A 1500 kg car is accelerated by a force of 2000 N for 15 s.

What is the acceleration of the car?


If the car started at rest, how fast is the car going after 15s? (1pts)


How far (∆x) would the car have traveled after the 15s? (1pts)

Answers

Answer 1

By Newton's second law, the car has acceleration a such that

2000 N = (1500 kg) a

a = (2000 N) / (1500 kg)

a = 4/3 m/s² ≈ 1.33 m/s²

After 15 s, the car will attain a speed of

(4/3 m/s²) (15 s) = 20 m/s

It will have traveled a distance ∆x such that

(20 m/s)² - 0² = 2 (4/3 m/s²) ∆x

x = (20 m/s)² / (8/3 m/s²)

x = 150 m


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heat food, warm water

Why was basketball invented?

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people were probably bored so they threw rocks into hoops

6th grade science I mark as brainliest.​

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Answer:

Solution:-

Distance =400m

Time=20s

We need to find speed

As we know that

[tex]{\boxed{\sf speed \dfrac {Distance {}_{(d)}}{Time {}_{(t)}}}}[/tex]

Substitute the values

[tex]\LARGE\leadsto\sf speed=\dfrac {400}{20}[/tex]

[tex]\LARGE\leadsto\sf 20m/s[/tex]

6th grade science I mark as brainliest.​

Answers

Answer:

2m 13[tex]\frac{1}{3}[/tex]s

Explanation:

1.5m = 1s

200m = [tex]\frac{200}{1.5}[/tex] × 1s

          = 133[tex]\frac{1}{3}[/tex]s

          = 2m 13[tex]\frac{1}{3}[/tex]s

Consider an elevator carrying Kermit the frog weighing 4000.0 N is held 5.00 m above a spring with a force constant of 8000.0 N/m. The elevator falls onto the spring while subject to a frictional force (brake) of 1000.0 N. Determine the maximum compression distance, x, of the spring.

Answers

Answer:

Maximum compression distance (x) = 2.236 m (Approx)

Explanation:

Given:

Weight of frog = 4,000 N

Height = 5 m

Constant force = 8,000 N/m

Frictional force = 1,000 N

Find:

Maximum compression distance (x)

Computation:

Using Law of conservation;

mgh = 1/2(k)(x)²

4,000(5) = 1/2(8,000)(x)²

Maximum compression distance (x) = 2.236 m (Approx)

A chemical bond resulting from the exchange of electrons between two atoms is a(n) ________________ bond.

a
covalent
b
ionic
c
Lewis
d
metallic

Answers

Answer:

a covalent

Explanation:

ionic bonding involves only the transfer of electrons but not exchange of electrons

A 0.050 kg ball starts from rest at some unknown height on a toy roller coaster.
At a later time, it travels through the top of a loop at 2 m/s and a height of 0.40 m.
Since this track is frictionless, what was the starting height of the ball?

Answers

Answer:

The starting height of the ball is approximately 0.604 m

Explanation:

The given parameters are;

The mass of the the ball = 0.050

The speed with which it travels through the top loop = 2 m/s

The given height at which the ball moves at 2 m/s = 0.40 m

Therefore, we have;

1/2·m·v² = m·g·h

1/2·v² = g·h

h = 1/2·v²/g = 1/2 × 2²/9.81 ≈ 0.204

The additional height = h = 0.204 m

Therefore;

The starting height of the ball  ≈ The given height at which the ball moves at 2 m/s + h

The starting height of the ball  ≈ 0.40 + 0.204 = 0.604 m

The starting height of the ball ≈ 0.604 m.

When any object in rest, then potential energy is present. But when object is in motion then object have kinetic energy.

Starting height of the ball is 0.604 m.

We know that, when any object is start from rest, then potential energy is converted into kinetic energy.

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

Where m is mass of object, g is gravitational acceleration , h is height and v is velocity of object. (value of g = 9.81 m/ second square)

from above equation,

we get,     extra height        [tex]h=\frac{v^{2} }{2g} \\\\h=\frac{4}{2*9.81}\\\\h=0.204[/tex] meter

The starting height of the ball will be sum of the height at which ball moves 2 m/s and extra height.

Starting height = 0.40 + 0.204 = 0.604 meter.

Learn more:

https://brainly.com/question/18963960

Which of the following charts correctly compares plant and animal cells?

Answers

Answer:

Wheres the charts??

Explanation:

I need the charts so I can help

A parallel-plate capacitor is constructed using adielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00X108V/m. The desired capacitance is 0.250 μF, and the capacitor must withstand a maximum potential difference of 4.00 kV. Find the minimum area of the capacitor plates.

Answers

Answer:

A = 0.188 m²

Explanation:

First we find the distance between the plates by using the formula of electric field intensity:

E = ΔV/d

d = ΔV/E

where,

d = distance between plates = ?

ΔV = Potential Difference = 4 KV = 4000 V

E = Electric Field = 2 x 10⁸ V/m

Therefore,

d = 4000 V/(2 x 10⁸ V/m)

d = 2 x 10⁻⁵ m

Now, we find the Area of Plates by using formula of capacitance:

C = A∈₀∈r/d

where,

C = Capacitance = 0.25 μF = 0.25 x 10⁻⁶ F

A = Area of Plates = ?

∈₀ = Permittivity of free space = 8.85 x 10⁻¹² C/N.m²

∈r = Dielectric Constant = 3

Therefore,

0.25 x 10⁻⁶ F = A(8.85 x 10⁻¹² C/N.m²)(3)/(2 x 10⁻⁵ m)

A = (0.25 x 10⁻⁶ F)(2 x 10⁻⁵ m)/(3)(8.85 x 10⁻¹² C/N.m²)

A = 0.188 m²

PLEASE HELP ME IM TIMED

Answers

At a boundary where plates slide past each other

Colette launches an air rocket in the upward, positive direction. It launches
with an initial velocity of 25.5 m/s. It accelerates in the downward, negative
direction at a rate of 9.81 m/s2. After 3.5 seconds, what is the magnitude of
the rocket's displacement?

A) 29 meters

B) 31 meters

C) –150 meters

D) 150 meters

Answers

Answer:

b

Explanation:

Answer:

The answer is A) 29 meters

Explanation:

I got this question right on the test! :)

what is the mass and volume of 1000kg/m3 of water?​

Answers

Answer: The mass would be 1000m3 and the volume would be 1000kg

Explanation:

How do u know that liquid exerts pressure ?​

Answers

Answer:

All fluids exert pressure like the air inside a tire. The particles of fluids are constantly moving in all directions at random. As the particles move, they keep bumping into each other and into anything else in their path. These collisions cause pressure, and the pressure is exerted equally in all directions.

Explanation:

Hope this helps!! Please consider marking brainliest! Have a good one!!

A car accelerates from rest at a constant acceleration of 25.0 m/s^2. At some point, it then turns off its engine, letting the car decelerate slowly from the force of friction at a constant deceleration of 3 m/s^2 until it is at rest again. The total speed the car moves in this time is 200 meters. What is the minimum time needed for the car to move 200 meters given that it both starts and ends at rest?

Answers

Answer:

t = 9.14 s

Explanation:

We first analyze the accelerating motion by applying first equation of motion:

Vf₁ = Vi₁ + a₁t₁

where,

Vf₁ = Final Speed of Car before turning off engine

Vi₁ = Initial Speed of Car = 0 m/s

a₁ = acceleration of car = 25 m/s²

t₁ = time taken in accelerating motion

Therefore,

Vf₁ = 25t₁   ---------- equation (1)

Now, we apply second equation of motion:

s₁ = Vi₁ t₁ + (1/2)a₁t₁²

where,

s₁ = distance covered during accelerating motion

Therefore,

s₁ = (0)t₁ + (1/2)(25)t₁²

s₁ = 12.5 t₁²   ----------- equation (2)

Now, we analyze the decelerating motion by applying first equation of motion:

Vf₂ = Vi₂ + a₂t₂

where,

Vf₂ = Final Speed of Car = 0 m/s

Vi₂ = Initial Speed of Car after turning off engine

a₂ = deceleration of car = - 3 m/s²

t₂ = time taken in decelerating motion

Therefore,

Vi₂ = 3t₂   ---------- equation (3)

Now, we apply second equation of motion:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₂ = distance covered during decelerating motion

Therefore,

s₂ = (Vi₂)t₂ + (1/2)(-3)t₂²

s₂ = Vi₂ t₂ - 1.5 t₂²  

using equation (3):

s₂ = 3 t₂² - 1.5 t₂²

s₂ = 1.5 t₂²   ------------ equation (4)

Now, we know that the Final Velocity of accelerating motion (Vf₁) is equal to the initial velocity of decelerating motion (Vi₂):

Vf₁ = Vi₂

using equation (1) and equation (3):

25 t₁ = 3 t₂

t₁ = 0.12 t₂   ------------ equation (5)

Also, we know that sum of the distances is 200 m:

s₁ + s₂ = 200

using equation (2) and equation (4):

12.5 t₁² + 1.5 t₂² = 200

using equation (5):

12.5 (0.12 t₂²) + 1.5 t₂² = 200

3 t₂² = 200

t₂² = 200/3

t₂ = 8.16 s

substitute this in equation (5):

t₁ = 0.12(8.16 s)

t₁ = 0.97 s

Hence, the minimum time required for this motion is:

t = t₁ + t₂ = 0.97 s + 8.16 s

t = 9.14 s

Which of the following equations accurately defines acceleration?

Plz help ASAP

Answers

Answer:

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.

Explanation:

Answer: i think its b

sorry if im wrong

Explanation:

A car decelerates uniformly from 54 m/s to 38 m/s in 8.0 seconds. How far does it travel dunng this time?

Answers

Answer:

368m

Explanation:

dx=vt

54+38/2=46

46*8=368m

When a 0.622 kg basketball hits the floor, its velocity changes from 4.23 m/s down to 3.85 m/s up. If the average force was 72.9 N, how much time was it in contact with the floor?
(Unit = s)
Remember: up is +, down is -

Answers

Answer:

t = 0.0689 s

Explanation:

Given that,

Mass of a basketball, m = 0.622 kg

Initial velocity, u = 4.23 m/s (downward or negative)

Final velocity, v = 3.85 m/s (up of positive)

Average force, F = 72.9 N

We need to find the time it was in contact with the floor. The force is given by :

[tex]F=ma\\\\F=m\dfrac{v-u}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-(-4.23))}{72.9}\\\\t=0.0689\ s[/tex]

So, the time of contact is 0.0689 s.

the specific heat of gold is 0.031 calories degrees Celsius and the specific heat of silver is 0.057 calories degrees Celsius so if equal. amounts of each metal are exposed to equal heating which will heat up faster​

Answers

Answer:

Explanation:

Given:

Specific heat of gold  = 0.031cal/°C

Specific heat of silver  = 0.057cal/°C

To know the metals that will heat up faster, we must understand the meaning of specific heat capacity.

It is the amount of heat required to raise the temperature of 1g of a substance by 1°C.

Now,

The higher the specific heat capacity the more energy it is required to heat up the substance.

So, Gold with a specific heat capacity of 0.031cal/°C  will heat up faster.

What is the mass of a rock that is accelerated to 1.4m/s^2 with a force of 350 N

Answers

F=m*a

350N=m*1.4m/s^2

m=350N/1.4m/s^2=250kg

Whet net force is required to accelerate a car at a rate of 10 m/s2 if the car
has a mass of 5,000 kg?

Answers

Answer:

[tex]\boxed {\boxed {\sf 50,000 \ Newtons }}[/tex]

Explanation:

Force can be found by multiplying the mass by the acceleration.

[tex]F=m*a[/tex]

The mass of the car is 5,000 kilograms and it's acceleration is 10 meters per square second.

[tex]m= 5,000 \ kg \\a= 10 \ m/s^2[/tex]

Substitute the values into the formula.

[tex]F= 5,000 \ kg * 10 \ m/s^2[/tex]

Multiply.

[tex]F= 50,000 \ kg*m/s^2[/tex]

1 kilogram meter per square second is equal to 1 Newton. So, our answer of 50,000 kg*m/s² is equal to 50,000 Newtons

[tex]F= 50,000 \ N[/tex]

A net force of 50,000 Newtons is required to accelerate a 5,000 kilogram car at 10 meters per square second.

a person covers equal half distance at speed V and remaining half at speed V1 and V2 in equal interval of time .find average speed.

Answers

Answer:

Average speed = ( 2V + V1 + V2)/4

Explanation:

Given that a person covers equal half distance at speed V and remaining half at speed V1 and V2 in equal interval of time .find average speed.

Since the distance is covered at equal intervals of time, and

Speed = distance/time

For the first half distance,

V = distance/t

Cross multiply

Distance = Vt

For the second half distance

(V1 + V2)/2 = distance/t

Distance = t(V1 + V2)/2

The average speed = total distance/ total time.

Average speed = [Vt + t( V1 + V2)/2] ÷ 2t

Average speed = (2Vt + V1t + V2t)/4t

Average speed = t( 2V + V1 + V2)/4t

Time t will cancel out

Average speed = ( 2V + V1 + V2)/4

State Newton's three laws of motion in your own words and give an example for each one that helps to explain it in everyday life.

Answers

Answer:

The first law is an object won’t change its place unless a force hits it

The second law is the force on an object equivalent to the mass times its acceleration

The third law is when two object meet, they both apply forces on each other making them go in opposite directions

Explanation:

A fish is 4.7 cm from the front surface of a fish bowl of radius 21 cm. Where does the fish appear to be to someone in air viewing it from in front of the bowl? Do not forget the proper sign. (Give your answer in cm.)


______ cm


Where does the fish appear to be when it is 38.9 cm from the front surface of the bowl? (Give your answer in cm.)


______ cm

Answers

Answer:

Explanation:

From the information given:

We can properly determine the distance where the fish appear in the air viewing it from in front of the bowl by using the formula:

[tex]\dfrac{n_i}{d_o}+\dfrac{n_2}{d_1}= \dfrac{n_2-n_1}{r}[/tex]

where;

[tex]n_1[/tex] = refractive index in the air;  = 1.33  &

[tex]n_2[/tex] = refractive index in water. = 1

[tex]\dfrac{n_2}{d_i}= \dfrac{n_2-n_1}{r}-\dfrac{n_1}{d_o}[/tex]

[tex]\dfrac{1}{d_i}= \dfrac{1-1.33}{-21 \ cm}-\dfrac{1.33}{4.7\ cm}[/tex]

[tex]\dfrac{1}{d_i}= - 0.26726 \ cm[/tex]

[tex]d_i =\dfrac{1}{ - 0.26726 \ cm}[/tex]

[tex]\mathbf{d_i }[/tex] = - 3.74 cm

2)

To determine where the fish appear to be when it is  38.9 cm from the front surface of the bowl by using the formula:

[tex]\dfrac{n_2}{d_i}= \dfrac{n_2-n_1}{r}-\dfrac{n_1}{d_o}[/tex]

[tex]\dfrac{1}{d_i}= \dfrac{1-1.33}{-21 \ cm}-\dfrac{1.33}{38.9\ cm}[/tex]

[tex]\dfrac{1}{d_i}=- 0.0184759 \ cm[/tex]

[tex]d_i = \dfrac{1}{- 0.0184759 \ cm}[/tex]

[tex]\mathbf{d_i = }[/tex] -54.12  cm

A car travels 3500 m in 200 seconds what is the car speed

Answers

Answer:

17.5 m/s

Explanation:

We can calculate the meters per second by dividing the distance by time. 3500 divided by 200 is 17.5, therefore the speed is 17.5 meters per second.

Calculate the average speed of a runner who runs to for 500 meters in 40 second

Answers

Answer:

12.5

Explanation:

(8th grade HELP)
1. Inertia causes a stationary object to...
a.stay still
b.move
c.have an increased velocity
d.change it’s speed or direction
2. Once an applied force causes an object to start moving, the object keeps moving because....
a.none of the above
b.the force continues to be applied to it
c.no other force is acting on it
d.it has inertia

Answers

What is the question your asking

plz help me with question ;-;

Answers

Answer:

C

Explanation:

Question 12 of 20
How does decreasing the length of a wire affect a circuit?
A. It reduces the resistance caused by the wire.
B. It reduces the voltage carried by the wire.
O C. It increases the resistance caused by the wire.
D. It increases the current passing through the wire.

Answers

If we decrease the length of the wire in a circuit,  It reduces the resistance.

What is resistance?

Resistance is the opposition offered to the flow of current by a circuit element. The factors that affect resistance include;

Length of the wireCross sectional area of the wire

Hence, if we decrease the length of the wire in a circuit,  It reduces the resistance caused by the wire since the resistance depends on the length of the wire.

Learn more about resistance: https://brainly.com/question/15067823

Answer: A

Explanation:

Select the correct answer.
Which two factors does the power of a machine depend on?
Α. work and distance
B. force and distance
C. work and time
D. time and distance

Answers

Answer: C. work and time

Explanation: hope this helps

The density of an object is dependent upon the object’s mass and ---

Answers

Answer:Volume

Explanation:

Density = mass/ Volume

Answer:

Volume

Explanation:

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