a sports event, the car starts from rest, in 5.0 s its acceleration is 5.0 m/s2.
culate the distance travelled by car.

Answers

Answer 1

Answer:

62.5m

Explanation:

Given parameters:

Time  = 5s

Acceleration  = 5m/s²

Unknown:

Distance traveled by car  = ?

Solution:

To solve this problem, we use one of the motion equations as given below:

     S  = ut + [tex]\frac{1}{2}[/tex] at²

S is the distance traveled

t is the time taken

u is the initial velocity  = 0m/s

a is the acceleration

Now insert the parameters and solve;

      S = (0 x 5) + ([tex]\frac{1}{2}[/tex] x 5 x 5²) = 62.5m


Related Questions

pls help me w this, I've been doing this since 5 minutes ago​

Answers

Answer:

t=0.0625s

Explanation:

F=number of swings/time taken

DATA

Frequency=4.0Hz

number of swings from Q to R

=1/4

time taken=?

Frequency=number of swings/time taken

make t the subject of the formula

t=n/f

substitute the given date

t=0.25/4.0

t=0.0625s

option A is collect

why watchman does no work but gets tired ?​

Answers

Answer: the body is trained to wake up when the sun rises and get the energy from the sun and sleep and it is night on when the sun is not there. So when a watchman works at night there is no sun first of all to give him any kind of energy and secondly the body clock resists him waking up thus making him tired.

Explanation:

A 1500 kg car has an applied forward force of 5000 N and experiences an air resistance of 1250 N. What is the car's acceleration?

Answers

Answer:

[tex]2.33\ m/s^2[/tex]

Explanation:

Net Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

Fn = m.a

Where a is the acceleration of the object.  The net force is the sum of the individual vector forces applied to the object.

The m=1500 Kg car has two horizontal forces applied: the forward force of 5000 N that causes the movement and the air resistance force of 1250 N that opposes motion.

The net force is Fn = 5000 N - 1500 N = 3500 N

To find the acceleration, we solve the equation for a:

[tex]\displaystyle a=\frac{Fn}{m}[/tex]

[tex]\displaystyle a=\frac{3500}{1500}[/tex]

[tex]\boxed{a = 2.33\ m/s^2}[/tex]

The car's acceleration is [tex]a = 2.33\ m/s^2[/tex]

A merry go round exerts a force of 1000 N on a rider on the
outer ring of animals when it takes 15 seconds to make a
complete revolution. If the person weighs 750 N, the radius
of the circle he is making is m. Round your answer to
the nearest tenth.

Answers

Answer:

The radius of the circle made by the person on the merry go round is 74.55 meters

Explanation:

The given parameters are;

The force the merry go round exerts on the rider = 1000 N

The time it takes the merry go round to make one complete revolution = 15 seconds

The weight of the person = 750 N

The radius of the circle made by the person on the merry go round = r

We have;

[tex]F_c = \dfrac{m \cdot v^2}{r} = m \cdot \omega ^2 \cdot r[/tex]

Where;

m = The mass of the person

v = The velocity of the person

[tex]F_c[/tex] = The centrifugal force acting on the person = 1,000 N

r = The radius of the circle made by the person on the merry go round

ω = Angular velocity = 2·π/15 rad/s

We have;

The mass of the person = The weight/(The acceleration due to gravity, g)

∴ The mass of the person = 750/9.81 ≈ 76.45 kg

By substituting the calculated and known values into the equation for  the centripetal force, we have;

[tex]F_c[/tex] = m × ω² × r

1000 = 76.45 × (2·π/15)² × r

r = 1000/(76.45 × (2·π/15)²) = 74.55 m

The radius of the circle made by the person on the merry go round = r = 74.55 m.

An object has a force of 29.43 N acting on it. Its acceleration is 3.5 m/s2. What is the mass of this object?

Answers

Answer:

The mass of the object is 8.41 Kg.

Explanation:

Mechanical Force

The second Newton's law states that the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object. The SI unit for the force is the Newton: [tex]1\ Nw = 1~Kg.m/s^2[/tex]

We are given the net force of F=29.43 N acting on an object and producing an acceleration of a=3.5~m/s^2.

To calculate the mass of the object, we solve the above equation for m:

[tex]\displaystyle m=\frac{F}{a}[/tex]

[tex]\displaystyle m=\frac{29.43}{3.5}[/tex]

Calculating:

m = 8.41 Kg

The mass of the object is 8.41 Kg.

Alex wants to learn how to surf, but he is not a strong swimmer. He knows he needs to increase his ability to paddle out in order to catch the best waves. Which piece of advice would you give to help him get started on reaching his goal? He should swim at least three times a week at the community pool to build stamina. He should tread water every day to get more comfortable in the water. He should purchase the best surfboard he can afford because it will help him paddle faster. He should watch your friend who is a competitive surfer practice to learn her technique.

Answers

Answer:

He should swim at least three times a week at the community pool to build stamina.

Explanation:

question in the photo

Answers

I think (50 joules ) I’m not sure

A child pushes a 100 kg refrigerator with a force of 50 N, but the refrigerator does not move. Suppose the coefficient of static friction between the floor and the refrigerator is 0.4. What is the force due to friction in this scenario

Answers

Answer:

50 N

Explanation:

Since the refrigerator doesn’t move, that means the force of friction equals the amount of force the child exerts on the fridge. If the friction force were greater than the force by the child, the fridge would start accelerating towards the child. If it were less than the force the child exerted, the fridge would start accelerating away from the child. Therefore, the net force must be 0, in this case, the friction force is equal to the force the child exerted, for it to stay at rest (as Newton’s First Law stated).

I hope this helps! :)

The fall of a body on the earth surface cannot be a complete free fall ? why

Answers

It can never be in a total free fall due to air resistance and terminal velocity due to mass

A train travels with a speed of 115km/hr. How much time does it take to cover a distance of 470km.​

Answers

It would take approximately 4 hours for the train to cover a distance of 470km.

4 hours 5 minutes 13.04 seconds

WILL MARK BRAINLY!! QUICK HELP PLEASE!
A 40.0 kg rocket produces a
764 N upwards force ("thrust").
What is the net force acting on
the rocket?

Answers

Answer:                                                                              

372                              

Explanation:(40.0)×(9.8)=392

764N-392=372

Directions: Summarize the main ideas of this lesson by answering the question below.
How might a person acquire a fear of or aversion to something even though it has no negative effect
on him or her?

Answers

Answer:

Explanation:

Trypophobia is a fear or disgust of closely-packed holes. People who have it feel queasy when looking at surfaces that have small holes gathered close together.A person may start by imagining what they fear, then looking at pictures of the fear object, and then finally being near or even touching the source of their anxiety. In the case of trypophobia, a person with symptoms may start by simply closing his eyes and imagining something such as a honeycomb or seed poThey can help you find the root of the fear and manage your symptoms. Last medically reviewed on July 20, 2017 Medically reviewed by Timothy J. Legg, Ph.D., CRNP — Written by Annamarya Scaccia ...

The force of attraction between two oppositely charged pith is 5mx 10 to the -6th power newtons. If the charge on the two is 6.7 x 10 to the -9th power coulombs what is the distance between the two charges

Answers

Answer:

0.28 m

Explanation:

The following data were obtained from the question:

Force (F) = 5×10¯⁶ N

Charge 1 (q₁) = 6.7×10¯⁹ C

Charge 2 (q₂) = 6.7×10¯⁹ C

Electrical constant (K) = 9×10⁹ Nm²C¯²

Distance apart (r) =?

Thus, the distance between the two charges can be obtained as follow:

F = Kq₁q₂/r²

5×10¯⁶ = 9×10⁹ × 6.7×10¯⁹ × 6.7×10¯⁹/r²

5×10¯⁶ = 4.0401×10¯⁷ / r²

Cross multiply

5×10¯⁶ × r² = 4.0401×10¯⁷

Divide both side by 5×10¯⁶

r² = 4.0401×10¯⁷ / 5×10¯⁶

Take the square root of both side

r = √(4.0401×10¯⁷ / 5×10¯⁶)

r = 0.28 m

Therefore, the distance between the two charges is 0.28 m

Why do women typically tend to have slightly greater stability than men?
A.
Women are shorter than men, and shorter people are more stable.
B.
Men have more muscle mass in their lower bodies that makes them stiff and less stable.
C.
Women have lower centers of gravity, and lower centers of gravity provide more stability.
D.
The increased muscle mass in their upper bodies makes their centers of gravity difficult to find.



Please select the best answer from the choices provided.


A
B
C
D

Answers

Answer:

C

Explanation:

I figured it sounded more accurate

Which of the following is NOT one of the essential components of an exercise program?

Answers

Answer:

i dont know dude but ask someone who is in your grade lol

Explanation:

Two skaters A and B. having masses 50 kg and
70 kg respectively, stand facing each other 6 m
apart on a horizontal smooth surface. They pull
on a rope stretched between them. How far does
each move before they meet?
(A) both move 3 m
(B) A moves 2.5 m and B moves 3.5 m
(C) A moves 3.5 m and B moves 2.5 m
(D) both move 4 m
(E) none of the above

Answers

Answer:

Explanation:

Hell im good dude

Which of the following does not illustrate a complete circuit?

Answers

Answer:

I think A

Explanation:

because it dosn't have enough tools

Draw a picture showing the possible phases of the moon in this binary star
system, do they look the same or different from those we see in the sky in
reality? (You might want to think about scale and how far the suns would be
when considering your answer.)

(Help ASAP)

Answers

Is this what your talking about if not sorry

26. For this table of data, which column is the independent variable?
radius of circle (m) area of circle (m2)
3.4
36.317
4.5
63.617
4.6
66.476
6.2
120.763
7.6
181.458
7.7
186.265
8.6
232.352
9.7
295.592
area of circle
o
radius of circle
O
None of these is correct.

Answers

Answer:

radius of a circle is measured by an instrument, so it is the independent variable

Explanation:

The independent variable corresponds to the magnitude controlled by the experimenter and in general can be measured by an instrument.

The dependent variable is the variable used to perform the calculations, obtained from a mathematical transformation of the independent variable.

The radius of a circle is measured by an instrument, so it is the independent variable and the area is calculated by

         A = π r²

Therefore the area corresponds to the dependent variable

A 10 kg object experiences an acceleration of 2 m/s squared. What net force was applied to the object?

Answers

F = ma

= 10kg × 2m/s

= 20N

Answer : 20N

Explanation :

F = Force (N)

m = mass (kg)

a = acceleration (m/s)

Asha walks 15 m west, then 20 m north, then 15 m east. Calculate the distance
covered by Asha.(a) (Numbering problem)_​

Answers

Distance walked to the west= 15m

" " " " north= 20m

" " " " east = 15m

Total Distance= (15+ 20+ 15) m

= 50m

Asha covered 50m distance in total.

n Olympic diver is on a diving platform 3.60 m above the water. To start her dive, she runs off of the platform with a speed of 1.3 m/s in the horizontal direction. What is the diver's speed, in m/s, just before she enters the water

Answers

Answer:

8.5m/s

Explanation:

Using the equation of motion

v² = u² + 2gH

v is the final speed of the diver

u  is the initial speed of the diver

g is the acceleration due to gravity

H is the height of the object

Given the following

V = 1.3m/s

H = 3.60m

g = 9.8m/s²

Required

Initial speed of the diver u

Substitute the given values into the formula:

v² = u² -2gH

1.3² = u²  - 2(9.8)(3.60)

1.69 = u²-70.56

u² = 1.69+70.56

u² = 72.25

u = √72.25

u = 8.5m/s

Hence the diver's speed, in m/s, just before she enters the water is 8.5m/s

How long will it take an object traveling at 7 m/s to reach a distance of 26 meters

Answers

Answer:

3 hours 71 minutes

Explanation:

As because Speed= distance/time taken

so time taken= 3.71

A sound wave passes from air into water and then into steel. With each change in medium, the velocity of this wave will a(increase,then decrease b(decrease c(increase d( decrease,then increase ?

Answers

Answer:

a

Explanation:

because it can increase anything


When the net force on an object is zero, what do we know about the motion of that object

Answers

Answer:

The objects speed or motion does not change if the net force is 0.

Explanation: There basically doesn't have anything acting on it so it stays put.

Calculate the kinetic energy of a 50 kg cart moving at a speed of 7.1 m/s.

ill give u 100 points if you give me the right answer.

Answers

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[tex]\huge\boxed{\fcolorbox{black}{yellow}{HOPE\: IT \: HELPS}}[/tex]

HOPE IT HELPS

[tex]\begin{gathered}\begin{gathered}\\\huge\underline{ \mathbb\blue{Please}\:\mathbb\green{Mark}\:\mathbb\orange{As}}\:\end{gathered}\end{gathered}[/tex]

Please Mark As

[tex]\huge\underline{ \mathbb\red{Brilliant}\:\mathbb\purple{Answers}}[/tex]

Brilliant Answers

Un automóvil se mueve a velocidad constante v = 60 [km/h]. Si el tiempo de reacción del conductor al ver la luz roja de una intersección es de 0.5 [s], y ´este desacelera a razón de 40 [m/s2 ] ¿Qué distancia recorre el automóvil desde el instante en que el conductor nota la se˜nal hasta detenerse por completo? Entiéndase por tiempo de reacción, el intervalo de tiempo transcurrido entre el instante en que se percibe la señal y el instante en que se lleva a cabo la acción (instante en que el conductor activa el sistema de frenado).

Answers

Answer:

El automóvil recorre una distancia de 11.806 metros antes de deternerse por completo.

Explanation:

De acuerdo con el enunciado, el conductor nota la luz roja, empieza a decelerar 0.5 segundos después y decelera hasta detenerse. La distancia total recorrida por el automóvil desde el instante en que el conductor nota la luz roja ([tex]\Delta s_{T}[/tex]), medida en metros:

[tex]\Delta s_{T} = \Delta s_{1}+\Delta s_{2}[/tex] (1)

Donde:

[tex]\Delta s_{1}[/tex] - Distancia recorrida a velocidad constante, medida en metros.

[tex]\Delta s_{2}[/tex] - Distancia recorrida hasta alcanzar el reposo, medida en metros.

Si suponemos que la segunda etapa describe un movimiento uniformemente acelerado, entonces la distancia recorrida total que representada por la siguiente fórmula:

[tex]\Delta s_{T} = v_{o}\cdot \Delta t_{o} + \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}[/tex] (2)

Donde:

[tex]v_{o}[/tex] - Velocidad inicial del automóvil, medida en metros por segundo.

[tex]v_{f}[/tex] - Velocidad final del automóvil, medida en metros por segundo.

[tex]\Delta t_{o}[/tex] - Tiempo de reacción del conductor, medido en segundo.

[tex]a[/tex] - Aceleración, medida en metros por segundo al cuadrado.

Si conocemos que [tex]v_{o} = 16.667\,\frac{m}{s}[/tex], [tex]v_{f} = 0\,\frac{m}{s}[/tex], [tex]\Delta t_{o} = 0.5\,s[/tex] y [tex]a = -40\,\frac{m}{s^{2}}[/tex], entonces la distancia recorrida total es:

[tex]\Delta s_{T} = \left(16.667\,\frac{m}{s} \right)\cdot (0.5\,s)+\frac{\left(0\,\frac{m}{s} \right)^{2}-\left(16.667\,\frac{m}{s}\right)^{2}}{2\cdot \left(-40\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\Delta s_{T} = 11.806\,m[/tex]

El automóvil recorre una distancia de 11.806 metros antes de deternerse por completo.

An object is falling from a height of d and reaches a final velocity of vf. Calculate velocity when the object has fallen a distance d/2.

Answers

Answer:

The final velocity, [tex]v_{f}[/tex], is gd.

Explanation:

The condition here is a free falling object. Thus from the third equation of motion under free fall, we have;

[tex]v_{f}[/tex] = [tex]v_{i}[/tex] + 2gs

where [tex]v_{f}[/tex] is the final velocity of the object, [tex]v_{i}[/tex] is the initial velocity of the object, g is the gravitational force and s is the height.

Since the object falls from a height of d, then [tex]v_{i}[/tex] = 0 m/s, and s = d.

So that;

[tex]v_{f}[/tex] = 0 + 2gd

   = 2gd

[tex]v_{f}[/tex] = 2gd

When the distance is [tex]\frac{d}{2}[/tex], [tex]v_{i}[/tex] = 0 m/s.

Then;

[tex]v_{f}[/tex] = 2g[tex]\frac{d}{2}[/tex]

[tex]v_{f}[/tex] = g x d

When the object falls through the height [tex]\frac{d}{2}[/tex], then the final velocity is gd.

Explain why the total positive charge in every atom of an element is always the
same.

Answers

Answer:

When an atom has an equal number of electrons and protons, it has an equal number of negative electric charges (the electrons) and positive electric charges (the protons). The total electric charge of the atom is therefore zero and the atom is said to be neutral.

Explanation:

An atom has an equal quantity of positive and negative electric charges (the protons and electrons) when its electron and proton counts are equal (the protons). Since the atom has a total electric charge of zero, it is referred regarded as being neutral.

What is an atom?

The atom is the smallest unit of matter that may be divided without resulting in electrically charged particles. Additionally, it is the tiniest substance with features like chemical elements. The atom is therefore the basic building block of chemistry.

Examine different electron arrangements in the electron shells that surround an atom's nucleus.

Examine different electron arrangements in the electron shells that surround an atom's nucleus. The rest is made up of a cloud of negatively charged electrons surrounding a positively charged nucleus made up of protons and neutrons. Compared to electrons, which are the lightest charged particles in nature, the nucleus is small and dense.

To get more information about atom :

https://brainly.com/question/13654549

#SPJ2

Which frequency would be the third harmonic in a series for an open-pipe resonator if the fundamental is 440 Hz

Answers

Answer:

1320 Hz

Explanation:

The third harmonic is given as 3fo, where fo is the fundamental frequency.

The fundamental frequency is the lowest frequency that can occur in a pipe. In an open pipe, both even and odd harmonics occur which are multiples of the fundamental frequency fo. Hence the harmonics in an open pipe are; 2fo, 3fo,4fo..... etc.

For the third harmonic; 3fo = 3 (440 Hz) = 1320 Hz

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