= -\sqrt{{u_{y}}^{2} + 2\, a_{y}\, x_{y}} \\ &= -\sqrt{0^{2} + 2\, (-9.8)\, (-460)}\; {\rm m\cdot s^{-1}} \\ &\approx (-94.953)\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].

(Negative since the supply would be travelling downwards.)

Let [tex]t[/tex] denote time it takes for the supply to land on the hill after being dropped from the plane. Make use of the SUVAT equation [tex]t = (v_{y} - u_{y}) / (a)[/tex] to find the value of [tex]t\![/tex]:

[tex]\begin{aligned} t &= \frac{v_{y} - u_{y}}{a} \\ &\approx \frac{(-94.953) - 0}{(-9.8)} \; {\rm s}\\ &\approx 9.6890 \; {\rm s} \end{aligned}[/tex].

Apply unit conversion and ensure that [tex]v_{x}[/tex], the horizontal speed of the plane is in the standard unit [tex]{\rm m\cdot s^{-1}}[/tex]:

[tex]\begin{aligned} v_{x} &= \frac{260\; {\rm km}}{1\; {\rm h}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \\ &\approx 72.222\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Under the assumptions, the horizontal speed of the supply will be the same as that of the plane- [tex]v_{x} \approx 72.222\; {\rm m\cdot s^{-1}}[/tex]- until it lands.

While in the air, the supply will travel a horizontal distance of:

[tex]\begin{aligned}x_{x} &= v_{x}\, t \\ &\approx 72.222\; {\rm m\cdot s^{-1}} \times 9.6890\; {\rm s} \\ &\approx 699.76\; {\rm m}\end{aligned}[/tex].

Hence, for the supply to land exactly at the top of the hill, the plane need to drop the supply while at a horizontal distance of approximately [tex]700\; {\rm m}[/tex] away from the hill.

The horizontal distance between the trees and the location where the plane dropped the supply would be approximately [tex](700\; {\rm m} - 30\; {\rm m}) = 670\; {\rm m}[/tex]. The time required for the the supply to reach that horizontal position would be:

[tex]\begin{aligned} t &= \frac{x_{x}}{v_{x}} \approx \frac{669.76\; {\rm m}}{72.222\; {\rm m\cdot s^{-1}}} \approx 9.2736\; {\rm s}\end{aligned}[/tex].

Let [tex]h_{0}[/tex] denote the initial height of the supply (relative to the sea level.) In this question, [tex]h_{0} = 810\; {\rm m}[/tex].

Let [tex]h(t)[/tex] denote the height of the supply (relative to the sea level) after being dropped from the plane for time [tex]t[/tex].

The SUVAT equation [tex]h(t) = (1/2)\, a\, t^{2} + u_{y}\, t + h_{0}[/tex] gives an expression for [tex]h(t)[/tex]. Make use of this equation to find the height of the supply (relative to the sea level) when the supply reach the horizontal position of the trees at [tex]t \approx 9.2736\; {\rm s}[/tex]:

[tex]\begin{aligned} h(t) &= \frac{1}{2}\, a\, t^{2} + u_{y}\, t + h_{0} \\ &= \frac{1}{2}\times (-9.8)\, (9.2736)^{2} + 0\times 9.2736 + 810 \\ &\approx 388.60\; {\rm m} \end{aligned}[/tex].

Note that the altitude of the top of the trees is [tex]350\; {\rm m} + 40\; {\rm m} = 390\; {\rm m}[/tex] relative to the sea level. Since [tex]388.90\; {\rm m} < 390\; {\rm m}[/tex], the supplies will run into the trees.

The vehicle traveling in the opposing direction travels the following distance in the same amount of time: 200/9 (2.5) equals 99.38 meters.   The safe separation between the two vehicles is therefore: 129.38 + 99.38 = 228.76 meters.

To obtain the velocity function, we must calculate the integral of the acceleration function:

V ( t ) = ∫ a ( t ) d t V ( t ) = ∫ 3 d t = 3 t + c 1

To find the integral's constant, we use our initial conditions as a guide. As of now

t = 0

The speed is 80 km per hour.

80 km/h is equal to (80)(1000)/3600, or 200/9 meters per second.

V ( 0 ) = 3 ( 0 ) + c 1

200/ 9 = 0 + c 1

c 1 = 200/ 9

The distance the vehicle travels in

the number of seconds required to travel the same distance as the truck plus 30 meters (15 meters in front of and 15 meters behind the truck).

The distance covered by the vehicle is

In the same number of seconds, 200/9 t.

This is the amount of time it takes for the car to overtake and pass the truck.

15 metres.

To do this, the automobile travels a distance of:

S ( t ) = 3 /2 ( 20 ) + 200/ 9 ( 2 5 ) = 129.38 meters

The vehicle traveling in the opposing direction travels the following distance in the same amount of time:

99.38 meters is equal to 200/9 ( 2 5).

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The velocity of the bird can be obtained as  6.38 m/s.

Let us recall that the acceleration is the rate at which the velocity is changed. In this case, we are told that; a bird flying 3.45 m / s directly north feels a wind directly East that accelerates at 0.558 m / S^2. Having seen the question we can pick the following parameters out from the question;

Initial velocity u = 3.45 m / s

Acceleration a = 0.558 m / S^2

Time taken =  5.25 s

Given that;

v = u + at

v = final velocity

u = initial velocity

a = acceleration

t = time taken

v = 3.45 + (0.558  *  5.25)

v = 6.38 m/s

We can see that as the bird tends to be flying as we can see in the question that have been shown above that the final velocity that it can have after 5.25 seconds is 6.38 m/s.

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The orange was at a height of 3.26 m above the ground.

The third equation of motion is

v² - u² = 2aS

Given is an orange dropped from a tree and had a velocity of 8 m/s just before it hits the ground.

We can write -

[v] = 8 m/s

[a] = 9.8 m/s²

[u] = 0 m/s

Using third equation of motion, we get -

v² - u² = 2aS

64 - 0 = 2 x 9.8 x S

64 = 19.6 x S

S = 3.26 m

Therefore, the orange was at a height of 3.26 m above the ground.

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If 10 light rays shined on a mirror than the 10 light rays light will converge or diverge.

A mirror is a wave reflector. Light consists of waves, and light waves reflect from the flat surface of a mirror.

There are two types of mirror concave and convex.

When parallel light beams strike a concave mirror, all of the rays that are reflected meet at the same location (focus). A concave mirror is also known as a converging mirror as a result.

Parallel light beams that strike a convex mirror's surface and are reflected appear to meet (diverge) at a single location, although they do not. so convex mirror is defined as a diverging mirror as a result.

So, rays of light will get converged or diverged.

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A mass of can 0.11 kg rests on a fence post 1.7 m high. a 0.0020 kg bullet is fired horizontally at a can. it hits the can at 900 m/s, passes through it, and emerges at 720 m/s.

when the can hits the ground, it is 1.93 m away from the fence post.

for vertical fall,

d = 0.5*g*t^2

1.7 = 0.5*9.8*t^2

t = 0.589 s

use conservation of momentum to calculate the initial horizontal velocity to tin

mb*vbi = mb*vbf + mt*vt

2*10^-3*900 = 2*10^-3*720 + 0.11*vt

vt = 3.273 m/s

horizontal distance = vt*time

= 3.273 m/s * 0.589 s

= 1.93 m

Answer= 1.93 m

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Vector values like displacement are aware of their direction. Distance and other scalar quantities are directionless. The different directions of motion can be disregarded when calculating the total distance traveled.

While distance and displacement appear to signify the same thing, they actually have very different definitions and meanings.

A scalar quantity known as distance measures "how much ground an object has traversed" while moving.

An object's total change in position is referred to as displacement, a vector variable that measures "how far out of place an object is."

Think about the motion shown in the following diagram as a way to test your grasp of this distinction. Walking 4 meters east, 2 meters south, 4 meters west, and then 2 meters north is a physics teacher.

The physics instructor has walked 12 meters in total, yet her displacement is 0 meters. She has "covered 12 meters of ground" (distance = 12 m) during the course of her move. However, when she has done moving, she is not "out of position"; that is, her motion has not caused any displacement (displacement = 0 m). Being a vector quantity, displacement must pay attention to direction.

In the same way as the 2 meters south cancels the 2 meters north, the 4 meters east cancels the 4 meters west. Vector values like displacement are aware of their direction. Distance and other scalar quantities are directionless. The different directions of motion of the physics teachers can be disregarded in calculating the overall distance traveled.

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The momentum of an object can be calculated with the formula below:

Where p is the momentum, m is the mass and v is the velocity.

The mass of the bus is greater than the mass of the car.

So, if they have the same velocity, the momentum of the bus will be greater.

Correct option: a.

Answer:

See below

Explanation:

acceleration = change in velocity / change in time

  Or      F = m * a

      So the first one   Accel = 40 m/s / 20 s = 2 m/s^2

   Or    100 N = 50 * a     A = 2 m/s^2       Same answer two different ways...

The others are similar ......

Answer:

See below

Explanation:

P1V1/T1 = P2V2/T2      IF  T1 = T2

Then

P1V1 = P2V2       P2 = 1/2 P1

P1V1 = 1/2 P1 V2        DIVIDE BOTH SIDES BY P1

V1 = 1/2 V2        MULTIPLY BOTH SIDES BY 2

2V1 = V2           OR     V2 = 2 V1     THE NEW VOLUME IS TWICE ORIGINAL

Answer:

The Answer is D. 300 s because in 5 minutes there are 300 seconds

Given:

Speed of light = 3 x 10⁸ m/s

Let's solve for the following:

• (a). How many miles does a pulse of light travel in a time interval of 0.1 s, which is about the blink of an eye?

Apply the formula:

Where:

v is the speed of light

t is the time.

Thus, we have:

Now let's convert the answer from meters to miles.

Where:

1 mile = 1609.34 meters

Δx = 18641.14 mi

• (b). Compare this distance to the diameter of Earth.

Apply the formula:

Where:

r = 6.38 x 10⁶ m.

Thus, we have:

ANSWER:

• (a). 18641.14 mi

• (b). 2.35

The Fundamental Theorem of Calculus is used to evaluate the definite integral. A important tool for simulating the dynamics of moving objects and other physical phenomena is the parametric form.

For a parameter t, the dynamics or trajectory of a moving object is given in parametric form. We determine the speed of the object over the range of its parameter using derivatives and a definite integral. The Fundamental Theorem of Calculus is used to evaluate the definite integral. A important tool for simulating the dynamics of moving objects and other physical phenomena is the parametric form.

The velocity function or vector's absolute value or norm can be used to calculate an object's speed.

We obtain the velocity v(t) as the first derivative of the position trajectory given as follows since the velocity is the rate of change of the distance traveled:

v ( t ) = r ′ ( t ) = ⟨ ( 2 t 3 ) ′ , ( − t 3 ) ′ , ( 3 t 3 ) ′ ⟩ = ⟨ 6 t 2 , − 3 t 2 , 9 t 2 ⟩

The speed is given as either its norm or absolute value in the velocity function above as follows:

Speed(t)=∥v(t)∥=∥⟨6t2,−3t2,9t2⟩∥

=√(6t2)2+(−3t2)2(9t2)2=√36t4+9t4+81t4=√126t2(2)

The speed at t = 7 will be (2) from (2).

Speed(7)=√126×72=49√126.

Using (2) and the Calculus Fundamental Theorem, the trajectory's length L can be determined as follows:

L = ∫ 7 0 ∥ r ′ ( t )∥ d t = ∫ 7 0 ∥ v ( t ) ∥ d t = ∫ 7 0 √ 126 t 2 d t = √ 126 [ t 3 3 ] 7 0 = √ 126 [ 7 3 3 − 0 3 3 ] = 7 3 √ 126 3. \s.

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In the psychology experiment in which Hakeem participates, UCS is click, UCR is blink and tear up, CS is blowing a puff of air, CR is thought of the puff of air and NS is the click before puff of air was introduced.

In psychology the abbreviations of the given terms are as follows:

Here the Unconditioned Response is the blinking and tearing up because he does that even though there is no puff of air. Since clicking provokes the Unconditioned Response, clicking is the Unconditioned Stimulus. The clicking is paired with puff of air for multiple times, that is why Unconditioned Response is stimulated. So the puff of air is the Conditioned Stimulus.

Therefore,

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The side the girl is sitting on will once tilt down.

The seesaw's left and right sides' opposing torques are balanced by the torque acting on the left side. The force will act on the seesaw itself at either the right or left side of the seesaw's center of gravity.

Formula: The expression for the amount of torque operating on a body is

torque =r F

where; r=position vector and F= force

if the force is applied at a right angle to the direction of r and the body is at a distance r from the pivot.

The seesaw's length is assumed to be l, and its left side's mass is assumed to be greater than its right side's mass.

[tex]m_{L}[/tex] > [tex]m_{R}[/tex]

Let the weight of the hefty body be [tex]M[/tex], and the mass of the girl be [tex]m[/tex].

The seesaw's left and right side rods' centers of gravity will be positioned [tex]\frac{l}{4}[/tex] of an inch apart from the center (or pivot). The girl and body are initially positioned [tex]\frac{l}{2}[/tex] apart from the pivot. as a result, initially

torque of left side= torque of right side

([tex]\frac{l}{4}[/tex])[tex]m_{L}[/tex][tex]g[/tex]+([tex]\frac{l}{2}[/tex])[tex]mg[/tex]=([tex]\frac{l}{4}[/tex])[tex]m_{R}[/tex][tex]g[/tex]+([tex]\frac{l}{2}[/tex])[tex]Mg[/tex]

we get,

[tex]\frac{m_{L} -m_{R} }{2}[/tex]= [tex]M-m[/tex]

Asking the body and the girl to move ([tex]\frac{l}{4}[/tex]) closer to the pivot will result in the following calculation of the net torque:

net torque= torque of left - torque of right

The seesaw is tilted to the left when we believe that the torque is greater on the left side.

τnet=[tex]\frac{l}{4} m_{L}[/tex][tex]g[/tex]+[tex]\frac{l}{4}[/tex][tex]mg[/tex]−[tex]\frac{l}{4} m_{R}[/tex][tex]g[/tex]−[tex]\frac{l}{4}[/tex][tex]Mg[/tex]

We obtain the following values by replacing the value of [tex]m_{L} -m_{R}[/tex] in terms of [tex]m[/tex] and [tex]M[/tex]

therefore,

 τnet=[tex]lg(\frac{M-m}{4} )[/tex]

Our presumption that the left side will tilt because M is heavy and m is tiny is valid since the torque on the left side will predominate.

The girl will again be sitting on the side that is tilting downward because she was previously sat on the left.

REMARK: We would have discovered negative net talk if we had thought that the right hand side was tilting or had more torque. As a result, in such scenario, we would have discovered that the left side tilts, contrary to what we had assumed. Both strategies are thus appropriate. But it is important to avoid guessing at all costs while answering these questions because doing so reduces the likelihood of getting the correct response.

The force that can cause an object to rotate along an axis is measured as torque. An object acquires angular acceleration due to torque.

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Using concepts of acceleration, velocity and position, it is found that:

A. The instantaneous velocity in 15 seconds is of -147 m/s².

B. The average velocity during this interval is of -9.8 m/s.

Cc. The distance fallen during this time is of 1102.5 meters.

The velocity after t seconds is given by the following equation:

v(t) = v(0) + at.

In which:

In the context of this problem, the values of these parameters are given as follows:

Then the velocity equation is given as follows:

v(t) = -9.8t.

In 15 seconds, the instantaneous velocity is given as follows:

v(15) = -9.8(15) = -147 m/s².

The average velocity during this interval is of given by the change in velocity divided by the change in time, hence:

Average velocity = (-147 - 0)/(15 - 0) = -9.8 m/s.

The position function is the integral of the velocity function, hence:

s(t) = -4.9t². (integral of -9.8t = -9.8t²/2 = -4.9t²).

The distance fallen in 15 seconds is:

s(t) = -4.9(15)² = -1102.5 (1102.5 meters fallen).

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Answer:

c. equal equal to the magnitude of f

An example of demonstrative evidence is the digital rendering of the crime scene that shows where the gun was fired from. That is option B.

A demonstrative evidence is the type of evidence that is being represented using visuals to help enhance the facts of a claim made against an opponent in the law court.

The components of demonstrative evidence include the following:

Based on the various evidence presented concerning the sudden death of Mr. Red Herring who was short dead in his backyard, the demonstrative evidence is when the forensics team also created a digital rendering of the crime showing where the gun was shot.

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Options of question:

In this story, which is an example of demonstrative evidence?

The fingerprint that showed that Mrs. Scarlet handled the gun

The digital rendering of the crime scene that shows where the gun was fired from

The report about bullet striations that prove the shot came from the discarded weapon

The DNA evidence gathered from the blood splatters

The average acceleration of the helicopter, given that it changes its velocity from 22.0 m/s to 10.0 m/s is –1.5 m/s²

We understood that acceleration is defined as the change in velocity with time i.e

a = (v – u) / t

With the above formula, we can determine the average acceleration of the helicopter. Details below

The following data were obtained from the question:

a = (10 – 22) / 8

a = –12 / 8

a = –1.5 m/s²

Thus, we can conclude that the average acceleration is –1.5 m/s²

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The temperature at which the resistance will be 22.8 Ω is 100 °C

The following data were obtained from the question:

The new temperature can be obtained as illustrated below:

α = R₂ – R₁ / R₁(T₂ – T₁)

0.0065 = 22.8 – 15 / 15(T₂ – 20)

0.0065 = 7.8 / 15(T₂ – 20)

Cross multiply

0.0065 × 15 (T₂ – 20) = 7.8

0.0975 (T₂ – 20) = 7.8

Divide both side by 0.0975

T₂ – 20 = 7.8 / 0.0975

T₂ – 20 = 80

Collect like terms

T₂ = 80 + 20

T₂ = 100 °C

Thus, the temperature is 100 °C

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The total tangential speed = 3.56 m/s

, r =2 m ,

t =44sw

=17rpm = 17*2*3.14/60

= 1.7793 rad/swo =

01) from rotational kinematic equation

w =wo+t1.7793 =0 + *44 =0.0404 rad/s^2

(2) from rotationla kinematic equation

w^2 -wo^2 = 21.7793*1.7793

= (2*0.0404*)=39.18 rad

3) v =rw =2*1.7793

tangential speed = 3.56 m/s

What is average angular acceleration?

A spinning object's change in angular velocity per unit of time is expressed quantitatively as angular acceleration, also known as rotational acceleration. It is a vector quantity with either one of two predetermined directions or senses and a magnitude component.

The formula for average angular acceleration is: a = change in velocity divided by change in time. a=(w2−w1)/(t2−t1) a = ( w 2 − w 1 ) / ( t 2 − t 1 ) . w2 represents the final velocity measured in radians per second. w1 represents the initial velocity measured in radians per second.

Therefore, tangential speed = 3.56 m/s

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The initial velocity of the projectile, Vo is 25√5 m/s.

What is initial velocity?

initial velocity is the velocity at time interval t=0 and it's representative as u.

It is velocity which the motion start.

Sol-as per the question

Sx=400 Vx=Vo cos 37°=4/5Vo

Sy= -100 Vy=Vo sin 37° 3/5Vo ay=-10 m/s^2

Let the time of flight be t

Sx=u^xt +1/2axt^2

=>400= 4/5Vot----------(eq1)

Sy=uyt +1/2 ay t^2

=>-100=3/5 Vot -5t^2-----(eq2)

By solving this we are get-

Vo=25√5 m/s

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Answer:

About 150 million kilometers

Explanation:

have good day♡

The diameter of the copper wire if its resistance is to be the same as that of an equal length of aluminum wire with diameter 1.94 mm is 1.568mm

The obstruction to current flow in an electrical circuit is measured by resistance. The Greek letter omega Ω represents resistance, which is measured in ohms.

The term "resistivity" refers to a quality that quantifies how much an object resists having current flow through it. It is a feature of the material itself, independent of the size or form of the sample, and is typically temperature dependent, though it could also be pressure-dependent.

by formula we have ;

R=ρL/A =ρL/πd²/4

where

R = resistance

ρ= resistivity

L = length ; and

A = area of cross section = πd²/4

we know that,

ρ₁ of aluminum = 2.63×10⁻⁸Ωm ; d₁ is diameter of aluminum

ρ₂ of copper=1.72×10⁻⁸Ωm ; d₂ is diameter of copper

given: L of copper= L of aluminum

also, R of copper= R of aluminum

on cancelling the equal terms we get,

ρ₁/d₁²=ρ₂/d₂²

here d₂=d₁[tex]\sqrt\frac{resistivity of copper}{resistivity of aluminum}[/tex]

d₂=1.94[tex]\sqrt \frac{1.72}{2.63}[/tex]

d₂=1.568mm

Hence the diameter of copper wire is 1.568mm

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Answer:

its upto your mind and magic

Explanation:

The likelihood that this star is not "venerable," or old, is low; but, we would need more details, particularly the star's spectral class, to be certain.

The relationship between the absolute magnitude M and the apparent magnitude m and luminosity distance DL is as follows:

M = m 5log10DL10 pc

The distance of 100 pc in this instance provides us a fairly straightforward relationship:

M=m−5=−1.0

Let's first calculate the star's luminosity in relation to the Sun's luminosity to understand why. This star is 5.7 magnitudes brighter than the Sun, which has an absolute magnitude of +4.7. Using the traditional scale where 5 magnitude equal a brightness factor of 100: LL=100(5.7)/5=190.5 where the subscript denotes the Sun.

Our star therefore has 200 times the brightness of the Sun. Depending on the spectral class, we might determine whether this star was a hot main sequence star or a low-temperature red giant. It's likely to be the latter if we were to attempt a guess. Smaller stars have not had the time to develop into red giants during the existence of our galaxy.

Larger star than the Sun will go through the red giant phase of their evolution much more quickly. In just 40 million years, even our Sun will increase from 100 to 2,000 times its current brilliance.

Therefore, if a red giant is visible to us at only 200 times the solar luminosity, we would have to observe it very quickly in cosmic time.

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