What is the acceleration of the the object during the first 4 seconds?

What Is The Acceleration Of The The Object During The First 4 Seconds?

Answers

Answer 1

Answer:

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

Explanation:

What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second


Related Questions

A container contains a number of marbles each having a mass 80 mg. If the total mass of the marbles in the container is 2 kg. find the numbers in the container​

Answers

Answer:

There are 25,000 marbles in the container

Explanation:

There is a certain number of marbles in a container.

We know each marble has a mass of 80 mg and we also know the total mass of the marbles in the container is 2 Kg.

Since the data is given in different units of mass, we convert them into one unit in common.

Let's convert 2 Kg to milligrams. There are 1,000 grams in a kilogram, and there are 1,000 milligrams in a gram, thus there are one million milligrams in one kilogram, that is:

1 Kg = 1,000,000 mg

And 2 Kg = 2,000,000 mg

The number of marbles can be found by dividing the total mass by the individual mass:

No. of marbles = 2,000,000 / 80 = 25,000

There are 25,000 marbles in the container

-I...................ok

Answers

Answer:

What?

Explanation:

Is this a real photo you took?

What is the name for family labeled #4 (Yellow)?
#3
#5
#2
#
341 sud-
lasa 1
17:55
Alkaline Earth Metals
Metalloids
Transition Metals
Alkali Metals

Answers

Answer:

transition metals im sorry if this was too late

a mass of 2.00 kg rest on a rough horizontal table. The coefficient of static friction between the block and the table is 0.60. The block is attached to a hanging mass by a string that goes over a smooth pulley,as shown in the diagram. Determine the largest mass that can hang in this way without forcing the block to slide.

Answers

Answer:

1.2 kg

__________________________________________________________

We are given:

Mass of the block = 2 kg

Coefficient of Static Friction = 0.6

__________________________________________________________

Friction Force on the Block:

Finding the Normal Force:

We know that the normal force will be equal and opposite to the weight of the 2 kg block

So, Normal Force = mg

replacing the variables with the given values

Normal Force = (2)(9.8)                    [Taking g = 9.8]

Normal Force = 19.6 N

Friction force on the Block:

We know that:

Coefficient of Static Friction =  Static Friction Force/Normal Force

replacing the variables

0.6 = Static Friction force / 19.6

Static Friction force = 0.6*19.6 N                 [Multiplying both sides by 19.6]

Static Friction force = 11.76 N

__________________________________________________________

Largest Mass that can Hang:

We know that the Static Friction force is 11.76 N, this means that a force of 11.76 N will be applied to keep the object at rest

So, if the weight of the second block is less than the static friction force, it will hang

Weight of the second block ≤ 11.76

We know that weight = mg

mg ≤ 11.76

m(9.8) ≤ 11.76                                                   [since g = 9.8]

m ≤ 1.2 kg                                                        [dividing both sides by 9.8]

From this, we can say that the maximum mass of the second block is 1.2 Kg

3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterwards, the 3.00 kg block moves 1.71 m/s left. find the velocity of the 2.22 kg block afterwards

Answers

Momentum is conserved, so the total momentum before collision is equal to the total momentum after collision. Take the right direction to be positive. Then

(3.00 kg) (2.09 m/s) + (2.22 kg) (-3.92 m/s) = (3.00 kg) (-1.71 m/s) + (2.22 kg) v

where v is the velocity of the 2.22 kg block after collision. Solve for v :

6.27 kg•m/s - 8.70 kg•m/s = -5.13 kg•m/s + (2.22 kg) v

(2.22 kg) v = 2.70 kg•m/s

v = (2.70 kg•m/s) / (2.22 kg)

v ≈ 1.22 m/s

i.e. a velocity of about 1.22 m/s to the right.

A. A piece of paper near a magnet
B. An aluminum nail near a magnet
C. An iron nail, not near a magnet
D. An iron nail near a magnet

Answers

Answer:

it’s c not d

Explanation:

took the test

Answer: D!!!

Explanation: jus got it wrong from the other answer.

Carter's favorite ride at playland amusement park is the rollercoaster. The roller coaster
car and passengers have a combined mass of 1620kg and they descend the first hill at
an angle of 45.0 degrees to the horizontal. With what force is the rollercoaster pulled
down the hill?

Answers

Answer:

F = 11226.02 N

Explanation:

The roller coaster  car and passengers have a combined mass of 1620kg.

They descend the first hill at  an angle of 45.0 degrees to the horizontal.

We need to find force is the rollercoaster pulled  down the hill.

We firstly find the rectangular component of the downward. The force acting in the downward direction is mgsinθ such that,

F = mgsinθ

= 1620 × 9.8 × sin(45)

= 11226.02 N

So, the roller coaster is pulled down the hill with a force of 11226.02 N.

convert 100 Newton into dyne​

Answers

Answer:10000000

Explanation:

It would actually be 10 million dyne

3) A 10kg object rests on a frictionless surface when it is struck by a 300N force. At what rate will it accelerate?

3m/s/s
30m/s/s
0.3m/s/s
300m/s/s

Answers

Answer: 0.3m/s/s

(i'm really sorry if i'm wrong)

:(

A car goes around a circular track at 30 m/s. If the radius of the curve is 90 m, what is the period of the car's revolution around the track?

Answers

Answer:

18.9s

Explanation:

Using the formula;

ω = v/r

Where;

ω = angular velocity (rad/s)

v = linear velocity (m/s)

r = radius of the circular track (m)

According to the given information, v = 30m/s, r = 90m

ω = v/r

ω = 30/90

ω = 3/9

ω = 0.3333 radians/seconds.

Since ω = 2π/T

Where;

π = 3.142

T = period (s)

ω = angular velocity

0.333 = 2 × 3.142/T

T = 2 × 3.142/0.333

T = 6.284/0.333

T = 18.87s

T = 18.9s

If two exactly the same cars are driving down a road, which one would have the most kinetic energy. The one that is moving faster, the one that is moving downhill, the one that is moving uphill, or the one that is moving slower.

Answers

Answer: the car that is moving downhill

Explanation:

There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N. The bell has ____________
energy.

Answers

Answer:

The bell has a potential energy of 8550 [J]

Explanation:

Since the belt is 45 [m] above ground level, only potential energy is available. And this energy can be calculated by means of the following equation.

[tex]E_{p}= W*h\\E_{p} = 190*45\\E_{p}=8550[J][/tex]

A metal box of weigh 20 N rests on its 1 m by 0.6m side on floor. How
much is the pressure exerted by the metal box on the floor? Take g=
10 m/s

Answers

Answer:

P = 33.33 [Pa]

Explanation:

The pressure can be calculated by the relationship of the force over the area.

[tex]P =F/A[/tex]

where:

F = force = 20 [N]

A = area = 1 x 0.6 = 0.6 [m²]

Now replacing:

[tex]P=20/0.6\\P=33.33 [Pa][/tex]

Compute the specific heat capacity at constant volume of nitrogen (N2) gas. The molar mass of N2 is 28.0 g/mol.

Answers

Answer:

724.3J/Kg.K

Explanation:

CHECK THE COMPLETE QUESTION BELOW

Compute the specific heat capacity at constant volume of nitrogen (N2) gas.and compare with specific heat of liquid water. The molar mass of N2 is 28.0 g/mol.

The specific heat capacity can be computed by using expression below

c= CV/M

Where c= specific heat capacity

M= molar mass

CV= molar hear capacity

Nitrogen is a diatomic element, the Cv can be related to gas constant with 5/2R

Where R= 8.314J/mol.k

Molar mass= 28 ×10^-3Kg/mol

If we substitute to the expression, we have

c= (5R/2)/(M)

=5R/2 × 1/M

=(5×8.314) /(2×28 ×10^-3)

=724.3J/Kg.K

Hence, the specific heat capacity at constant volume of nitrogen (N2) gas is

724.3J/Kg.K

The specific heat of liquid water is about 4182 J/(K kg) which is among substance with high specific heat, therefore specific heat of Nitrogen gas is 724.3J/Kg.K which is low compare to that of liquid water.

a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop

Answers

Complete Question

a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop? Assume that the car does not skid and that each tire has a radius of 0.307 m. Answer in units of rev.

Answer:

The value is  [tex]N = 109 \ rev[/tex]      

Explanation:

From the question we are told that

    The speed of the car is  [tex]u = 28.4 \ m/s[/tex]

     The constant deceleration experienced is  [tex]a = 1.92 \ m/s^2[/tex]

      The radius of the tire is  [tex]r = 0.307 \ m[/tex]

     

Generally from kinematic equation we have that

      [tex]v^2 = u^2 + 2as[/tex]

Here  v is the final velocity which is  0 m/s

   So

         [tex]0^2 = 28.4^2 + 2 * 1.92 * s[/tex]

=>      [tex]s = 210.04 \ m[/tex]

Generally the circumference of the tire is mathematically represented as

         [tex]C = 2 \pi r[/tex]

=>      [tex]C = 2 * 3.142 * 0.307[/tex]    

=>      [tex]C = 1.929 \ m[/tex]

Generally the number of revolution is mathematically represented as

         [tex]N = \frac{ s}{C}[/tex]    

=>     [tex]N = \frac{210.04}{1.929}[/tex]

=>     [tex]N = 109 \ rev[/tex]      

d. If a dog has a mass of 12 kg, what is its weight on Neptune?
11.7N/kg

Answers

Answer:

133.8 N

Explanation:

Recall that the acceleration of gravity in Neptune is estimated as 11.15 m/s^2

Therefore, the weight of the dog on this planet would be:

Weight = mass x acceleration of gravity = 12 kg x 11.15 m/s^2 = 133.8 N

A 30 N force toward the west is applied to an object. The object moves 50 m east during the time the force is applied. What is the change in kinetic energy of the object?
a) 1.0 J
b) 750 J
c) 1.7 J
d) -1500 J

Answers

Answer:

D.-1500Joules

Explanation:

The change in kinetic energy of the object s equivalent to the workdone by the body in the west direction (negative x direction)

Workdone = Force * Distance

Given

Force = 30N

Distance moved by the object = 30m

Required

Kinetic energy

Kinetic energy = 30 * 50

Kinetic energy = 1500Joules

Since the body moves in the negative  direction, hence the kinetic energy will be -1500Joules

A vertical tube one meter long is open at the top. It is filled with 50 cm of water. If the velocity of sound is 344 m/s, what will the fundamental resonant frequency be (in Hz)?

Answers

Answer:

The fundamental resonance frequency is 172 Hz.

Explanation:

Given;

velocity of sound, v = 344 m/s

total length of tube, Lt = 1 m = 100 cm

height of water, hw = 50 cm

length of air column, L = Lt - hw = 100 cm - 50 cm = 50 cm

For a tube open at the top (closed pipe), the fundamental wavelength is given as;

Node to anti-node (N ---- A) : L = λ / 4

λ = 4L

λ = 4 (50 cm)

λ = 200 cm = 2 m

The fundamental resonance frequency is given by;

[tex]f_0 = \frac{v}{\lambda}\\\\f_0 = \frac{344}{2}\\\\f_0 = 172 \ Hz\\\\[/tex]

Therefore, the fundamental resonance frequency is 172 Hz.

A student throws a baseball upwards at an angle of 60 degrees to the horizontal. The initial
horizontal and vertical components are 12.5 m/s and 21.7 m/s, respectively.
60 degrees
Refer to the above information and diagram. What position will have the smallest magnitude
of vertical velocity?

Answers

Answer:

The initial horizontal and vertical components are 12.5 m/s and 21.7 m/s, respectively?

Explanation:

An unknown radioactive sample is observed to decrease in activity by a factor of two in a one hour period. What is its half-life?

Answers

Answer:

The half-life is [tex] t_{1/2} = 1.005 h[/tex]

Explanation:

Using the decay equation we have:

[tex]A=A_{0}e^{-\lambda t}[/tex]

Where:

λ is the decay constantA(0) the initial activityA is the activity at time t

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that [tex]A = \frac{A_{0}}{2}[/tex]

[tex]\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}[/tex]

[tex]0.5=e^{-\lambda*1 h}[/tex]

Taking the natural logarithm on each side we have:

[tex]ln(0.5)=-\lambda[/tex]

[tex]\lambda=0.69 h^{-1}[/tex]

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

[tex]\lambda = \frac{ln(2)}{t_{1/2}}[/tex]

[tex] t_{1/2} = \frac{ln(2)}{\lambda}[/tex]

[tex] t_{1/2} = \frac{ln(2)}{0.69}[/tex]

[tex] t_{1/2} = 1.005 h[/tex]

I hope it helps you!

what is the force of an egg that is thrown at a brick wall if the egg has a mass of 0.3 kg and an acceleration of 50 m/s/s

Answers

Answer:

15N

Explanation:

F=ma so F=.3*50 therefore F=15N

The force of an egg that is thrown at a brick wall is equal to 15 N.

What is force?

Force can be defined as the influence or effect that changes the state of the body of from motion to rest or vice versa. The S.I. unit of force is Newton (N) as well as force is a vector quantity. Force can change the direction or the speed of the moving object.

The force acting on an object can be calculated from the multiplication of the mass(m) and acceleration(a). The mathematical form of the second law of motion for force can be written as follows:

F = ma

Given, the mass of the egg, m = 0.3 Kg

The acceleration of the egg with which it is thrown on the wall, a = 50 m/s²

The force of an egg that is thrown at a brick wall can be calculated as:

F = ma = 50 ×0.3 = 15 N

Learn more about force, here:

brainly.com/question/13191643

#SPJ2

Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.311 rad/s2 for 4.13 s. What is the drill's angular displacement during that time interval?

Answers

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, [tex]\omega _i[/tex] = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

[tex]\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad[/tex]

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

n
Question 4
1 pts
A bus travels on an interstate highway at an average speed of 90 km/hrs. How far does it take to travel
in 30 mins? The distance equals speed times time, or d = st.
O 45 km
O 98 Km
O 56 km
O 432 Km

Answers

[tex]d = s \times t \\ d = 90 \times \frac{30}{60} \\ d = 90 \times \frac{1}{2 } \\ d = 45km[/tex]

Determine the acceleration that results when a 12 N net force is applied to a 3 kg object.

Answers

Heya!!

For calculate aceleration, let's applicate second law of Newton:

[tex]\boxed{F=ma}[/tex]

⇒ Being:

→ F = Force = 12 N

→ m = Mass = 3 kg

→ a = aceleration = ?

Lets replace according formula and leave the "a" alone:

[tex]12\ N = 3\ kg * \textbf{a}[/tex]

[tex]\textbf{a} = 12\ N / 3\ kg[/tex]

[tex]\textbf{a} = 4\ m/s^{2}[/tex]

Result:

The aceleration of the object is of 4 m/s²

(1-dimension) A fish has a mass of 6 kg and is moving at a speed of 4m/s to the right. What is its momentum?

Answers

Answer:

24 kg m/s

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 6 × 4

We have the final answer as

24 kg m/s

Hope this helps you

Allure of the seas is one of the most expensive cruise ships around the world with a length of 362 meters(1,187 ft) and a height of 72 meters(236 ft) above water line. On her first day of operation she moves with a uniform acceleration of 83.5 km/hr2 from rest has gone 10 nautical miles. How many seconds she is in motion? Note: 1 nautical mile = 1.852 km (help 3 mins left​

Answers

Answer:

  about 2398 seconds

Explanation:

The relation between time, distance, and acceleration is ...

  d = (1/2)at²

  t = √(2d/a) = √(2·10·1.852 km/(83.5 km/h²)) ≈ √0.4436 h ≈ 0.6660 h

That is about ...

  (0.6660 h)(3600 s/h) ≈ 2397.7 s

The cruise ship takes about 2397.7 seconds to cruise 10 nautical miles, accelerating all the way.

Describe the motion of an object as it accelerates. IN YOUR OWN WORD!! ASAP

Answers

Answer:

The aceleration of an object is in the direction of the net force. If you push or pull an object in a particular direction, it accelerates in that direction. The aceleration has a magnitude directly proportional to the magnitude of the net force.

Explanation:

Hope this helps Plz mark brainliest

in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s about its vertical axis. the radius of the disk is 1 meter. What is the magnitude of the friction?

Answers

Answer:

25

Explanation:

A hare can run at a rate of 15 m/s, while a turbocharged tortoise can now crawl at a rate of 3 m/s, how much of a head-start (time-wise) does the tortoise need in order to tie the hare in a 250 meter race?

A.
16.7 seconds

B.
66.7 seconds

C.
83.3 seconds

D.
100 seconds

Answers

Answer:

t = 66.7 s

Explanation:

Given that,

Speed of a hare, v = 15 m/s

Speed of a turbocharged tortoise, v' = 3 m/s

The hare in a 250 meter race

Let the Hare takes time t. It can be calculated as follows :

[tex]t=\dfrac{250}{15}=16.67\ s[/tex]

Let a turbocharged tortoise takes t'. It can be calulated as follows :

[tex]t'=\dfrac{250}{3}= 83.33\ s[/tex]

To tie the race, required time is given by :

[tex]\Delta t = t'-t\\\\=83.33-16.67\\\\=66.66\ s\\\\\approx 66.7\ s[/tex]

Hence, the correct option is (b) i.e. 66.7 seconds.

A rocket will move upward as long as which condition applies?

Answers

The force of thrust is greater than the force if gravity !
Answer found on quizlet !
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