the change in position of an object is

So because trajectories including both universes are more or less inclined, Mars' orbit is not far from this same ecliptic.

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v₀ = initial velocity of the car before brakes are applied

v = final velocity of the car after it comes to a stop = 0 mph

d = stopping distance

a = acceleration caused due to braking force

Using the kinematics equation

v² = v²₀ - 2 a d

0² =  v²₀ - 2 a d

d = v²₀ /(2a)

Since the acceleration is same , the stopping distance is directly proportional to the square of the initial speed of car before brakes are applied

hence

d₁/d₂ = v²₀₁/v²₀₂

Given that : v₀₁ = 60 mph   and v₀₂ = 15 mph

inserting the values

d₁/d₂ = (60)²/(15)²

d₁ = 16 d₂

hence distance increases by a factor of 16 times.

A car skidding from 60 mph will skid 16 times farther than a car from 15 mph because the kinetic energy is proportional to the square of the velocity, and the work done by friction to stop the car is equal to its initial kinetic energy.

The question asks how much farther a car will skid if it locks its brakes at 60 mph compared to a skid from 15 mph. When a car locks its brakes, the distance it skids before coming to a stop is directly related to the kinetic energy it had when the brakes were applied. Kinetic energy is given by the formula [tex]KE = \frac{1}{2}mv^2[/tex]. Because the kinetic energy depends on the square of the velocity, a car moving at 60 mph (which is four times faster than 15 mph) will have 16 times the kinetic energy ([tex]4^2[/tex]).

Assuming that the frictional force acting on the car is the same in both situations, we know that the work done by the frictional force to stop the car is equal to the car's initial kinetic energy. Hence, the car skidding from 60 mph will skid 16 times farther than the car skidding from 15 mph, because work is directly proportional to distance when force is constant (Work = Force × Distance). Therefore, the distance increases by a factor of 16 times.

As it is given that 2 kg mass is suspended by 12 cm long thread and then a horizontal force is applied on it so that it remains in equilibrium at 30 degree angle

So here we can use force balance in X and Y directions

now for X direction or horizontal direction we can use

[tex]F = Tsin30[/tex]

for vertical direction similarly we can say

[tex]mg = T cos30[/tex]

so here we first divide the two equations

[tex]\frac{F}{mg} = \frac{sin 30}{cos 30}[/tex]

[tex]\frac{F}{mg} = tan 30[/tex]

[tex]F = mg tan30[/tex]

now plug in all values in the above equation

[tex]F = 2 * 9.8 * tan30[/tex]

[tex]F = 11.3 N[/tex]

Part b)

now in order to find the tension in the thread we can use any above equation

[tex]F = T sin30[/tex]

[tex]11.3 = T sin30[/tex]

[tex]T = \frac{11.3}{sin30}[/tex]

[tex]T = 22.6 N[/tex]

so tension in the thread will be 22.6 N

Final answer:

To find the horizontal force and tension in a string supporting a displaced metal ball, one must use equilibrium principles along with trigonometric relations from Newton's second law, focusing on the components of tension.

Explanation:

The question involves finding a) the horizontal force and b) the tension in the thread that supports a metal ball displaced by 30 degrees to the vertical. This situation can be analyzed using principles of equilibrium and Newton's second law. When the ball is displaced, it creates an angle with the vertical, leading to a horizontal component of the tension acting as the horizontal force, and a vertical component of the tension balancing the weight of the ball.

To calculate these forces, we can use trigonometric relations and Newton's second law. The tension in the string can be resolved into two components: the horizontal component (Thorizontal) and the vertical component (Tvertical). The vertical component balances the weight of the ball (mg), and the horizontal component is the force needed to keep the ball in equilibrium.

For a ball of mass 2 kg displaced at a 30-degree angle, assuming g = 9.8 m/s2, the calculations would involve using the equations Tvertical = mg and Thorizontal = Tvertical * tan(θ) where θ is the angle of displacement. However, specific calculations are not provided here without the complete equations and values.

The answer is B because the bowling ball is the heaviest.

Let the directions are defined by vectors as following

East = + x direction

West = - x direction

North = + y direction

South = - y direction

upwards = + z direction

downwards = - z direction

So here we are given that it moves

1). 10 m East

2). 9 m South

3). 3 m North

4). 2 m down

So we can write it as

[tex]d_1 = 10 \hat i[/tex]

[tex]d_2 = 9 (-\hat j)[/tex]

[tex]d_3 = 3 \hat j[/tex]

[tex]d_4 = 2 (-\hat k)[/tex]

Now total displacement will be given as

[tex]d = d_1 + d_2 + d_3 +d_4[/tex]

[tex]d = 10 \hat i - 9 \hat j + 3 \hat j - 2 \hat k[/tex]

[tex]d = 10 \hat i - 6 \hat j - 2 \hat k[/tex]

now the magnitude of the displacement will be

[tex]d = \sqrt{10^2 + 6^2 + 2^2}[/tex]

[tex]d = 11.8 m[/tex]

so the magnitude of displacement will be 11.8 m

The magnitude of the mole's displacement from its original position is approximately 11.83 meters.

To find the magnitude of the mole's displacement, you can break down the motion into its components and use vector addition. Here's the step-by-step calculation:

1. The mole runs 10 meters due East, which we'll represent as +10 meters in the horizontal (X) direction.

2. Then, the mole runs 9 meters due South, which we'll represent as -9 meters in the vertical (Y) direction. This is because it's in the opposite direction of the positive Y-axis.

3. After that, the mole runs 3 meters due North, which we'll represent as +3 meters in the vertical (Y) direction.

4. Finally, the mole burrows 2 meters down a hole, which we'll represent as -2 meters in the depth (Z) direction. This is because it's in the opposite direction of the positive Z-axis.

Now, we can calculate the total displacement vector by adding the individual components:

Displacement in the X direction: 10 meters (East)

Displacement in the Y direction: (3 meters - 9 meters) = -6 meters (South)

Displacement in the Z direction: -2 meters (Down)

To find the magnitude of the displacement, we can use the Pythagorean theorem in three dimensions:

[tex]\[|D| = \sqrt{(Dx^2 + Dy^2 + Dz^2)}\][/tex]

[tex]\[|D| = \sqrt{(10^2 + (-6)^2 + (-2)^2)}\][/tex]

[tex]\[|D| = \sqrt{(100 + 36 + 4)}\][/tex]

[tex]\[|D| = \sqrt{140}\][/tex]

The magnitude of the mole's displacement from its original position is approximately 11.83 meters.

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Explanation:

Given that,

The dummy is thrown forward with a force of 130 N, [tex]F_1=130\ N[/tex]

Side force acting on the dummy, [tex]F_2=4500\ N[/tex]

We need to find the force acting on the sensor report. It can be calculated using Pythagoras theorem as :

[tex]F_{net}=\sqrt{F_1^2+F_2^2}[/tex]

[tex]F_{net}=\sqrt{130^2+4500^2}[/tex]

[tex]F_{net}=4501.87\ N[/tex]

So, the net force acting on the sensor report is 4501.87 N. Hence, this is the required solution.

The net force acting on the dummy is 4502 N.

Force is a vector, the resultant force (net force) is that single force that has the same effect in magnitude and direction as two or more forces acting together. The resultant of a vector must take into cognizance, the geometry of the problem.

The dummy is thrown forward with a force of 130.0N while simultaneously being hit from the side with a force of 4500.0N. The net force must now be obtained by Pythagoras theorem.

Fnet^2 = F1^2 + F2^2

F1 = 130.0N

F2 = 4500.0N

Fnet = √(130.0N)^2 + (4500.0N)^2

Fnet = 4502 N

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If the acceleration applied is constant, then it is the same as the average acceleration throughout the duration of the stop.

[tex]a=\dfrac{\Delta v}{\Delta t}\implies-6.4\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{0-28\,\frac{\mathrm m}{\mathrm s}}{\Delta t}\implies\Delta t=4.4\,\mathrm s[/tex]

Since the hippocampus' involvement with the brain is to do with memories, I would go with B.) The Hippocampus acts as a "memory index" of sorts, and if you send information to it, it would make connections to, in a sense, help remember it.

Answer:

it is b

Explanation:

As E= mc²

E = (4.36 × 10^ -5) ×( 3 × 108 m/s.)²

E= 3.924×10^12J

E= (3.924×10^12)KJ/ 1000

E =3.92 × 109 kJ

Answer:

Released energy, [tex]E=3.92\times 10^6\ kJ[/tex]

Explanation:

It is given that,

Loss in mass in a nuclear reaction, [tex]m=4.36\times 10^{-5}\ g=4.36\times 10^{-8}\ g[/tex]

The relation between the mass and energy is given by Einstein mass energy equivalence equation :

[tex]E=mc^2[/tex]

c is the speed of light

So, [tex]E=4.36\times 10^{-8}\times (3\times 10^8)^2[/tex]

[tex]E=3.92\times 10^9\ J[/tex]

[tex]E=3.92\times 10^6\ kJ[/tex]

The energy released in a nuclear reaction is [tex]3.92\times 10^6\ kJ[/tex]. Hence, the correct option is (B)

As per the question there are three physical quantities named as position,distance and displacement.

Before coming into a conclusion first we have to understand a vector and a scalar.

A scalar quantity is a quantity which has only magnitude for it's complete specifications.

A vector is a quantity which has magnitude as well as direction and at the same time it is in accordance with the paraellogram law of vector addition.

Out of the three options displacement and position are vector quantities.It is because it is the minimum distance between two points .It has magnitude as well as direction.

Distance can not be considered as a vector quantity as it has only magnitude.There is no specific directions of distance travelled.

Position vector is a vector which provides location of an object in a plane or space.It is nothing else except the point which has x,y,z coordinates with origin is taken as the reference point.

Hence position and displacement are vectors

Answer:

position and displacement

Explanation:

T= The time it takes for the flower pot to pass the top of my window.  

V= The velocity of the flower pot at the moment it is passing the top of my window.  

X= The height above the top of my window that the flower pot was dropped.  

h = Lw + X  

Lw = (1/2)*g*t^2 + V*t  

V*t = Lw - (1/2)*g*t^2  

V= Lw/t - (1/2)*g*t , On the other hand we know : V=gT.  

Therefore we will have: Tg= Lw/t - (1/2)*g*t  

T= Lw/(tg) - t/2  

Now substitute for T in the following equation: X = (1/2)*g*T^2  

X= (1/2)*g*(Lw/(tg) - t/2)^2  

Now substitute for X in the very first equation I mentioned: h = Lw + X  

h = Lw + (1/2)*g*(Lw/(tg) - t/2)^2  

In case you wanted the answer to be simplified, then:  

h= (Lw^2)/(2*g*t^2) + (g*t^2)/8 + Lw/2

Answer:

3.26m

Explanation:

Using one of the equation of motion to get the distance of the pot from the window and the ground;

v² = u²+2as where

v is the final velocity = 8m/s

u is the initial velocity = 0m/s

a =+g = acceleration due to gravity (this acceleration is positive since the body is falling downwards)

g = 9.81m/s

s is the distance between the object and the window from which it dropped.

Substituting this values to get the distance s we have;

8² = 0²+2(9.81)s

64 = 19.62s

s = 64/19.62

S = 3.26m

let the magnitude of force applied by each worker be "F"

consider east-west direction along X-axis and north-south direction along Y-axis

In unit vector form, force vector by worker pushing in east direction is given as

[tex]\underset{A}{\rightarrow}[/tex] = F [tex]\hat{i}[/tex] + 0  [tex]\hat{j}[/tex]

In unit vector form, force vector by worker pushing in north direction is given as

[tex]\underset{B}{\rightarrow}[/tex] = 0 [tex]\hat{i}[/tex] + F  [tex]\hat{j}[/tex]

resultant force is given as the vector sum of two vector forces as

[tex]\underset{R}{\rightarrow}[/tex] = [tex]\underset{A}{\rightarrow}[/tex] + [tex]\underset{B}{\rightarrow}[/tex]

[tex]\underset{R}{\rightarrow}[/tex] = (F [tex]\hat{i}[/tex] + 0  [tex]\hat{j}[/tex] ) + (0 [tex]\hat{i}[/tex] + F  [tex]\hat{j}[/tex] )

[tex]\underset{R}{\rightarrow}[/tex] = F [tex]\hat{i}[/tex] +  F  [tex]\hat{j}[/tex]

direction of the force is hence given as

θ = tan⁻¹(F/F)

θ =  tan⁻¹(1)

θ = 45 degree north of east

hence the direction is north-east

Answer:

true

Explanation:

1 h 13 min and 50 s it will take.

Force of gravitation between two balls is given by the formula

[tex]F = \frac{Gm_1 m_2}{r^2}[/tex]

here we know that

[tex]m_1[/tex] = mass of ball 1

[tex]m_2[/tex] = mass of ball 2

[tex]r[/tex] = distance between two balls

So here the two factors that will affect the force of gravitation is

1. Distance between two balls

2. mass of two balls

Here balls do not move due to gravitation attraction force because here the force of gravitation is very small as it is compared with other forces like frictional force between balls and ground.

So this weak gravitational force is balanced by frictional force on balls

SO all balls remains at rest

Gravitational force is directly affected by the masses of the objects involved and the distance between them. Despite this, objects do not simply move towards each other due to gravity because other forces such as friction and air resistance often counteract this attraction.

The question essentially asks about two factors that affect gravity and why the balls do not move towards each other unless acted upon by an external force. Here's what we know from Newton's Law of Universal Gravitation: the gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. In simple terms, this means that the larger the masses of the objects, and the closer they are to each other, the stronger the gravitational pull between them.

Now, onto the second part of your question. In a perfect vacuum where no other forces exist, the balls would indeed move towards each other. However, in the real world, there are other forces at play, such as friction and air resistance, that counteract the gravitational pull. Therefore, unless these balls are acted upon by a stronger force (like a push or a pull), they will not move towards each other.

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Solution with explanation is given below in attachment.

Answer:

Objects in free-fall do not experience air resistance.

Explanation:

This comes from the definition of free-fall: an object in free fall is an object which is falling and the only force acting on it is gravity. Therefore, air resistance is not present for an object in free-fall. As a result, the acceleration of any object in free-fall is always equal to g (gravitational acceleration), and so all the objects fall with the same velocity and same time, regardless of their mass.

If the object is massive then the acceleration of gravity is faster, such objects will free fall more quickly than lighter objects. The objects in free fall do not experience air resistance are all true statements because the force of gravity which depends upon weight.

It is directly proportional to the weight. Yet the mass and weight are not same because mass is only value in kilograms but the weight is the value in which mass and gravitational acceleration are included. However, its unit is kilograms - 2 which is equal to Newton law of formula which is also unit of force.

When car is at rest the steaks makes makes vertical lines

which means the rain is falling in vertically downward direction

Now when car is moving with some speed v

Now the steaks makes and an angle 45 degree

So here we can say that relative velocity of rain with car is 45 degree

Now this is the resultant speed of rain in car frame

[tex]V_{rc} = V_r - V_c[/tex]

now if relative velocity makes 45 degree angle so this vector must have same components in vertical and horizontal direction

Since we know that relative velocity is resultant of rain velocity and car velocity so we can say here its two components are rain velocity and car velocity

So these two components must be of same magnitude

as it makes 45 degree

because when two vector are of same magnitude then the resultant vector always makes 45 degree with them if these two vectors are perpendicular to each other

car is moving at same speed as the speed of rain

Final answer:

When the rain makes a 45-degree angle streak on a moving car's window, the speed of the car is equal to the speed of the falling rain.

Explanation:

The student's question involves understanding relative velocities and their relationship to observed angles. Specifically, if rain makes 45-degree streaks on a moving vehicle's window, we want to know how fast the car is moving compared with the speed of the falling rain.

To solve this problem, we make use of trigonometry, particularly the tangent function which relates opposite and adjacent sides in a right-angled triangle. If the streaks are at a 45-degree angle, the vertical and horizontal speeds (i.e., speed of the rain and speed of the car) are equal. Therefore, in this scenario, the car is moving at the same speed as the rain is falling. Using the formula tan(θ) = vhorizontal / vvertical, where θ is the angle of the rain relative to the vertical, we find that at 45 degrees, tan(45) = 1 which implies the car's speed (vhorizontal) is equal to the rain's speed (vvertical).

Intially, we have:

Pressure P1 = 400KPa

Temperature T1= 110Kelvin

When,

Temperature T2= 235 Kelvin

Pressure P2= ?

We have gas equation:

PV= nRT

P/T= nR/V

Considering nR/T as constant, we have:

P1/T1 = P2/T2

400/110= P2/235

P2= 854.5 KPa

So the new temperature will be 854.5 KPa

Focal distance of an ellipse is given by the formula

[tex]f = ae[/tex]

here a = length of semi major axis

e = eccentricity of the path

now here we know that

length of major axis for the path of planet is given as 32 units

so here we can say

[tex]2a = 32 units[/tex]

[tex]a = 16 units[/tex]

so length of semi major axis is 16 units

focal distance for the planet path is given as 8 units

now from the above formula we can write

[tex]f = a*e[/tex]

[tex]8 = 16*e[/tex]

[tex]e = \frac{8}{16}[/tex]

[tex]e = \frac{1}{2} = 0.5[/tex]

so eccentricity for the path of planet will be 0.5

You've managed somehow to post the mirror image of the circuit diagram, including the numbers and values of the resistors.  I'm curious to know how you did that.

The three resistors at the left end of the diagram are  3Ω ,  2Ω , and  1Ω  all in series.  They behave like a single resistor of  (3+2+1) = 6Ω .

That  6Ω  resistor is in parallel with the  2Ω  drawn vertically in the middle of the diagram.  That combination acts like a single resistor of  1.5Ω  in that position.

Finally, we have that  1.5Ω  resistor in series with  1Ω  and  4Ω  .  That series combination behaves like a single resistor of  6.5Ω  across the battery V.  

We need to find the average speed of the ball during the motion of 1 m

In order to find that we took several reading and found following times to cover the distance of 1 m

t1 = 2.26 s

t2 = 2.38 s

t3 = 3.02 s

t4 = 2.26 s

t5 = 2.31 s

Now in order to find the average time we can write

[tex]T_{mean} = \frac{t_1 + t_2 + t_3 + t_4 + t_5}{5}[/tex]

[tex]T_{mean} = \frac{2.26 + 2.38 + 3.02 + 2.26 + 2.31}{5}[/tex]

[tex]T_{mean} = 2.45 s[/tex]

So average time to cover the distance of 1 m by ball will be 2.45 s

here 3.02 s is not the average time but we can say it is the median of the readings of all possible values which we can not use in our calculation as average time

Potential Kinetic Kinetic Potential Potential

answer: you stand in a lab experiment because if you sit during the lab, you have much reach of the materials on the table. and also, you might have a risk on some chemical spill on your clothes. the chemical might be flammable and it might set your clothes on fire.

so that's why you have to stand during lab experiments.

hope this helps! ❤ from peachimin

Standing during a lab experiment is important for safety and accuracy. It allows better control over equipment and materials and minimizes the risk of hazards.

When conducting a lab experiment, it is important to stand to ensure safety and accuracy of the experiment. Standing allows you to have better control and stability over the equipment and materials you are working with. Additionally, it helps minimize the risk of accidental spills, breakage, or other hazards.

For example, if you are working with chemicals or glassware, standing can prevent the equipment from tipping over and causing injury. It also allows you to observe the reaction or process more closely, making it easier to record accurate data and observations.

Overall, standing during a lab experiment promotes safety, precision, and optimal results.

We have that from the Question, it can be said that

From the Question we are told

If a weight lifter holds a 200 kg barbell in place over his head, he _______

A:does no work because the force he applies does not cause the barbell to move.

B: does work because the barbell is heavy to hold.

Generally the equation for work is mathematically given as

w=f*d

Therefore

For an instance where distance traveled is zero and force exerted is a 1000N  the work done will still be zero because

1000*0 =0

Therefore

does no work because the force he applies does not cause the barbell to move.

Option A

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The volume of a cube is given by s^3. So the volume of this block is 3cm x 3cm x3 cm = 27 cm^3.

density = mass/volume =27 g / 27 cm^3 = 1 g/cm^3.

the answer is 1 g/cm^3. I hope this helps!

initial height of the boy when he jump or dive is 5 meter

[tex]h_1 = 5 m[/tex]

now his final position is 2 yards above the surface

[tex]h_2 = 2 yards[/tex]

as we know that

[tex]1 yard = 0.9144 m[/tex]

[tex]2 yards = 1.83 m[/tex]

now by energy conservation we can say

change in potential energy = gain in kinetic energy

[tex]mg(h_1 - h_2) = \frac{1}{2} mv^2[/tex]

divide both sides by mass "m"

[tex]g*(5 - 1.83) = \frac{1}{2}*v^2[/tex]

[tex]v^2 = 2*9.8*(5 - 1.83)[/tex]

Now kinetic energy will be given as

[tex]KE = \frac{1}{2} mv^2[/tex]

[tex]KE = \frac{1}{2}*70 * 2*9.8*( 5 - 1.83)[/tex]

[tex]KE = 2175 J[/tex]

so his kinetic energy will be 2175 J