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?

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

Explanation:

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

Answer:

48kg

Explanation:

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.

Answer:

height

weight of ball

time of ball falling

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.

Answer:

static force

Explanation:

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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:

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.

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

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

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

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:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, The Potential energy of the spring is 2.25 J

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

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.

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

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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.

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

Answer:

72m

12m x 6s=72m

there ya go

Answer:

F = 56 N

Explanation:

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

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.

Answer:

biology

Explanation:

Animals and plants

Answer:

35 N

Explanation:

F = ma

centripetal force = 10(3.5) = 35 N

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.

Answer:

The answer should be D

Explanation:

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

Answer:

true

Explanation:

force or powerbecause he pushes a disk

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

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: