At the base of a hill, a 90 kg cart drives at 13 m/s toward it then lifts off the accelerator pedal). If the cart just barely makes it to the top of this hill and stops, how high must the hill be?

At The Base Of A Hill, A 90 Kg Cart Drives At 13 M/s Toward It Then Lifts Off The Accelerator Pedal).

Answers

Answer 1

Answer:

8.45 m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 90 Kg

Initial velocity (u) = 13 m/s

Final velocity (v) = 0 m/s

Height (h) =?

NOTE: Acceleration due to gravity (g) = 10 m/s²

The height of the hill can be obtained as follow:

v² = u² – 2gh (since the cart is going against gravity)

0² = 13² – (2 × 10 × h)

0 = 169 – 20h

Rearrange

20h = 169

Divide both side by 20

h = 169/20

h = 8.45 m

Therefore, the height of the hill is 8.45 m


Related Questions

Force exerted on a body changes it's

Answers

Momentum is the answer

As a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 4.12 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W.

Answers

Answer:

102.5N

Explanation:

Given that  a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 4.12 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W

The power = 4.12 × 746 = 3073.52 W

Using the formula

Power = force × velocity

3073.52 = force × 30

Force = 3073.52 / 30

Force = 102.5 N

Since most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road, therefore,

the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s is 102.5 N

What is the weight of a 48kg rock?

Answers

Answer:

48kg

Explanation:

Two resistors have resistances R1 and R2. When the resistors are connected in series to a 12.6-V battery, the current from the battery is 2.07 A. When the resistors are connected in parallel to the battery, the total current from the battery is 8.98 A. Determine R1 and R2. (Enter your answers from smallest to largest.)

Answers

Answer:

When R1 = 2.193, R2 = 3.894

When R1 = 3.894, R2 = 2.193

Explanation:

We are told that when R1 and R2 are connected in series, the voltage is 12.6 V and the current is 2.07 A.

Formula for resistance is;

R = V/I

R = 12.6/2.07

R = 6.087 ohms

Since R1 and R2 are connected in series.

Thus; R1 + R2 = 6.087 ohms

R1 = 6.087 - R2

We are also told that when they are connected in parallel, the current is 8.98 A.

Thus, R = 12/8.98

R = 1.403 ohms

Thus;

(1/R1) + (1/R2) = 1/1.403

Let's put 6.087 - R2 for R1;

(1/(6.087 - R2)) + (1/R2) = 1/1.403

Multiply through by 1.403R2(6.087 - R2) to get;

1.403R2 + 1.403(6.087 - R2) = R2(6.087 - R2)

Expanding gives;

1.403R2 + 8.54 - 1.403R2 = 6.087R2 - (R2)²

(R2)² - 6.087R2 + 8.54 = 0

Using quadratic formula, we have;

R2 = 2.193 ohms or 3.894 ohms

Thus,

R1 = 6.087 - 2.193 or R1 = 6.087 - 3.894

R1 = 3.894 or 2.193

When R1 = 2.193, R2 = 3.894

When R1 = 3.894, R2 = 2.193

Solve the below problems being sure to provide the correct significant figures.

1) 1000 ÷ 4.886 = __________

2) 240 ÷ 12.3 = __________

3) 80 x 4.6 = __________

4) 4.527 x 30 = __________

5) 86 x 63.855 x 8000 = __________

6) 700 x 91.186 = __________

7) 7.1 x 348 = __________

8) 50 ÷ 29.1 = __________

9) 98.773 x 24.891 x 409 = __________

10) 0.065 x 3 x 3007 = __________

Answers

Answer:

1) 204.6663938

2) 19.51219512

3) 368

4) 135.81

5) 43932240

6) 63830.2

7) 2470.8

8) 1.718213058

9) 1005550.526

10) 586.365

Most of the questions you asked were in repeating decimal form.

Explanation:

1
Select the correct answer.
Which type of energy is thermal energy a form of?
A
chemical energy
B.
kinetic energy
C. magnetic energy
D. potential energy
Reset
Next

Answers

Answer:

B. kinetic energy

Explanation:

Thermal energy (It’s a low form of energy ) is a form of kinetic energy as it is produced as a result of motion of particles either if they vibrate at their position or they move along longer paths.

If it takes 560 joules of work to move an object 10 meters what force was needed

Answers

Answer:

F = 56 N

Explanation:

Assuming that the applied force is parallel to the displacement, according to the definition of work, we can write the following expression for the force required:

       [tex]F = \frac{W}{\Delta x} =\frac{560J}{10m} = 56 N (1)[/tex]

a boy throws a ball straight up into the air it reaches the highest point of its flight after 4 seconds how fast was the ball going when it left the boy's hand

Answers

Answer:

Gravity pulls down on the ball at g=-9.81 m/s^2. Up is positive, down is negative. The ball started at a certain initial velocity of Vi m/s. Time it took is t=4s. Final velocity is Vf=0 m/s, because at the highest point the ball stops moving.

A planet with a mass one-half that of Earth has a radius that is 3 times that of Earth's radius. What is the gravitational field strength of the planet?

Answers

The gravitational acceleration of a planet is proportional to the planet's mass, and inversely proportional to square of the planet's radius.

So when you stand on the surface of this particular planet, you feel a force of gravity that is

(1/2) / (3²)

of the force that you feel on the surface of the Earth.

That's (1/18) as much as on Earth.

The acceleration of gravity there would be about 0.545 m/s².  

This is about 12% less than the gravity on Pluto.

Calculate the first and second order angles for light of wavelength 400. nm and 700. nm of the grating contains 1.00 x 104 lines/cm.

Answers

Answer:

[tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex]

[tex]44.43^{\circ}[/tex], second order does not exist

Explanation:

n = Number of lines grating = [tex]1\times10^4\ \text{Lines/cm}[/tex]

[tex]\lambda[/tex] = Wavelength

m = Order

Distance between slits is given by

[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{1\times 10^4}\\\Rightarrow d=10^{-6}\ \text{m}[/tex]

[tex]\lambda=400\ \text{nm}[/tex]

m = 1

We have the relation

[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}\\\Rightarrow \theta=\sin^{-1}\dfrac{1\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=23.58^{\circ}[/tex]

m = 2

[tex]\theta=\sin^{-1}\dfrac{2\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=53.13^{\circ}[/tex]

The first and second order angles for light of wavelength 400 nm are [tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex].

[tex]\lambda=700\ \text{nm}[/tex]

m = 1

[tex]\theta=\sin^{-1}\dfrac{1\times 700\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=44.43^{\circ}[/tex]

m = 2

[tex]\theta=\sin^{-1}\dfrac{2\times 700\times 10^{-9}}{10^{-6}}[/tex]

Here [tex]\dfrac{2\times 700\times 10^{-9}}{10^{-6}}=1.4>1[/tex] so there is no second order angle for this case.

The first order angle for light of wavelength 700 nm are [tex]44.43^{\circ}[/tex].

Second order angle does not exist.

If we always drop the balls from 1-m
height in each trial, what type of variable
is this in this experiment?
Constant Variable
Dependent Variable
Independent Variable

Answers

Answer:

height

weight of ball

time of ball falling

If soldiers march across the bridge with a cadence equal to the bridge’s natural frequency and impart 1x104 J of energy each second, how long does it take for the bridge’s oscillations to go from 0.1 m to 0.5 m amplitude?

Answers

Answer: Hello, Mark me as Brainliest! :)

If soldiers march across the bridge with a cadence equal to the bridge's natural frequency and impart $$1.00 × 10^4 J$$ of energy each second, how long does it take for the bridge's oscillations to go from 0.100 m to 0.500 m amplitude. $ 5 \times 10^7 \text{J} $ . \\ b) $ 12 \times 10^4 \text{s}$ .

Your Welcome!

Explanation:

(will give brainliest to whoever answers first and explains reasoning) A 10kg object is spun around in a circle with a centripetal acceleration of 3.5m/s^2. What is the centripetal force acting on the object?

Answers

Answer:

35 N

Explanation:

F = ma

centripetal force = 10(3.5) = 35 N

an iron Tyre of diameter 50cm at 288k is to be shrank on to a wheel of diameter 50.35cm.To what temperature must the tyre be heated so that it will slip over the wheel with a radial gap of 0.5mm.Linear expansivity of iron is 0.000012k-1

Answers

Answer:

The answer should be D

Explanation:

what colors of light are absorbed when white light falls on a green object?

Answers

It’s blue because it makes it go in to it and sink

a cohesive force between the liquids molecules is responsible for the fluids is called​

Answers

Answer:

static force

Explanation:

mark me brainliest

The amount of light that enters the pupil is controlled by the:
retina.
lens.
inis.

Answers

Answer: The amount of light that enters the pupil is controlled by the Iris

Explanation:

What is the scientific study of how animals are classified?

Answers

Answer:

biology

Explanation:

Animals and plants

A 50 N force causes a spring to compress 0.09 m. What is the spring constant? What is the potential energy of the spring?​

Answers

Initial considerationsApplied force = 50NDistance to equilibrium position = 0.09mSpring constant = ?

Resolution

using hooke's relation

F = K . d

50N = k . 0.09m

k = 50N / 0.09m

k = 5555.56 N/m

Calculating the potential energy of the spring

Ep = 1/2 k . x²

Ep = 1/2 (5555.56 N/m) (0.09m)²

Ep = 22.5 Joules

Answer

the spring constant? =

k =  5555.56 N/m

potential energy of the spring?​

Ep = 22.5 Joules

The Potential energy of the spring is 2.25 J

What is the Potential energy of spring?

This is the energy stored in spring due to its elastic properties.

To calculate the potential energy of the spring, we use the formula below.

Formula:

E = Fe/2................ Equation 1

Where:

E = Potential energy of the springF = Force applied to the springe = compression.

From the question,

Given:

F = 50 Ne = 0.09 m

Substitute these values into equation 1

E = 50(0.09)/2E = 2.25 J.

Hence, The Potential energy of the spring is 2.25 J

Learn more about the potential energy of spring: https://brainly.com/question/3510901

I don’t understand this

Answers

What was the question? I can help I just need the question !

Answer:

true

Explanation:

force or powerbecause he pushes a disk

A bump should primarily be used in which situation?

A. when the ball is going out of bounds
B. when returning a serve
C. to start game play

Answers

The answer is b to the question

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength ?

Answers

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

[tex]E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0[/tex]

Therefore, the electric field strength at the mid-point between the two rings is zero.

A ball rolls along the floor with a constant velocity of 12 m/s. How far will it have
gone after 6 seconds?

Answers

Answer:

72m

12m x 6s=72m

there ya go

You would do 12x6 to get the velocity. After 6 seconds it would be 72m/s

prove that d1=R(d1-d2) in relative density​

Answers

You know I’m going to be honest I don’t feel like it

which energy conservation takes place when a toaster is switched on?​

Answers

Answer:

A toaster usually takes in electrical energy.

Two types of energy are created.

In the first step, all of the input electrical energy is transformed into heat energy. That heat goes first into the coils or heating elements.

The heating elements get hot and glow.

The hot elements then transfer thermal energy (heat energy) into the air inside the toaster.

Explanation:

Answer: electrical energy

Explanation: An electric toaster takes in electrical energy from the power outlet and converts it into heat, very efficiently. If you want your toast to cook quickly, you need a toaster that radiates as much heat as possible each second onto your bread. hope this helps. Can u pls give me brainliest

A small town has decided to forego the use of electrical power and send energy through town via mechanical waves on ropes. They use rope with a mass per length of 1.50 kg/m under 6000 N tension. If they are limited to a wave amplitude of 0.500 m, what must be the frequency of waves necessary to transmit power at the average rate of 2.00 kW

Answers

Answer:

the required frequency of waves is 2.066 Hz

Explanation:

Given the data in the question;

μ = 1.50 kg/m

T = 6000 N

Amplitude A = 0.500 m

P = 2.00 kW = 2000 W

we know that, the average power transmit through the rope can be expressed as;

p = [tex]\frac{1}{2}[/tex]vμω²A²

p = [tex]\frac{1}{2}[/tex]√(T/μ)μω²A²

so we solve for ω

ω² = 2P / √(T/μ)μA²

we substitute

ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²

ω² = 4000 / 23.71708

ω² = 168.65

(2πf)² = ω²

so

(2πf)² = 168.65

4π²f² = 168.65

f² = 168.65 / 4π²

f² = 4.27195

f = √4.27195

f = 2.066 Hz

Therefore, the required frequency of waves is 2.066 Hz

please answer asap! will give brainliest if correct! ^^

A train of waves is moving at a speed of 30 m/s. The frequency of the waves is 10 Hz. What is the wavelength?
a. 300 m
b. 30 m
c. 3 m
d. 0.1 m

Answers

Answer:

the answer is c, 3m......

When a train is moving at the speed of 30 m/s. The wavelength of the wave is 3 m.

What is wavelength?

Wavelength is the characteristic of the wave which represents the distance between the adjacent crest or trough of the wave.

Given is the speed of the train of waves 30 m/s and frequency of the wave is 10 Hz, then the wavelength is related to speed and wavelength as

v = fλ

30 = 10 × λ

λ = 3 m

Thus, the wavelength of the wave is 3 m.

Learn more about wavelength.

https://brainly.com/question/13533093

#SPJ2


While a boulder is on top of a hill, it has kinetic energy.
True
False

Answers

True Boulder is a kinetic energy

A uniform-density 7 kg disk of radius 0.20 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 42 N through a distance of 0.9 m. Now what is the angular speed

Answers

Answer:

The angular speed is 23.24 rad/s.

Explanation:

Given;

mass of the disk, m = 7 kg

radius of the disk, r = 0.2 m

applied force, F = 42 N

distance moved by disk, d = 0.9 m

The torque experienced by the disk is calculated as follows;

τ = F x d = I x α

where;

I is the moment of inertia of the disk = ¹/₂mr²

α is the angular acceleration

F x r = ¹/₂mr² x α

The angular acceleration is calculated as;

[tex]\alpha = \frac{2Fr}{mr^2} \\\\\ \alpha = \frac{2F}{mr}\\\\\alpha = \frac{2 \times 42 }{7 \times 0.2} \\\\\alpha = 60 \ rad/s^2[/tex]

The angular speed is determined by applying the following kinematic equation;

[tex]\omega _f^2 = \omega_i ^2 + 2\alpha \theta[/tex]

initial angular speed, ωi = 0

angular distance, θ = d/r = 0.9/0.2 = 4.5 rad

[tex]\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta} \\\\\omega _f = \sqrt{2 \times 60 \times 4.5} \\\\\omega _f = 23.24 \ rad/s[/tex]

Therefore, the angular speed is 23.24 rad/s.

A ball on a string in uniform circular motion has a velocity of 8 meters per second, a mass of 2 kilograms, and the radius of the circle is 0.5 meters. What is the centripetal force keeping the ball in the circle?

Answers

Answer:

256 N

Explanation:

formula of centripetal force = mv²/r

m= 2kg

v= 8m/s

r= 0.5m

mv²/r = 2×8²/0.5 = 256N

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