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3 years ago

In the diagram below, a 10-kilogram ball is fired with a

Physics

1 answer:

Tcecarenko [31]3 years ago

3 0

Answer:

A) 5m/s

Explanation:

Momentum of ball in one direction=momentum of cannon in other direction

mv = mv

(10)(500)=(1000)v

5000=1000v

∴v=5m/s

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What is the standard unit of velocity

Bond [772]

Answer:

Explanation:

Meters per second

7 0

3 years ago

Read 2 more answers

Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep

SIZIF [17.4K]

Answer:

1) iii i= 1m, 2) iii and iv, 3) i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

f = r / 2

f = 2/2

f = 1 m

The builder's equation

1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

1 / f = 0 + 1 / i

i = f

i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

R₂ = infinity

R₁ > 0

Cas2 Flat-concave (divergent) lens

R₂ = infinity

R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

1 / f₂ = 1 / (L-f₁) + 1 / i

1 / i = 1 / f₂ - 1 / (L-f₁)

1 / i = (L-f₁-f₂) / f₂ (L-f₁)

i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

6 0

4 years ago

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0

Murrr4er [49]

Complete question:

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.

Answer:

The value of its capacitance is 1.027 x 10⁻¹² F

Explanation:

Given;

area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²

separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m

voltage of the battery, V = 18 V

The value of its capacitance is calculated as;

$C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F$

Therefore, the value of its capacitance is 1.027 x 10⁻¹² F

8 0

3 years ago

A river flows with a speed of 0.600 m/s. A student first swims upriver 0.500 km, then turns around and returns to his starting p

DerKrebs [107]

Answer:

a) 1111.0 seconds

b) 833.3 s

c) Because of proportions

Explanation:

a) Total time of round trip is the sum of time upriver and time downriver

$t_{total}=t_{up}+t_{down}$

Time upriver is calculated with the net speed of student and 0.500 km:

$t_{up}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=v_{relative to river}+v_{river}=-1.2+0.6=-0.6 m/s\\t_{up}=\frac{500 m}{0.6 m/s}=833.3 s$

(Becareful with units 0.5 km= 500m) Similarly of downriver:

$t_{down}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=1.2+0.6=1.8 m/s\\t_{down}=\frac{500 m}{1.8 m/s}=277.7 s$

So the sum is:

$t_{total}=1111.0s$

b) Still water does not affect student speed, so total time would be simply:

$t_{total}=\frac{1000 m}{1.2 m/s}=833.3 s$

c) For the upriver trip, student moved half the distance in half speed of the calculation in b), so it kept the same ratio and therefore, same time. So the aditional time is actually the downriver.

6 0

3 years ago

A muon is a short-lived particle. It is found experimentally that muonsmoving at speed 0.9ccan travel about 1500 meters between

Monica [59]

Answer:

a) $t=5.5\mu s$

b) $\Delta t'=12.5\mu s$

Explanation:

a) Let's use the constant velocity equation:

$v=\frac{\Delta x}{\Delta t}$

v is the speed of the muon. 0.9*c
c is the speed of light 3*10⁸ m/s

$t=\frac{\Delta x}{v}=\frac{1500}{0.9*(3*10^{8})}$

$t=5.5\mu s$

b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.

$\Delta t'=\gamma\Delta t$