Information givenknown: mass of Christine=60 kgmass of cart= 22 kg mass of hailey=69The two girls on the cart to the left pushing off of Conner take .3833 s to reach a distance of 0.3m. Conner reaches the same distance in 0.2333s. What is the mass of Conner?

The processes that help Divya and Melissa identify the spicy taste of the chicken sandwiches with sriracha sauce is sensation.

Sensation is the process in which one feels something after being in physical contact with something. As Divya and Melissa eats the chicken sandwiches with sriracha sauce, their senses tells that the food is spicy.

Acculturation is the process in which one adopts and adjusts to a new cultural environment. Homogenization is the process in which two non-soluble liquids are mixed uniformly. Adaptation is the process of changing one's trait to a new environment.

Therefore,  the processes that helps them identify the spicy taste is sensation.

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Answer:

ΔS = 3.5 m

Explanation:

First, apply the 2º Newton's Law to find the resulting acceleration.

∑F = m × a

F = m × a

14 = 32 × a

a = 14 /32 = 7/16

a = 0.4375 m/s²

Then, we must apply the equation of accelerated movement:

ΔS = V₀*t + (1/2)*a*t²

Starting from the rest and using the value of acceleration:

ΔS = 0*4 + (1/2)*(0.4375)*(4)²

ΔS = (1/2)*(0.4375)*(16)

ΔS = 3.5 m

The time taken for the javelin thrown at 28.5 m/s from flat ground at a 43.2 degree angle to reach the maximum height is 2.0 s

From projectile motion, we understand that the time taken to reach maximum height is given as:

t = uSineθ / g

Where

Now, using the above formua, we can obtain the time taken for the javelin to reach the maximum height as follow:

Initial velocity (u) = 28.5 m/s

Angle of projection (θ) = 43.2 degrees

Acceleration due to gravity (g) = 9.8 m/s²

Time taken to reach the peak (t) = ?

t = uSineθ / g

t = (28.5 × Sine 43.2) / 9.8

t = 2.0 s

Thus, the time taken to reach maximum height is 2.0 s

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The age of the Universe is  13.8 ± 0.1 billion years.

The ‘hubble constant, h0’ is used to determine the size of the universe. It is a  measurement that is used in describing the universe's expansion. It was a constant introduced by Edwin Hubble. He found out that the farther the distance of the galaxies from the Earth, the faster they would appear to be moving.

One of the most important numbers in cosmology Hubble's constant because it is needed to estimate the size and age of the universe which is  13.8 ± 0.1 billion years.

The age of the universe using Hubble's constant which is  13.8 ± 0.1 billion years.

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The cargo while it was accelerating at a distance of  0.602 meters.

Acceleration is the name we give to any system wherein the speed changes. on the grounds that speed is a pace and a direction, there are only two approaches so that you can boost up.

Velocity is the rate of change of displacement. Acceleration is the price of alternate velocity. velocity is a vector quantity because it consists of both significance and direction. Acceleration is likewise a vector quantity as it's far simply the charge of alternate velocity.

initial velocity = 55 m/hr

                   = 55/60× 60 m/sec

                    = 0.0152 m/sec

final velocity = 16 mi/hr

                      = 16/3600

                     = 0.0044 m/sec

time = 3.2 sec

V = u -  at

a = ( U - V )/t

  =  (0.0152 - 0.0044)/t

  = 0.0108 m/sec²

S = ut + 1/2at²

   =  0.0152 × 3.2 sec + 1/2 × 0.0108 m/sec² ×  (3.2)²

   = 0.0486 + 0.553

   = 0.602 meter.

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The equation which represent the velocity, if the satellite is moved to an orbital radius of 5r is V = V/[tex]\sqrt\\5[/tex].

Orbital velocity or orbital speed is the minimum velocity a body must maintain to stay in an orbit. Due to the inertia of a moving body, the body has the tendency to move on in a straight path.

It is given that the radius is 5r

To find out the equation which represent the velocity if the satellite is moved to an orbital radius of 5r.

Mathematically, the orbital velocity is calculated by the formula:

V = [tex]\sqrt{GM}[/tex]/r

where, V = orbital velocity,

G = gravitational constant

M = mass of a satellite

r = radius

Substituting the value of r in this equation, we have:

V = [tex]\sqrt{GM}[/tex]/r

V = 1/ [tex]\sqrt{5}[/tex] ×  [tex]\sqrt{GM}[/tex]/r

V = [tex]\sqrt{GM}[/tex]/r

Therefore,

V = 1/ [tex]\sqrt{5}[/tex]  × V

V = V/[tex]\sqrt{5}[/tex]

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The correct answer is option C, 0.50 N to the left.

Given,

The charges,

q₁=5×10⁻⁶ C

q₂=4.0×10⁻⁶ C

q₃=10.0×10⁻⁶ C

The distance between the 1st charge and the second charge, r₁=20 cm=0.20 m

The distance between 2nd charge and the 3 rd charge, r₂=30 cm=0.30 m

As all the charges have the same sign, the nature of the force is repulsive. Thus the force applied by the q₂ is directed to the left. And the force applied by the charge q₃ is pointed to the right.

From Coulomb's law, the net electrostatic force on q₂ is given by,

On substituting the knwon values,

Thus the magnitude of the net electrostatic force on the charge q₂ is 0.5 N

As the force to the left is greater than the force to the right, the net force will be directed to the left.

Therefore the correct answer is option A, 0.50 N to the left.

Linear velocity (V) = r* w

where:

r = radius = diameter/2= 0.35 /2 = 0.175 m

w= angular velocity = rpm = 2500 = 2500/60 = 125/3 rps

1 revolution = 360° = 2pi rad/s = 125/3 2pi rad/s

= 261.8 rad/s

v = 0.175 m * 261.8 m/s = 45.8 m/s

Answer:

1.22 m/s² (3 s.f.)

Explanation:

Draw a diagram modelling the given situation (see attachment).

Given values:

As the pulling force is at an angle to the plane (ground), resolve the force into components parallel and perpendicular to the plane.

Resolving vertically (↑) to find the Normal Reaction, R:

[tex]\begin{aligned}\implies R+82.2 \sin 64^{\circ}&=14g\\R&=14g-82.2 \sin 64^{\circ}\end{aligned}[/tex]

The frictional force takes its maximum value when an object starts to move (or is on the point of moving):

[tex]\boxed{F_{\text{max}}= \mu R}[/tex]

where R is the Normal Reaction and μ is the coefficient of friction.

Using F = μR to find the Frictional Force, F:

[tex]\begin{aligned}\implies F &= 0.3\left(14g-82.2 \sin 64^{\circ}\right)\\ & =0.3\left(14(9.8)-82.2 \sin 64^{\circ}\right)\\& =0.3\left(137.2-82.2 \sin 64^{\circ}\right)\\ & = 41.16-24.66\sin 64^{\circ}\end{aligned}[/tex]

Newton's second law states that the overall resultant force acting on a body is equal to the mass of the body multiplied by the body’s acceleration:

[tex]\boxed{F_{\text{net}}=ma}[/tex]

Resolving horizontally (→) using Newton's second law of motion to find acceleration:

[tex]\begin{aligned}\textsf{Using} \quad F_{\text{net}}&=ma\\\\\implies 82.2 \cos64^{\circ}-(41.16-24.66\sin 64^{\circ})&=14a\\17.0383694...&=14a\\a&=\dfrac{17.0383694...}{14}\\a&=1.21702638...\\a&=1.22\;\sf m/s^2\;(3\;s.f.)\end{aligned}[/tex]

Therefore, the acceleration of the box is 1.22 m/s² (3 s.f.).

The magnification of first lens can be given as,

Here, m1 is the magnification of first lens, v1 is the distance of image from first lens and u1 is the distance of object from first lens.

Substitute the known values in the equation,

The magnification of second lens can be given as,

Here, m2 is the magnification of second lens, v2 is the distance of image from second lens and u2 is the distance of object from second lens.

Substitute the known values in the equation,

The combined magnification of two lenses can be given as,

Substitute the known values,

Thus, the combined magnification of two lenses is 1.5.

We will have that the value will be:

Answer:

Vx = 150 * cos 53 = 90.3 m/s     horizontal speed

Vy = 150 * sin 53 = 120 m/s      vertical speed

t1 = 120 / 9.8 = 12.2 sec       time to reach vertical height  (vy = 0 at apex)

H1 = 1/2 g t^2 = 9.8 / 2 * 12.2^2 = 729 m    height risen above cliff

729 + 405 = 1134 m      total height above striking point

H = 1/2 g T^2          where T is time to fall 1134 m

T = (2 * 1134 / 9.8)^1/2 = 15.2 sec

T + t1 = 15.2  + 12.2 = 27.4 sec total time in air

Sx = Vx * 27.4 = 90.3 * 27.4 = 2470 m   total horizontal distance

The area of the small piston, a, is 0.1963 sq in. and the area of the large piston, A, is 6.28 sq in.​ The force applied to the small piston (f)= 31.21 lb that is required to deliver a working force (F)= 1000 lb

Force is a physical quantity that means if some kind of external or internal affair is happen on a object and the object is move a bit that means a force is act on it. Its a vector quantity.

To calculate the force of the small piston f we are using the formula,

p=P

Or, f/a=F/A

Here we are given,

a= The area of the small piston = 0.1963 sq in

A= The area of the big piston = 6.28 sq in.

F= The force of the big piston = 1000 lb.

We have to calculate the force of the small piston = f

Now we put the values in above equation we get,

f / a = F / A

Or, f = a F/A

Or, f = [tex]\frac{0.1963 \times 1000}{6.28}[/tex]

Or, f = 31.21 lb

Now from the calculation we can say that, The force applied to the small piston (f)= 31.21 lb

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Answer:

III only

Explanation:

The net vertical force is equal to 0 because the block is not moving up. Then, we can write the following equation

Fnet = N + Fsinθ - mg = 0

Where Fsinθ is the vertical component of the force F. Solving for the normal force N, we get:

N = mg - Fsinθ

Therefore, the normal force is not equal to the weight and the normal force is not mgcos(θ)

However, the horizontal component of F is Fcos(θ). So, the answer is

III only.

Set of equation which is true for the situation is

c) 2SA = SB  and 2BA=BB

Force applied perpendicular to the surface of an object per unit area is called pressure

SI unit of pressure is one newton per square meter (N/m2).

Given, length = 3 foot, width= 1 foot and height = 1 foot

1foot = 0.305m

Volume of tank A = 3*0.305*0.305*0.305

= 0.085

density of water= 1000kg/m^3

Pressure= 1000kg/m^3 *gravity * height

=1000kg/m^3 * 9.8 m/s^2* 0.305

=2989 N

Force exerted by water at bottom= P * A

                                                        = 2989*0.085

                                                         =254.065 Nm^2

similarly, for tank B

Given ,length = 6 foot, width =1 foot and height= 1foot

Volume of tank B = 6*0.305*0.305*0.305

                               =0.17

pressure= 1000kg/m^3 * 9.8m/s^2 * 0.305

               = 2989 N

Force exerted by water at bottom = P*A

                                                         = 2989 * 0.17

                                                          =508.13 Nm^2

Hence proved that pressure exerted by tank B is twice that of tank A.

And same goes for pressure exerted on the side of the tank.

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Given:

The lamp draws the current is

The voltage supplied by the generator is

The time for which energy converted is

Explanation:

(a)

The power delivered by the generator is given by

Plugging all the values in the above equation, we get

The power delivered by the generator is

(b)

Energy is converted by the electric lamp is given by

Plugging all the values in the above relation, we get

Thus, the electric energy convert by the lamp is

Answer:

4500 N

Explanation:

Bc f is equal to mass times acceleration and the acceleration is 60

If a 75 kg student drives a 650 kg car, then the car starts from rest and accelerates to 60 m/s in 8 seconds, then the applied force on the car would be 5437.5 Newtons.

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.

The mathematical expression for Newton's second law is as follows

F = m × a

As given in problem an If a 75 kg student drives a 650 kg car, the car starts from rest and accelerates to 60 m/s in 8 seconds.

The acceleration of the car = 60 / 8

                                             = 7.5 m / s²

The force applied on the car = 725 × 7.5

                                                = 5437.5 Newtons

Thus,  the applied force on the car would be 5437.5 Newtons.

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Answer:

Therefore the friction Force is 65 N.

Explanation:

Look at the image above.

The frictional force of the rubber block is 63.7N.

Friction is defined as a force that opposes the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are many types of friction such as dry friction which is defined as the force that opposes the relative lateral motion of two solid surfaces in contact.

This friction force is directed in the opposite direction to the motion of the object itself because the friction described so far between surfaces in relative motion is called kinetic friction.

For above given information,

mass of the block, m =10kg

coefficient of friction, µ=0.65

f=coefficient of the friction ×normal= µ*R

R= normal =mg = 9.8×10=98N

So, f =0.65×98=63.70N

≈65N(when g is taken as 10m/s²)

Thus, the frictional force of the rubber block is 63.7N.

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Just because there is a relationship between two behaviors does not mean that one causes the other. Option A.

Natural observation is a research method that observes objects in their natural environment. This approach is commonly used by psychologists and other social scientists. It is a process of deep reflection, questioning, and evidence evaluation. A study in which researchers look for relationships between people's demographic neighborhood and their level of prejudice.

Psychologists use this because it is the most objective method known to produce reliable insights. The science of behavior and mental processes. It attempts to describe and explain aspects of human thinking feeling perception and behavior. Naturalistic observation is a method of observation that involves observing people's behavior in environments in which they normally occur. Laboratory Observation observing behavior in a more artificial environment focusing on a small number of carefully defined behaviors.

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If an electron travels 2.68 m in 6.11 × 10^-8 s, it will travel at a speed of 43.86 * [tex]10^{8}[/tex] cm / s

v = d / t

v = Velocity

d = Distance

t = Time

d = 2.68 m

t = 6.11 * [tex]10^{-8}[/tex]

v = 2.68 / 6.11 * [tex]10^{-8}[/tex]

v = 0.4386 * [tex]10^{8}[/tex] m / s

v = 0.4386 * [tex]10^{8}[/tex] * 10² cm / s

v = 43.86 * [tex]10^{8}[/tex] cm / s

One meter is equal to 100 centimeters. So in order to convert a meter value into centimeter, multiply the meter value by 100 or 10².

Therefore, the electron travels at 43.86 * [tex]10^{8}[/tex] cm / s

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W0e are asked to determine the frequency of a sound given its velocity and wavelength. To do that we will use the following formula:

Where:

Now, we solve for the frequency by dividing both sides by the wavelength.

Now, we substitute the values:

Solving the operations:

Therefore, the frequency is 40 Hz.

Increasing his acceleration will impact his velocity and rate of displacement covered in that as the speed increases due to the increased rate of acceleration, the rate of air resistance also increases.  

Air resistance is a force created by air. When an item moves through the air, the force operates in the opposite direction.

When a diver descends, the force of air resistance acts to counteract the force of gravity. As the skydiver falls faster and faster, the quantity of air resistance grows until it equals the magnitude of gravity's force.

A balance of forces is achieved when the force of gravity equals the force of air resistance, and the skydiver no longer accelerates. The skydiver reaches what is known as terminal velocity.

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When an electron moves between the ends of a 9.00-volt battery, the work done on the electron is 1.44 * 10⁻¹⁸ Jouoles; option C

The work done in moving an electron between two points is defined as the product of the potential difference between the points and the charge on an electron.

Mathematically;

where;

ΔV is the potential difference between the points

q is the charge on the electron.

The potential difference between the two points is the work done in moving a unit positive charge from one point to the other in the circuit.

The charge, q on an electron is 1.609 * 10⁻¹⁹ Coulombs.

For the given electron that moves between the ends of a 9.00-volt battery;

ΔV = 9.000 Volts

q =  1.609 * 10⁻¹⁹ Coulombs

Work done = 9.000 * 1.609 * 10⁻¹⁹

Work done = 1.44 * 10⁻¹⁸ Jouoles

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We know that the components of any vector along the x and y axes are given by:

In this case we know the tha magnitude of the vector is 6.96 m/s; we also know that the angle from the x-axis to the vector is 51.5° (the sum of angles θ and the 26.5° angle); then we have:

Therefore, the components of the vector are:

The time taken for the bullet moving with a speed of 600 m/s to hit a target  200 meters away is 3.33 s.

Time is the measured or measurable period during which an action, process, or condition exists or continues.

To calculate the time it takes the bullet to hit the target, we use the formula below.

Formula:

Where:

Make t the subject of the equation

From the question,

Given:

Substitute these values into into equation 2

Hence, the time it will take to hut the target is 3.33 seconds.

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The x-component of the vector is determined as -16.24 m.

The x-component of a vector is the vector component measured along x - direction. It is the vector acting in the x-coordinate.

The magnitude of x -component of a vector is given by the following formula as shown below;

Vₓ = V cosθ

where;

Vₓ = 148 m x cos(-96.3)

Vₓ = -16.24 m

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Answer:

Y= -147

X= -16.24

Explanation:

The spring constant is  400 N/m and The mass which would produce an extension of 3.0 cm is 0.0459 kg.

Here we are dealing with a problem related to hook's law which states that the force required to expand or compress a spring by a few distances is corresponding to that distance. This can be expressed scientifically as F= -kx, where F is the force connected to the spring and x is the displacement of the spring, with negative esteem illustrating the displacement of the spring once it is extended; and here k is the spring constant.

For the first case, we are given the force applied which is 20 N, and the extended length which is 5cm =0.05 m,

So the spring constant will be, k= F/x=  20/0.05 = 400 N/m

For the second  case we are given that the spring constant which is 15 N/m  and the extension is 3.0 cm =0.03 m.

Since we know F= mg ( where g is the acceleration due to gravity )

So the hanging mass, will be

=>F= kx

=>mg=kx

=>m=kx/g

      =15 x0.03/9.8

     = 0.0459 kg

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Given

Initial speed, u=45 m/s

Acceleration

Time taken, t=4 sec

To find

The final speed

Explanation

Let the speed he be going in be v

Thus,

Conclusion

The velocity he is going in is 13 m/s.