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1 year ago

Compare and contrast camera obscura with what you know about modern digital photography, including cell phones.

Physics

2 answers:

Sonja [21]1 year ago

5 0

Answer:

It captures images but does not preserve them.

Ugo [173]1 year ago

5 0

A camera obscura is a darkened room. For making pictures there was a small hole in one wall which allowed an image of the outside scene to project onto the opposite wall, where an artist would draw the outlines of the scenery objects by tracing around the significant lines in the image.

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A 100 N force is applied to a 50 kg crate resting on a level floor. The coefficient of kinetic friction is 0.15. What is the acc

Nookie1986 [14]

The acceleration of the crate after it begins to move is 0.5 m/s²

We'll begin by calculating the the frictional force

Mass (m) = 50 Kg

Coefficient of kinetic friction (μ) = 0.15

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

Normal reaction (N) = mg = 50 × 10 = 500 N

<h3>Frictional force (Fբ) =?</h3>

Fբ = μN

Fբ = 0.15 × 500

Next, we shall determine the net force acting on the crate

Frictional force (Fբ) = 75 N

Force (F) = 100 N

<h3>Net force (Fₙ) =?</h3>

Fₙ = F – Fբ

Fₙ = 100 – 75

Finally, we shall determine the acceleration of the crate

Mass (m) = 50 Kg

Net force (Fₙ) = 25 N

<h3>Acceleration (a) =?</h3>

a = Fₙ / m

a = 25 / 50

Therefore, the acceleration of the crate is 0.5 m/s²

Learn more on friction: brainly.com/question/364384

8 0

1 year ago

A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th

son4ous [18]

Answer:

Part a)

$v = 7.57 km/h$

Part b)

$\theta = 67.5 degree$ North of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

$d_1 = 40 km$

now it turns towards 50 degree East of North

so its distance is given as

$d_2 = 20 km(sin50 \hat i + cos50\hat j)$

$d_2 = 15.3 \hat i + 12.8 \hat j$

then finally it moves towards west for 50 min

$d_3 = -50 \hat i$

Now the total displacement of the train is given as

$d = d_1 + d_2 + d_3$

$d = (40 + 15.3 - 50)\hat i + 12.8 \hat j$

$d = 5.3\hat i + 12.8 \hat j$

now total time duration of the motion is given as

$T = 40 min + 20 min + 50 min$

$T = 1.83 h$

now average velocity is given as

$v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}$

$v_{avg} = 2.89\hat i + 6.99\hat j$

Part a)

magnitude of the average velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{2.89^2 + 6.99^2}$

$v = 7.57 km/h$

Part b)

Direction of the velocity is given as

$tan\theta = \frac{v_y}{v_x}$

$tan\theta = \frac{6.99}{2.89}$

$\theta = 67.5 degree$ North of East

6 0

3 years ago

A car with mass 1500 kg moves with constant velocity of 36 m/s. The driver sees a group of cows in front and he immediately step

Crazy boy [7]

From laws of motion:

$S = ( \frac{v + u}{2} ) \times t$
Where S is the distance/displacement (as you would call it) which is unknown
v = final velocity which is 0m/s (this is because the car stops)
u = initial velocity which is 36m/s (from the data given)
t = time taken for the distance to be covered and it is 6s

Substitute the values, hence:
$S = ( \frac{0 + 36}{2} ) \times 6$
$S = (18) \times 6 \\ \\ S = 108m$

But this is merely the distance he travelled in the 6 seconds he was trying to stop the car.

Therefore, the distance between the car and the cows = 160-108
Distance = 52m

6 0

2 years ago

a dump truck travels 70mph for 3 hours, then for 20mph for another 2 hours in the same direction. what is the average speed the

Dmitry_Shevchenko [17]

Answer:

the average speed of the car is 170 mph.

Explanation:

Given;

initial speed, u = 70 mph

time of motion, t₁ = 3 hours

final speed, v = 20 mph

time of motion, t₂ = 2 hours

The average speed of the car is calculated as;

$v' = \frac{70(3) - 20(2)}{3-2} \\\\v' = 170 \ mph$

Therefore, the average speed of the car is 170 mph.

8 0

2 years ago

A sailboat is heading directly north at a speed of 20 knots (1 knot 50.514 m/s). the wind is blowing toward the east with a spee

stellarik [79]

(a) The magnitude of the wind as it is measured on the boat will be the result of the two vectors. Since they are at 90°, the resultant can be determined by the Pythagorean theorem.

R = sqrt ((20 knots)² + (17 knots)²)
R = sqrt (400 + 289)
R = 26.24 knots

The direction of the wind will have to be angle between the boat and the resultant.
cos θ = (20 knots)/(26.24 knots)
θ = 40.36°

Hence, the direction is 40.36° east of north.

(b) As stated, the wind is blowing in the direction that is to the east. This means that it only has one direction. Parallel to the motion of the boat, the magnitude of the wind velocity will have to be zero.

5 0

3 years ago