PHYSICS HIGH SCHOOL

At what speed must a 950-kg subcompact car be moving to have the same kinetic energy as a 3.2×104−kg truck going 20 km/h?

Answers

Answer 1

Answer:

950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr

Explanation:

Kinetic energy of body with mass m and moving at velocity v is given by

$KE=(1)/(2) mv^2$

KE of truck $= (1)/(2)*3.2*10^4*20^2$

KE of subcompact car $= (1)/(2)*950*v^2$

Equating

$(1)/(2)*3.2*10^4*20^2= (1)/(2)*950*v^2\n \n v=116.08 km/hr$

So 950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr

Answer 2

Answer:

Final answer:

To have the same amount of kinetic energy as a 3.2x10^4kg truck moving at 20km/h, a 950-kg subcompact car would need to be moving at an approximate speed of 224 m/s.

Explanation:

The problem requires us to calculate the speed that a 950-kg car must achieve to have the same kinetic energy as a 3.2×104-kg truck moving at 20 km/h. The kinetic energy of an object is represented by the equation K = 0.5mv² where m signifies mass and v speed.

The truck's kinetic energy can be calculated using this formula: Ktruck=0.5(3.2×104kg)(20 km/hr * 1000 m/km / 3600 s/hr)² = 1.96 x 107 J.

For the car to have the same kinetic energy, we set up the following equation and solve for v: 1.96 x 107 J = 0.5(950 kg)v², yielded v = 224 m/s.

Thus, the subcompact car needs to move at a speed of 224 m/s to have the identical kinetic energy as the truck moving at 20 km/h.

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The number of electrons in the outer ring of the elements phosphorus,iodine,vanadium,cobalt,sodium,nitrogen,helium,gold,zinc

Answers

helium will have 2 electrons
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Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 3.7 m/s2 for 4.4 seconds. It then continues at a constant speed for 8.3 seconds, before applying the brakes such that the carâs speed decreases uniformly coming to rest 216.0 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop. 1. How fast is the blue car going 1.8 seconds after it starts? 2. How fast is the blue car going 10.0 seconds after it starts? 3. How far does the blue car travel before its brakes are applied to slow down? 4. What is the acceleration of the blue car once the brakes are applied? 5. What is the total time the blue car is moving? 6. What is the acceleration of the yellow car?

Answers

1. How fast is the blue car going 1.8 seconds after it starts?

Recall this kinematic equation:

Vf = Vi + aΔt

Vf is the final velocity.

Vi is the initial velocity.

a is the acceleration.

Δt is the amount of elapsed time.

Given values:

Vi = 0 m/s (the car starts at rest)

a = 3.7 m/s² (this is the acceleration between t = 0s and t = 4.4s)

Δt = 1.8 s

Substitute the terms in the equation with the given values and solve for Vf:

Vf = 0 + 3.7×1.8

Vf = 6.66 m/s

2. How fast is the blue car going 10.0 seconds after it starts?

The car stops accelerating after t = 4.4s and continues at a constant velocity for the next 8.3 seconds. This means the car is traveling at a constant velocity between t = 4.4s and t = 12.7s. At t = 10s the car is still traveling at this constant velocity.

We must use the kinematic equation from the previous question to solve for this velocity. Use the same values except Δt = 4.4s which is the entire time interval during which the car is accelerating:

Vf = 0 + 3.7×4.4

Vf = 16.28 m/s

The constant velocity at which the car is traveling at t = 10s is 16.28 m/s

3. How far does the blue car travel before its brakes are applied to slow down?

We must break down the car's path into two parts: When it is traveling under constant acceleration and when it is traveling at constant velocity.

Traveling under constant acceleration:

Recall this kinematic equation:

d = $(Vi+Vf)/(2)$ ×Δt

d is the distance traveled.

Vi is the initial velocity.

Vf is the final velocity.

Δt is the amount of elapsed time.

Given values:

Vi = 0 m/s (the car starts at rest).

Vf = 16.28 m/s (determined from question 2).

Δt = 4.4 s

Substitute the terms in the equation with the given values and solve for d:

d = $(0+16.28)/(2)$ ×4.4

d = 35.8 m

Traveling at constant velocity:

Recall the relationship between velocity and distance:

d = vΔt

d is the distance traveled.

v is the velocity.

Δt is the amount of elapsed time.

Given values:

v = 16.28 m/s (the constant velocity from question 2).

Δt = 8.3 s (the time interval during which the car travels at constant velocity)

Substitute the terms in the equation with the given values:

d = 16.28×8.3

d = 135.1 m

Add up the distances traveled.

d = 35.8 + 135.1

d = 170.9 m

4. What is the acceleration of the blue car once the brakes are applied?

Recall this kinematic equation:

Vf²=Vi²+2ad

Vf is the final velocity.

Vi is the initial velocity.

a is the acceleration

d is the distance traveled.

Given values:

Vi = 16.28 m/s

Vf = 0 m/s

d = 216 m - 170.9 m = 45.1 m (subtracting the distance already traveled from the total path length)

Substitute the terms in the equation with the given values and solve for a:

0² = 16.28²+2a×45.1

a = -2.94 m/s²

5. What is the total time the blue car is moving?

We already know the time during which the car is traveling under constant acceleration and traveling at constant velocity. We now need to solve for the amount of time during which the car is decelerating.

Recall again:

d = $(Vi+Vf)/(2)$ ×Δt

Given values:

d = 45.1 m

Vi = 16.28 m/s (the velocity the car was traveling at before hitting the brakes).

Vf = 0 m/s (the car slows to a stop).

Substitute the terms in the equation with the given values and solve for Δt:

45.1 = $(16.28+0)/(2)$ ×Δt

Δt = 5.54s

Add up the times to get the total travel time:

t = 4.4 + 8.3 + 5.54 =

t = 18.24s

6. What is the acceleration of the yellow car?

Recall this kinematic equation:

d = ViΔt + 0.5aΔt²

d is the distance traveled.

Vi is the initial velocity.

a is the acceleration.

Δt is the amount of elapsed time.

Given values:

d = 216 m (both cars meet at 216m)

Vi = 0 m/s (the car starts at rest)

Δt = 18.24 s (take the same amount of time to reach 216m)

Substitute the terms in the equation with the given values and solve for a:

216 = 0×18.24 + 0.5a×18.24²

a = 1.3 m/s²

What is nonexample of distance

Answers

a non example of distance is capacity , volume , temperature , time , weight , power , velocity , energy , pressure , and angle

Final answer:

A nonexample of distance refers to statements or information that do not provide quantitative information about the length between two points.

Explanation:

A nonexample of distance would be something that does not involve measuring the length between two points. For example, when talking about the distance between two cities, saying 'I like the weather there' would be a nonexample because it does not provide any quantitative information about the distance. Similarly, if we are measuring the distance a car travels, saying 'It is a red car' would also be a nonexample as it does not relate to the measurement of distance.

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Some organelles are present in both plant and animal cells. Each organelle performs a different function. Which of the functions listed below are performed by organelles found in both plants and animals?

Answers

Organelles that are present in both forms of eukaryotic cells are the following :

Nucleus

Endoplasmic Reticulum

Ribosomal units

Golgi apparatus

Lysosomes

Perixosomes

Mitochondrion

Cytoskeleton/Cytosol

Vacuole

Nucleolus

Plasma membrane

Microtubules/Microfilaments

All celluar functions corresponding to the organelles can be found in both plant and animal cells

The cell is made up of the cell membrane and the cell organelles.

There are two types of cells on the basis of an organism. These are as follows:-

Plant cell
Animal cell

Plants cells are different from animal cells in the following ways:-

Cell wall
Ribosome
Vacolues
Plasmodesmata

These are the organelles that are found in both cells

Nucleus
Endoplasmic Reticulum
Ribosomal units
Golgi apparatus
Lysosomes
Perixosomes
Mitochondrion
Cytoskeleton/Cytosol
Vacuole
Nucleolus
Plasma membrane
Microtubules/Microfilaments

For more information, refer to the link:-

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Charge q is 1 unit of distance away from the source charge S. Charge p is two times further away. The force exerted between S and q is _____ the force exerted between S and p. a. 1/2
b. 2 times
c. 1/4
d. 4 times

Answers

Answer : The correct option is, (d) 4 times

Solution :

According to the Coulomb's law, the electrostatic force of attraction or repulsion between two charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between the the charges.

Formula used :

$F=k_e(q_1q_2)/(r^2)$

where,

F = electrostatic force of attraction or repulsion

$k_e$ = Coulomb's constant

$q_1$ and $q_2$ are the charges

r = distance between two charges

First we have to calculate the force exerted between S and q when the distance between the charge is 1 unit and let us assumed that the charge be 'q'

$F_(sq)=k_e(qq)/(1^2)=k_e* q^2$ ..........(1)

Now we have to calculate the force exerted between S and p when the distance between the charge is 2 unit at the same charge.

$F_(sp)=k_e(qq)/(2^2)=k_e(q^2)/(4)$ ...........(2)

Equation equation 1 and 2, we get

$(F_(sq))/(F_(sp))=(1)/(4)$

$F_(sq)=4* F_(sp)$

Therefore, the force exerted between S and q is 4 times the force exerted between S and p.

I believe the answer is "4 times".

A projectile is fired in such a way that its horizontal range is equal to 13.5 times its maximum height. What is the angle of projection?

Answers

Maximum height: h = v0² sin² α / 2 g
Horizontal range: x = v0² sin 2 α / g
13.5 v0² sin² α /2 g = v0² sin 2 α /g / : v0²/g
6.75 sin² α = 2 sin α cos α / : sin α
6.75 sin α = 2 cos α / : cos α
6.75 tan α = 2
tan α = 0.2963
α = tan^(-1) 0.2963
α = 16.504° ≈ 16.5°
The angle of projection is 16° 30`.

Answer:

Maximum height: h = v0² sin² α / 2 g

Horizontal range: x = v0² sin 2 α / g

13.5 v0² sin² α /2 g = v0² sin 2 α /g / : v0²/g

6.75 sin² α = 2 sin α cos α / : sin α

6.75 sin α = 2 cos α / : cos α

6.75 tan α = 2

tan α = 0.2963

α = tan^(-1) 0.2963

α = 16.504° ≈ 16.5°

The angle of projection is 16° 30`.

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