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3 years ago

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Newton's 1st law of motion states that a moving object will go in a straight line unless an outside force changes this motion. W

hat outside force prevents the planets from moving in a straight line into space?
gravity
meteorite impacts
supernovas
nuclear fusion

1 answer:

navik [9.2K]3 years ago

4 0

The gravity of the sun is keeping all the planets in orbit

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A force of 1.150×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 795 N. If he starts

nalin [4]

Answer:

He has a speed of 16.60m/s after 35.0 meters.

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

$v_{f}^{2} = v_{i}^{2} + 2ad$

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$ (1)

The acceleration can be found by means of Newton's second law:

$\sum F_{net} = ma$

Where $\sum F_{net}$ is the net force, m is the mass and a is the acceleration.

$Fx + Fy = ma$ (2)

All the forces can be easily represented in a free body diagram, as it is shown below.

Forces in the x axis:

$F_{x} = F - F_{air}$ (3)

Forces in the y axis:

$F_{y} = 0$ (4)

Solving for the forces in the x axis:

$F_{x} = F - F_{air}$

Where $F = 1.150x10^{3} N$ and $F_{air} = 795 N$ :

$F_{x} = 1.150x10^{3} N - 795 N$

$F_{x} = 355 N$

Replacing in equation (2) it is gotten:

$Fx + Fy = ma$

$355 N + 0 N = (90.0 Kg)a$

$355 N = (90.0 Kg)a$

$a = \frac{355 N}{90.0Kg}$

$a = \frac{355 Kg.m/s^{2}}{90.0Kg}$

$a = 3.94 m/s^{2}$

So the acceleration for the cyclist is $3.94 m/s^{2}$ , now that the acceleration is known, equation (1) can be used:

$v_{f} = \sqrt{v_{i}^{2} + 2ad}$

However, since he was originally at rest its initial velocity will be zero ( $v_{i} = 0$ ).

$v_{f} = \sqrt{2ad}$

$v_{f} = \sqrt{2(3.94m/s^{2})(35.0m)}$

$v_{f} = 16.60m/s$

He has a speed of 16.60m/s after 35.0 meters

8 0

3 years ago

A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that r

kaheart [24]

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

$\Phi = \frac{2\pi \delta}{\lambda}$

Where

$\delta =$ Horizontal distance between two points

$\lambda =$ Wavelength

From our values we have,

$\lambda = 500nm = 5*10^{-6}m$

$\theta = 1\°$

The horizontal distance between this two points would be given for

$\delta = dsin\theta$

Therefore using the equation we have

$\Phi = \frac{2\pi \delta}{\lambda}$

$\Phi = \frac{2\pi(dsin\theta)}{\lambda}$

$\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}$

$\Phi= 1.096 rad \approx = 1.1 rad$

Therefore the correct answer is C.

6 0

3 years ago

A standing wave has a frequency of 471 Hz and a wavelength of 1.9. What is the speed of the

Brums [2.3K]

Answer:

c = 894.90 m/s

Explanation:

Given data:

Frequency of wave = 471 Hz

Wavelength of wave = 1.9 m

Speed of wave = ?

Solution:

Formula:

Speed of wave = frequency × wavelength

c = f×λ

c = 471 Hz × 1.9 m

Hz = s⁻¹

c = 471s⁻¹ × 1.9 m

c = 894.90 m/s

The speed of wave is 894.90 m/s.

6 0

3 years ago

If a ball is tossed straight up into the air, at what position is its potential energy the greatest? Question 2 options: a) when

yarga [219]

Q1: At the highest point

3 0

3 years ago

8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.

mart [117]

Answer:

The resultant tension of the two ropes is approximately 42.4 N

The length of the line representing the resultant tension is approximately 8.48 cm

Please find included with the answer the scale drawing created with Microsoft Word

Explanation:

The given parameters are;

The tension in rope P, $T_P$ = 30 N

The tension in rope Q, $T_Q$ = 30 N

The angle the rope, 'P', makes with the horizontal = 45°

The angle the rope, 'Q', makes with the horizontal = 45°

The scale factor of the scale diagram, S.F. = 5.0 N/cm

By the resolution of forces at equilibrium, we have;

The sum of the vertical forces, $\Sigma F_y$ = $T_P_y$ + $T_Q_y$ + W = 0

∴ W = -( $T_P_y$ + $T_Q_y$ )

W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712

The weight of the heavy ball, W ≈ 42.4 N acting downwards

The sum of the horizontal forces, $\Sigma F_x$ = $T_P_x$ + $T_Q_x$ = 0

The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm

The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included

3 0

3 years ago