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3 years ago

All of the following are true about scientific evidence except: *

Physics

1 answer:

goldenfox [79]3 years ago

4 0

Answer:

Your answer is D It does not need to be repeatable.

the reason for this is because C is correct.

You need to be able to experiment on something multiple times so that you can gather further data to imbedded your evidence in facts.

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A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force

VikaD [51]

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 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

An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2

Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

$M=-\frac{q}{p}$

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

$M=-\frac{200 cm}{50 cm}=-4.0$

7 0

3 years ago

how to A golf ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 3.5

Alexandra [31]

Answer: a) 12857.1 m/s/s b) 578.6 N

Explanation:

Impulse = change in momentum

Ft = mV2 - mV1

V = AT, 45 / .0035 = 12857.1 m/s/s

(b) .045 x 12857.1 = 578.6 N

4 0

3 years ago

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An astronaut is moving in space when a big explosion occurs about 50 metres BEHIND him. How will the astronaut come to know abou

kakasveta [241]

The matter from the explosion can reach him, hitting him. He should be able to feel that.

6 0

3 years ago

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