What type of simple machine can pinball machine flippers can be explained?

A truck travels 1430 m uphill along a road that makes a constant angle of 5.76◦ with the horizontal.

Alex73 [517]

Answer:

The horizontal component of displacement is d' = 1422.7 m

Explanation:

Given data,

The distance covered by the truck, d = 1430 m

The angle formed with the horizontal, Ф = 5.76°

The displacement is a vector quantity.

The horizontal component of displacement is given by,

d' = d cos Ф

= 1430 cos 5.76°

= 1422.7 m

Hence, the horizontal component of displacement is d' = 1422.7 m

5 0

3 years ago

A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?

OLga [1]

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

7 0

3 years ago

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A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w

frosja888 [35]

Answer:

(a) $n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) $n_{2} = 1.349$

(c) $v_{1} = 2.04\times 10^{8}\ m/s$

(d) $v_{2} = 2.22\times 10^{8}\ m/s$

Solution:

As per the question:

Refractive index of medium 1, $n_{1} = 1.47$

Angle of refraction for medium 1, $\theta_{1} = 59^{\circ}$

Angle of refraction for medium 2, $\theta_{1} = 69^{\circ}$

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$

where

$n_{2}$ = Refractive Index of medium 2

Now,

$n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

$n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}$

$n_{2} = 1.349$

(c) To calculate the velocity of light in medium 1:

We know that:

$Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}$

Thus for medium 1

$n_{1} = \frac{c}{v_{1}$

$v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s$

(d) To calculate the velocity of light in medium 2:

For medium 2:

$n_{2} = \frac{c}{v_{2}$

$v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s$

5 0

2 years ago

Read 2 more answers

A positively charged particle is in the center of a parallel plate capacitor that has charge +/- Q on it's plates. Suppose the d

ivolga24 [154]

Answer:

the force remains constant if the charge does not change

Explanation:

In a capacitor the capacitance is given by

C = ε₀ A / d

Where ε₀ is the permissiveness of emptiness, A is about the plates and d the distance between them.

The charge on the capacitor is given by the ratio

Q = C ΔV

Let's apply these expressions to our problem, if the load remains constant

C = Q / ΔV = ε₀ A / d

ΔV / d = Q / ε₀ A

If the distance increases the capacitance should decrease, therefore if the charge is a constant the voltaje difference must increase

Now we can analyze the force on the test charge in the center of the capacitor

ΔV = E d

E= ΔV/d

F = q E

F = q ΔV / d

Let's replace

F = q Q /ε₀ A

From this expression we see that the force is constant since the voltage increase is compensated by increasing the distance, therefore the correct answer is that the force remains constant if the charge does not change

8 0

3 years ago

A cyclotron (Fig. 29.16) designed to accelerate protons has an outer radius of 0.350 m . The protons are emitted nearly at rest

cupoosta [38]

The time interval for which the proton accelerated can be calculated using -

$$t=\frac{N}{f}=$ $$\frac{3.3\times 10^{25} }{f}$

Using the above formula you can find the time interval for any frequency.

We have a Cyclotron designed to accelerate protons.

We have to determine for what time interval does the proton accelerate.

<h3>What is a Cyclotron? On what principle it works?</h3>

Cyclotron is a device used to accelerate charged particles to high energies. It works on the principle that a charged particle moving normal to a magnetic field experiences magnetic Lorentz force due to which the particle moves in a circular path. The magnetic Lorentz force is given by-

F = qvB sinθ

According to the question -

In a cyclotron the force by a magnetic field is equal to the centrifugal force.

Now, for r(max) - ( Since v = rω)

q v(max) B sin (90) = $$\frac{m\times v(max)^{2} }{r(max)}$

$$v(max)=\frac{qBr(max)}{m}$

Now -

r(max) = 0.35 m

B = 0.8 Tesla

Therefore -

v(max) = $$\frac{1.6\times 10^{-19} \times 0.8 \times 0.35}{1.6\times 10^{-27} }$

v(max) = 0.28 x $10^{8}$ m/sec

Therefore, the Maximum kinetic energy = Kinetic energy at which they leave the cyclotron-

E(max) = $\frac{1}{2} m\times v(max)^{2}$

E(max) = $\frac{1}{2} \times 1.6\times 10^{-27} \times 0.28\times 10^{8} \times 0.28\times 10^{8}$

E(max) = 0.063 x $10^{11}$ joules.

Now -

The kinetic energy K is built up from 2N passes through Voltage V = 600 volts. Therefore -

$$N = \frac{E(max)}{2qV}$ = $$\frac{0.063\times 10^{11} }{2\times 1.6\times 10^{-19} \times 600}$ = 3.3 x $10^{25}$ revolutions.

The total number of revolutions are 3.3 x $10^{25}$ revolutions.

Now - The time interval for which the proton accelerated can be calculated using -

$$t=\frac{N}{f}=$ $$\frac{3.3\times 10^{25} }{f}$

Using the above formula you can find the time interval for any frequency.

To solve more questions on Cyclotron, visit the link below-

brainly.com/question/14555284

#SPJ4

5 0

1 year ago