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3 years ago

Why couldn't the Royal Society publish Newton's Principia?

Physics

1 answer:

Gemiola [76]3 years ago

8 0

Answer:

The Principia was eventually published in 1687. After publishing the work, the Royal Society told Halley it could no longer afford his salary and offered to pay him in unsold copies of the Historia Piscium instead.

Explanation:

But the Royal Society refused to fund the book because it had suffered huge financial losses publishing 'The History of Fish', an indigestible tome of piscine anatomy which barely sold and nearly bankrupted the institution.

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Two forces of 83 pounds and 56 pounds act simultaneously on an object. The resultant forms an angle of 52° with the 56-pound for

Nikitich [7]

Answer:

95.9°

Explanation:

The diagram illustrating the action of the two forces on the object is given in the attached photo.

Using sine rule a/SineA = b/SineB, we can obtain the value of B° as shown in the attached photo as follow:

a/SineA = b/SineB,

83/Sine52 = 56/SineB

Cross multiply to express in linear form

83 x SineB = 56 x Sine52

Divide both side by 83

SineB = (56 x Sine52)/83

SineB = 0.5317

B = Sine^-1(0.5317)

B = 32.1°

Now, we can obtain the angle θ, between the two forces as shown in the attached photo as follow:

52° + B° + θ = 180° ( sum of angles in a triangle)

52° + 32.1° + θ = 180°

Collect like terms

θ = 180° - 52° - 32.1°

θ = 95.9°

Therefore, the angle between the two forces is 95.9°

3 0

2 years ago

If two temperatures differ by 25 degrees on Celsius scale, the difference of temperature on Fahrenheit scale is?

Mekhanik [1.2K]

Answer:

25 x 9/5 = 45 degrees Fahrenheit

Explanation:

4 0

2 years ago

Blue light of wavelength λ passes through a single slit of width d and forms a diffraction pattern on a screen. If we replace th

ololo11 [35]

Answer:

We can retain the original diffraction pattern if we change the slit width to d) 2d.

Explanation:

The diffraction pattern of a single slit has a bright central maximum and dimmer maxima on either side. We will retain the original diffraction pattern on a screen if the relative spacing of the minimum or maximum of intensity remains the same when changing the wavelength and the slit width simultaneously.

Using the following parameters: y for the distance from the center of the bright maximum to a place of minimum intensity, m for the order of the minimum, λ for the wavelength, D for the distance from the slit to the screen where we see the pattern and d for the slit width. The distance from the center to a minimum of intensity can be calculated with:

$y\approx\frac{m\lambda D}{d}$

From the above expression we see that if we replace the blue light of wavelength λ by red light of wavelength 2λ in order to retain the original diffraction pattern we need to change the slit width to 2d:

$y\approx\frac{m\lambda D}{d} =\frac{m2\lambda D}{2d}$

7 0

3 years ago

A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500

kenny6666 [7]

Answer:

6 m/s

Explanation:

Given that :

mass of the block m = 200.0 g = 200 × 10⁻³ kg

the horizontal spring constant k = 4500.0 N/m

position of the block (distance x) = 4.00 cm = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e $\frac{1}{2} kx^2 = \frac{1}{2} mv^2$

$kx^2 = mv^2$

$4500* 0.04^2 = 200*10^{-3} *v^2$

$7.2 =200*10^{-3}*v^{2}$

$v^{2} =\frac{7.2}{200*10^{-3}}$

$v =\sqrt{\frac{7.2}{200*10^{-3}}}$

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is 6 m/s

5 0

3 years ago

A 332 kg mako shark is moving in the positive direction at a constant velocity of 2.30 m/s along the bottom of a sea when it enc

Digiron [165]

To solve this problem we will apply the concepts related to the conservation of momentum. By definition we know that the initial moment must be equivalent to the final moment of the two objects therefore

$p_1 = p_2$

$m_1u_1+m_2u_2 = m_1v_1+ m_2v_2$

Here,

$m_{1,2} =$ Mass of each object

$u_{1,2} =$ Initial velocity of each object

$v_{1,2}$ = Final velocity of each object

Since the initial velocity relative to the metal tank is at rest, that velocity will be zero. And considering that in the end, the speed of the two bodies is the same, the equation would become

$m_1u_1 = (m_1+m_2)v_f$

Rearranging to find the velocity,

$v_f = \frac{m_1u_1}{ (m_1+m_2)}$

Replacing we have that,

$v_f = \frac{(332)(2.3)}{ (332+19.5)}$

$v_f = 2.17 m/s$

Therefore the velocity of the shark immediately after it swallows the tank is $2.17m/s$

4 0

3 years ago