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lakkis [162]
3 years ago
14

Galileo performed an experiment to measure the speed of light by timing how long it took light to travel from a lamp he was hold

ing to an assistant about 1.5 km away and back again. Why was Galileo unable to conclude that light had a finite speed?
Physics
1 answer:
krok68 [10]3 years ago
6 0

Answer:

The time it takes light to cover 1.5 km was too short to be measured by Galileo's instruments.

Explanation:

The speed of light is c=3*10^8m/s, which means the time it takes to cover a distance of 1.5 km (or 1,500m) will be

t= \dfrac{1500m}{3*10^8m/s}

t= 0.000005s

which is \dfrac{1}{200000} of a second! This time delay could in no way be measured by Galileo considering the fact that he was using his heartbeat to measure time!

You might be interested in
A hammer exerts 49.8 N of force on the head (r=0.00510 m) of a nail. How much pressure does it exert on the nail?
Kisachek [45]

Answer:

609547.12 Pa ≈ 6.10×10^5 Pa

Explanation:

Step 1:

Data obtained from the question. This include the following:

Force (F) = 49.8 N

Radius (r) = 0.00510 m

Pressure (P) =..?

Step 2:

Determination of the area of the head of the nail.

The head of a nail is circular in nature. Therefore, the area is given by:

Area (A) = πr²

With the above formula we can obtain the area as follow:

Radius (r) = 0.00510 m

Area (A) =?

A = πr²

A = π x (0.00510)²

A = 8.17×10^-5 m²

Therefore the area of the head of the nail is 8.17×10^-5 m²

Step 3:

Determination of the pressure exerted by the hammer.

This is illustrated below:

Force (F) = 49.8 N

Area (A) = 8.17×10^-5 m²

Pressure (P) =..?

Pressure (P) = Force (F) /Area (A)

P = F/A

P = 49.8/8.17×10^-5

P = 609547.12 N/m²

Now, we shall convert 609547.12 N/m² to Pa.

1 N/m² = 1 Pa

Therefore, 609547.12 N/m² = 609547.12 Pa.

Therefore, the pressure exerted by the hammer on the nail is 609547.12 Pa or 6.10×10^5 Pa

8 0
4 years ago
Two identical trucks have mass 5100 kg when empty, and the maximum permissible load for each is 8000 kg. the first truck, carryi
Oksanka [162]
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg. The key here is the conservation of momentum. For the first truck, the momentum is 0(5100 + 4300) The second truck has a starting momentum of 60(5100 + x) And finally, after the collision, the momentum of the whole system is 42(5100 + 4300 + 5100 + x) So let's set the equations for before and after the collision equal to each other. 0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x) And solve for x, first by adding the constant terms 0(5100 + 4300) + 60(5100 + x) = 42(14500 + x) Getting rid of the zero term 60(5100 + x) = 42(14500 + x) Distribute the 60 and the 42. 60*5100 + 60x = 42*14500 + 42x 306000 + 60x = 609000 + 42x Subtract 42x from both sides 306000 + 18x = 609000 Subtract 306000 from both sides 18x = 303000 And divide both sides by 18 x = 16833.33 So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums. 60(5100 + 16833.33) = 60(21933.33) = 1316000 42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000 They match. The 2nd truck was definitely over loaded.</span>
6 0
4 years ago
A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul
Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2

also we know that

I = \frac{2}{5}mR^2

w= \frac{v}{R}

Now kinetic energy is given by

KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2

KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2

KE = \frac{7}{10}mR^2w^2

KE = \frac{7}{10}*150*(0.200)^2(50)^2

KE = 10500 J

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0
3 years ago
Describe an object's velocity when an acceleration-time graph is zero?
svp [43]
Anything times zero is zero
7 0
3 years ago
Read 2 more answers
A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.
GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is 8.75 m/s^2 in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, F_f = 15 N, pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, W=mg, where m is the mass of the book and g=9.8 m/s^2 is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

W=(4)(9.8)=39.2 N

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

N-W=ma

where a is the acceleration; however, since N = W, this becomes

a=0

And since the book is initially at rest on the desk, this means that there is no motion.

c)

We said there are two forces acting in the horizontal direction:

- The applied force, F = 50 N, forward

- The frictional force, F_f = 15 N, backward

Since they act along the same line, we can calculate their resultant as

\sum F = F - F_f = 50 - 15 = 35 N

and therefore the net force is 35 N in the forward direction.

d)

The net force is obtained as the resultant  of the net forces in the horizontal and vertical direction. However, we have:

- The net force in the horizontal direction is 35 N

- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction

Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.

e)

The acceleration of the book can be calculated by using Newton's second law:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

Here we have:

\sum F = 35 N (in the forward direction)

m = 4 kg

Therefore, the acceleration is

a=\frac{\sum F}{m}=\frac{35}{4}=8.75 m/s^2 (forward)

Learn more about forces, weight and Newton's second law:

brainly.com/question/8459017

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brainly.com/question/12978926

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

8 0
4 years ago
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