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3 years ago

Who discovered laws of gravity and motion? He discovered aspects of light and color, leading to today's spectrum analysis

Physics

1 answer:

Vitek1552 [10]3 years ago

5 0

Answer:

Sir Isaac Newton

Explanation:

The laws according to which bodies move and how one body attracts another body were developed by Newton. The three laws of motion and the law of universal gravitation were developed by Newton.

In the seventeenth century almost all the refracting telescopes exhibited color distortion. Newton believed that white light had a spectrum which was the reason for this distortion. The lens of a refracting telescope used to act like a prism which caused the dispersion of light proving Newton's hypothesis.

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How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?

charle [14.2K]

The change in internal energy is only gravitional PE because the tube is being drug up at a constant speed. Since it is at a constant speed, the change in KE is 0. Change in PE = m*g*h = 78 kg * 10 m/s^2 * 30 m = 23400 J Work done on the system is from the force Work = force * distance = 350 N * 120 m = 42000 J So, work added 42000 J to the system, but the rider's energy only increased 23400 J. Therefore, friction took up the difference. Friction is where the thermal energy comes from Q = 42000 J - 23400 J = 18600 J. Therfore, friction generated 18600 J of heat to the surroundings.

7 0

3 years ago

"The International Space Station (ISS) orbits at a distance of 350 km above the surface of the Earth. (a) Determine the gravitat

vagabundo [1.1K]

Answer:

(a) g = 8.82158145 $m/s^2$ .

(b) 7699.990192m/s.

Explanation:

(a) Strength of gravitational field 'g' by definition is

$g = \frac{M_{(earth)} }{r^2} G$ , here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.

r = 6721,000 meters, putting this value in above equation gives g = 8.82158145 $m/s^2$ .

(b) We have to essentially calculate centripetal acceleration that equals new 'g'.

$a_{centripetal}=\frac{V^2}{r} =g$ here g is known, r is known and v is unknown.

plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.

(c) S = vT, here T is time period or time required to complete one full revolution.

S = earth's circumfrence , V is calculated in (B) T is unknown.

solving for unknown gives T = 5484.3301s = 1.5234hours.

3 0

3 years ago

Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with

Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

Part A)

Let's use the parabolic motion equation to solve it. Let's define the variables:

y(i) is the initial height, it is a constant.
y(f) is the final height, in our case is 0
v(i) is the initial velocity (v(i)=16 m/s)
θ1 is the first angle, 30°
θ2 is the first angle, -30°

For the first stone

$y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}$

$0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}$

$0=y_{i1}+8t_{1}-4.905*t_{1}^{2}$ (1)

For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

$0=y_{i2}-8t_{2}-4.905t_{2}^{2}$ (2)

If we solve the equation (1) we will have:

$t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can do the same procedure for the equation (2)

$t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

Part B)

We can use the equation of the horizontal position here.

First stone

$x_{f1}=x_{i1}+vcos(30)t_{1}$

$x_{f1}=0+13.86*t_{1}$

$x_{f1}=13.86*t_{1}$

Second stone

$x_{2}=x_{i2}+vcos(-30)t_{2}$

$x_{1}=0+13.86*t_{1}$

$x_{1}=13.86*t_{2}$

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

3 0

3 years ago

In a shipping yard, a crane operator attaches a cable to a 1,400 kg shipping container and then uses the crane to lift the conta

Molodets [167]

Answer:

Explanation:

Given

mass of crane $m=1400\ kg$

distance moved $d=40\ m$

Since it is moving with a constant velocity therefore net force on it is zero

Tension force=weight

T=mg

Work done by Tension T is

$W_T=T\cdot d$

$W_T=1400\times 9.8\cdot 40$

$W_T=548.8\ KJ$

Work done by Gravity will be equal in magnitude but opposite in sign and can be obtained by work energy theorem which states that change in kinetic energy of object is equal to work done by all the forces

$W_T+W_g=0$

$W_g=-548.8\ KJ$

4 0

3 years ago

What distance is covered by an airplane traveling at a velocity of 660 miles per hour in 3.5 hours?

N76 [4]

As per the question, the velocity of the airplane [v] = 660 miles per hour.

The total time taken by airplane [t] = 3.5 hours.

We are asked to determine the total distance travelled by the airplane during that period.

The distance covered [ S] by a body is the product of velocity with the time.

Mathematically distance covered = velocity × total time

S = v × t

= 660 miles/hour ×3.5 hours

= 2310 miles.

Hence, the total distance travelled by the airplane in 3.5 hour is 2310 miles.

4 0

3 years ago