Two balls were dropped at the same time from the top of a ladder. The balls are of equal size but ball A has more mass than Ball B. Which best describes the time it would take for the two balls to hit the ground

A .Not enough information
B. Ball B would hit the ground first
C. Ball A and B will hit the ground at the same time.

Answers

Answer 1

Answer:

A

Explanation:

If we ignore air resistance or friction, the balls would hit the ground at the same time. The question does not state whether or not there is air resistance.

If there was air resistance, ball A would hit the ground first.

Again, there is no information provided on whether is is air resistance or not, so there is not enough information.


Related Questions

How to Convert 20000N/m2 to N/cm2 ? Explain it Step by Step.​

Answers

Answer:

2N/cm²

Explanation:

1m=100cm so, 1m²= 100×100 cm² = 10000cm².

therefore , 20000N/m²= 20000/10000 N/cm².

= 2N/cm².

hope this helps you.

Correct answer is by. The equiv 48 - (-12) = ​

Answers

Answer:

60

Explanation:

48-(-12)=60

the -(- is like a big plus sign really if you see that then just add the numbers together so do 48 plus 12 it equals 60.

An object with an initial velocity of 2 m/s moves in a straight line under a constant acceleration.
Three seconds later, its velocity is 16 m/s.
(a) How far did the object travel during this time?
(b) What was the acceleration of the object?

Answers

Answer:

(a) distance =27m

(b)acceleration =14/3ms-2

Explanation:

s=(u+v)t/2

= (2+16)3/2

=27m

v=u+at

16=2+a*3

a= 14/3ms-2

A 1200-kg car is moving at 17.0 m/s due north. A 1600-kg car is moving at 21.0 m/s due east. The two cars simultaneously approach an icy intersection where, with no brakes or steering, they collide and stick together.

1)
Determine the speed of the combined two-car wreck immediately after the collision. (Express your answer to two significant figures.)

2)
Determine the direction of the combined two-car wreck immediately after the collision. (Express your answer to two significant figures.)

Answers

Answer:

f

Explanation:

The speed of the two-car wreck right after the crash is 13.9 m/s, and the direction is 58.4° east of north.

Momentum conservation can address this problem. Before and after the collision, momentum is equal. Calculate the two-car wreck's speed and direction after the collision:

Step 1: Calculate automobile beginning momentum.

Mass (m) times velocity (v) equals momentum (p).

Northbound 1200-kg car:

Initial momentum of the 1200-kg automobile (p₁) = mass (m₁) × velocity (v₁).

p₁ = 1200 kg × 17.0 m/s = 20,400 kg/s (North direction).

1600-kg eastbound car:

The 1600-kg car's initial momentum (p₂) = m₂ × v₂.

p₂ = 1600 kg × 21.0 m/s = 33,600 kg/s (East direction).

Step 2: Calculate the pre-collision momentum.

The vector sum of the two automobiles' momentum before the accident is p[tex]_total[/tex].

p[tex]_total[/tex] = p₁+p₂.

p[tex]_total[/tex] = 20,400 kg m/s (North) + 33,600 (East)

Step 3: Calculate the two cars' combined mass after the collision.

Combined mass (m[tex]_combined[/tex]) = 1200-kg + 1600-kg cars.

m[tex]_combined[/tex] = 1200 + 1600 = 2800 kg

Step 4: Determine the two-car wreck's velocity after the accident.

Total momentum after the collision (p[tex]_combined[/tex]) is the same as before, but it belongs to the combined mass (m[tex]_combined[/tex]) travelling at the velocity (v[tex]_combined[/tex]) in a given direction:

p[tex]_combined[/tex] = m[tex]_combined[/tex] × v[tex]_combined[/tex]

Since the collision is inelastic (cars stick together), momentum remains the same.

= p[tex]_total[/tex]

m[tex]_combined[/tex] × v[tex]_combined[/tex] = 20,400 kg/s (North) + 33,600 kg/s (East).

Step 5: Determine the two-car wreck's speed and direction.

Calculating the resultant vector of the North and East momentum components gives the combined two-car wreck's velocity (speed) (v[tex]_combined[/tex]).

(v[tex]_combined[/tex][tex]_north²[/tex] + v[tex]_combined[/tex][tex]_east²[/tex])

Where:

North velocity is v[tex]_combined[/tex][tex]_north[/tex].

East velocity is v[tex]_combined[/tex][tex]_east[/tex].

v[tex]_combined[/tex][tex]_north[/tex] = 20,400 kg m/s, 2800 kg, 7.29 m/s (North).

v[tex]_combined[/tex][tex]_east[/tex] = 33,600 kg m/s < 2800 kg × 12.0 m/s (East).

13.9 m/s = (7.29² + 12.0²)

Step 6: Determine the two-car wreck's direction.

Trigonometry can determine direction:

tan() = v[tex]_combined[/tex][tex]_east/north[/tex]

arctan(v[tex]_combined[/tex][tex]_east/north[/tex]) =

arctan(12.0 / 7.29 m/s) arctan(1.645) 58.4°

The two-car crash faces east at 58.4°.

Results summary:

The two-car disaster shortly after the accident is 13.9 m/s.

The two-car crash lies 58.4° east of north following the collision.

To kow more about speed

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If you were in a spaceship travelling at 0.80 c away from a star, at what speed would the star's light pass you?

Still travelling at 0.80 c you see another spaceship approaching you, and the star. You measure its speed to be 0.84 c.

What is the other spaceship's speed in the star's frame of reference?

Answers

(1) The speed with which the star's light will pass you is 3 x 10⁸ m/s.

(2) The speed of the other spaceship is -2.943 x 10⁸ m/s.

The given parameters;

speed of the start light, u' = cspeed of spaceship, v = 0.8c

(1)

Since you are travelling away from the star (same direction), relative to a third observer, the speed at which the star's light will pass you is calculated using Lorentz velocity addition.

[tex]u = \frac{u'+ v}{1 + \frac{v}{c^2} u'} \\\\u = \frac{c+(-0.8c)}{1 + \frac{-0.8c}{c^2} (c)}\\\\u = \frac{c-0.8c}{1- \frac{0.8c}{c} } \\\\u = \frac{0.2c}{0.2} = c = 3\times 10^{8} \ m/s[/tex]

Thus, the speed with which the star's light will pass you is 3 x 10⁸ m/s.

(2) The speed of the spaceship approaching = 0.84 c

Since you are travelling in opposite direction to second spaceship, relative to a third observer (frame of reference), the speed of the other spaceship is calculated using Lorentz velocity addition.

let the speed of the approaching spaceship u', = 0.84c

[tex]u = \frac{u' + v}{1 + \frac{v}{c^2}u' } \\\\u = \frac{-0.84c - 0.8c}{1 + \frac{(-0.8c)}{c^2}(-0.84c) } \\\\u = \frac{-1.64c}{1.672} = -0.981c \\\\u = -0.981 \times 3\times 10^{8} = -2.943 \times 10^{8} \ m/s[/tex]

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In a five paragraph essay, the ________ is the main idea you're trying to prove.

Answers

What is the story about?

Which statement describes the law of conservation of energy?
A. There is only one form of energy.
B. When an energy transformation happens, no energy is destroyed
or created.
C. The total energy in a system can only increase.
D. Energy can only change from nuclear energy to chemical energy.
SUBMIT

Answers

Answer:

B. When an energy transformation happens, no energy is destroyed

or created.

Explanation:

Study the venn diagram given below carefully. Which of the following can be placed at B? Options: Clay _ Paper _ Water_ Nitrogen. ​

Answers

Answer:

water and nitrogen can be place at B

The B loop is the element that can be liquid, cas or solid. Water can have three states : solid (ice), liquid, or gas (as a vapor). The same can be said with nitrogen, it's a gas at room temperature, and it also can be a liquid or solid.

a car, which has a mass of 2000kg traveled a distance of 200 meters in 5 seconds. After 20 seconds the car was raveling at a speed of 60 m/s what is the force of the car?

Answers

Answer:

Explanation:

vi = 200/5 = 40 m/s

a = (vf - vi)/t = (60 - 40)/20 = 1 m/s²

F = ma = 2000(1) = 2000 N

What happens to the direction of the line joining when the object slows down ?Explain the observations​

Answers

Answer:

it will be curved as in deceleration

Explanation:

A car initially traveling at a speed of 15.0 m/s accelerates uniformly to a speed of 20.0 m/s over a distance of 40.0 meters. What is the magnitude of the car's acceleration?

Answers:
1.1 m/s^2
2.0 m/s^2
2.2 m/s^2
9 m/s^2

Answers

Answer:

[tex]\boxed {\boxed {\sf 2.2 \ m/s^2}}[/tex]

Explanation:

We are asked to solve for the magnitude of the car's acceleration.

We are given the initial speed, final speed, and distance, so we will use the following kinematic equation.

[tex]{v_f}^2={v_i}^2+2ad[/tex]

The car is initially traveling at 15.0 meters per second and accelerates to 20.0 meters per second over a distance of 40.0 meters. Therefore,

[tex]v_f[/tex]= 20.0 m/s[tex]v_i[/tex]= 15.0 m/s d= 40.0 m

Substitute the values into the formula.

[tex](20.0 \ m/s)^2= (15.0 \ m/s)^2 + 2 a (40.0 \ m)[/tex]

Solve the exponents.

(20.0 m/s)² = 20.0 m/s * 20.0 m/s = 400.0 m²/s² (15.0 m/s)² = 15.0 m/s * 15.0 m/s = 225.0 m²/s²

[tex]400.0 \ m^2/s^2 = 225.0 \ m^2/s^2 + 2 a(40.0 \ m)[/tex]

Subtract 225.0 m²/s² from both sides of the equation.

[tex]400.0 \ m^2/s^2 - 225.0 m^2/s^2 = 225.0 \ m^2/s^2 -225 \ m^2/s^2 +2a(40.0 \ m)[/tex]

[tex]400.0 \ m^2/s^2 - 225.0 m^2/s^2 = 2a(40.0 \ m)[/tex]

[tex]175 \ m^2/s^2 = 2a(40.0 \ m)[/tex]

Multiply on the right side of the equation.

[tex]175 \ m^2/s^2 =80.0 \ m *a[/tex]

Divide both sides by 80.0 meters to isolate the variable a.

[tex]\frac {175 \ m^2/s^2}{80.0 \ m}= \frac{80.0 \ m *a}{80.0 \ m}[/tex]

[tex]\frac {175 \ m^2/s^2}{80.0 \ m}=a[/tex]

[tex]2.1875 \ m/s^2 =a[/tex]

Round to the tenths place. The 8 in the hundredth place tells us to round the 1 up to a 2.

[tex]2.2 \ m/s^2=a[/tex]

The magnitude of the car's acceleration is 2.2 meters per second squared.

Two cars drive on a straight highway. At time t = 0 , car 1 passes ad marker 0 traveling due east with a speed of 20.0 m/s. At the same time, car 2 is 1.0 km east of road marker 0 traveling at 30.0 m/s due west. Car 1 is speeding up, with an acceleration of 2.5 m/s², and car 2 is slowing down, with an acceleration of -3.2 m/s².

(a) Write position-time equations for both cars. Let east be the positive direction.

(b) At what time do the two cars meet?​

Answers

Answer: (a) Write position-time equations for both cars. Let east be the positive direction.

Explanation:

Hope this helps!

An airplane travels 250 km due east, then turns and travels 180 km [E 60°S]. Determine the resulting travel displacement for this plane.​

Answers

Answer:

388.56km[E13degS]

Explanation:

Δdx = Δd1x + Δd2x

Δdx = 250km [E]+ 180sin(60) [E]

Δdx = 405.88 km [E]

Δdy = Δd1y + Δd2y

Δdy = 0 + 160cos(60) [S]

Δdy = 90km [S]

Δdt = √(Δdx)^2+(Δdy)^2

Δdt = √164738.57+8100

Δdt = 398.85km

tanθ = Δdy/Δdx

tanθ = 90/388.56

θ = tan-1(90/388.56)

θ = 13.04 deg

How far will a bus travel if it averages a speed of 65 km/h for 7 hours?

Answers

Answer:

the bus travles 65x7=455 km

Explanation:

hope this helps you

Answer:

455 km/h

Explanation:

65km/h x 7 hours = 455 km/h

Una varilla de 5m de longitud y 1.5 cm^2 de sección transversal se alarga 0.10 cm al someterla a una tensión de 700 N. Determinar el Módulo de Young de la varilla

Answers

bzzhbzaijsb she’s aha ha ha s she’s has shah a sus abuse she d dvhbx

What is the tension in the string?
(I’m really having a hard time understanding this topic, and my professor isn’t helping much, so any extra explanation are greatly appreciated). The angles are 54.9205°.

Answers

If you look at the sketch I drew for the earlier part of this question, you'll see that, with respect to the positive horizontal (i.e. directly to the right), T₁ makes an angle of 180° - 54.9205° ≈ 125°, while T₂ makes an angle of 54.9205° ≈ 55°.

Split up the force acting on the block into vertical and horizontal components. We have

• net vertical force

F = T₁ sin(125°) + T₂ sin(55°) - mg = 0

where m = 0.56 kg, and

• net horizontal force

F = T₁ cos(125°) + T₂ cos(55°) = 0

Both net forces are 0 because the block is suspended in equilibrium.

Notice that cos(125°) = -cos(55°), so the second equation tells you that T₁ = T₂ and that the tensions in either string are the same. Also, sin(125°) = sin(55°).

Then in the first equation, we have

T₁ sin(125°) + T₂ sin(55°) - mg = 0

2 T₁ sin(55°) = mg

T₁ = mg/(2 sin(55°))

T₁ = (0.56 kg) (9.8 m/s²)/(2 sin(55°))

T₁ ≈ 3.35 N

A small bar of pure gold whose density is 19.3g/cm. Displaces 80 cm

Answers

Answer:

The mass of the gold bar is 1,544 g

Explanation:

Flag An alpha particle (consisting of two protons and two neutrons) is moving in a circle at a constant speed, perpendicular to a uniform magnetic field applied by some current-carrying coild. The alpha particle makes an clockwise revolution every 88 ns. If the speed is small compared to the speed of light, what is the magnitude B of the magnetic field made by the coils

Answers

Answer:

m = 4 At Mass units = 4 * 1.66E-27 kg = 6.64E-27 kg  mass of particle

q = 2 * 1.60E-19 C = 3.2E-19 C     total charge of alpha particle

T = 8.8E-8 s     period of 1 revolution

F = q V B      magnetic force

F = m V^2/ R     centripetal acceleration

q V B = m V^2 / R

B = m V / (q * R)

T = 2 pi R / V     time for 1 revolution

R = V T / (2 pi)

B = (m V / q) * 2 pi / (V T) = 2 pi m / (q T)

B = 2 pi * 6.64E-27 / (3.2E-19 * 8.8E-8) = 1.48

B = 1.48 W/m^2

Which group of the periodic table consists of elements that share similar
properties and have a single electron in their outer shells?
A. 18
B. 2
C. 13
D. 1

Answers

Answer:

c 13

Explanation:

I took test

Group 18 of the periodic table, also known as the noble gases or Group 18 elements, consists of elements that share similar properties and have a single electron in their outer shells. The correct option is option (A).

These elements include helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). They are characterized by having full outer electron shells, making them stable and unreactive under normal conditions.

This stability is due to the fact that they have achieved a full complement of electrons in their outermost energy level, except for helium, which has only two electrons in total.

Group 18 of the periodic table, also known as the noble gases or Group 18 elements, The correct option is option (A).

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How can people help stop erosion on beach

Answers

Answer:

since erosion is unavoidable the problem becomes discovering ways to prevent it. present beach erosion prevention methods include sand bags,vegetation,seawalls,sand dunes,and sand fences.

What is the rotational speed of the hour hand on a clock?

Answers

Answer:

π/21600 rad /s.

A racket causes a tennis ball to rebound at a speed of 25 m/s The tennis ball is 0.4 kg and is initially moving at a speed of 12 m/s. A high-speed movie film determines that the racket and ball are in contact for 0.05 s. What is the average net force exerted on the ball by the racket

Answers

Hi there!

We know that:

I = Δp = mΔv = m(vf - vi)

Let the positive direction be towards the wall. Plug in the given values:

Δp = 0.4(-25 - 12) = -14.8 Ns

Δp = Ft, so:

Δp/t = F

-14.8/0.05 = -296 N (depending on which direction you assign positive/negative, the answer would be positive/negative)

A student is planning an experiment to find
out how the height from which he drops a ball
affects how high the ball bounces. What is the control variable?
a) The ball used
b) The height the ball bounces
c) Height of the ball being dropped

Answers

Answer:

b) the height the ball bounces

Explanation:

the control variable is the variable that you change yourself. since you change the height that the ball bounces from we know this is the answer

The height of the bar being dropped is there an answer. Because It’s a variable that is not of interest to the studies in but it can be controlled and can influence the outcome of the experiment.

an object down, but this is not true. If you place a box of mass 8 kg on a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed of the belt, which is 5 m/s. The coefficient of kinetic friction between box and belt is 0.6. (a) How much time does it take for the box to reach this final speed

Answers

Answer:

Explanation:

ASSUMING the belt is horizontal

kinetic friction force is μmg = 0.6(8)(9.8) = 47.04 N

Horizontal acceleration is

a = F/m = 47.04 / 8 = 5.88 m/s²

t = v/a = 5.0 / 5.88 = 0.85034...

t = 0.85 s

a. of water a. of bread a. of soap a. of juice a. of packet a. of sand option_piece packet grain bottle cake slice bale tank pane note tube bar can sack pile sheaf​

Answers

Answer:

is it a poem or recipe :(

don't know

21)
uses a hypothesis to describe the relationship between dependent and independent variables and
results in advances in scientific knowledge.
A)
Engineering method
1
B)
Architectural design
Eliminate
C)
Technological design
D
Scientific investigation

Answers

im not sure if im right or not but maybe its "D scientific investigation" but ldk so

a boat travel due East with at 40metre per second across a river flowing due South 30metre per Second. what is the resultant speed of the boat​

Answers

Answer:

Explanation:

Use vector addition

v = √(30² + 40²) = 50 m/s as viewed from the shore

Trust me when I say I would never venture onto a river moving at 30 m/s no matter how powerful the boat was.

30 m/s(3600 s/hr / 1000 m/km) = 108 km/hr

The turbulence in water flowing over ground at 108 kilometers per hour would capsize anything ever built...including aircraft carriers.

These numbers are more suitable for a plane flying in air. Even that could be a very rough ride.

A _ system is one in which one subsystem provides services to another subsystem.

Answers

A large system is one in which one subsystem provides services to another subsystem.

What is a large system?

This is a type of system that contains many subsystems. The subsystem provides independent services to the larger by communicating, or interacting with each other.

What is a subsystem?

A subsystem is smaller units of a large system, designed to perform specific task in the large system.

Thus, a large system is one in which one subsystem provides services to another subsystem.

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One strategy in a snowball fight is to throw
a snowball at a high angle over level ground.
While your opponent is watching this first
snowball, you throw a second snowball at a
low angle and time it to arrive at the same
time as the first.
Assume both snowballs are thrown with
the same initial speed 15.1 m/s. The first
snowball is thrown at an angle of 69◦
above
the horizontal. At what angle should you
throw the second snowball to make it hit the
same point as the first? Note the starting and
ending heights are the same. The acceleration
of gravity is 9.8 m/s^2

How many seconds after the first snowball
should you throw the second so that they
arrive on target at the same time?
Answer in units of s.

Answers

Answer:

Range formula:  R = v^2 sin (2 theta) / g

If theta = 69 deg and v = 15.1

R = 15.1^2 sin 138 / 9,8 = 15.6 m

sin 138 = .669 = sin 42

So a snowball thrown at 21 deg will travel

R = 15.1 * .669^2 / 9.8 = 15.6 m

The second snowball can be thrown at 21 deg to travel the same distance

Vx = V cos theta = 15.1 * cos 69 = 5.41     first snowball

t1 = 15.6 / 5.41 = 2.88 sec

Vx = V cos theta = 15.1 cos 21 = 14.1 m/s

t2 = 15.6 / 14.1 = 1.11 sec

Difference = t1 - t2 = 1.77 sec     time delay for second snowball

is everyone lowkey turning goth?

Answers

Answer:

Yeah it's the new trend lol it's kinda weird

Answer:

it is marked by mystery and ambiguity. If to this fascination for creating tension is added a feeling of fear

Explanation:

do I help you

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