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Rudik [331]
3 years ago
7

Which situation is most likely to lead to improving a scientific theory?

Physics
1 answer:
Nonamiya [84]3 years ago
4 0

Answer:

A. Repeating an experiment to verify results

Explanation:

A scientific theory may be defined as the theory of science that is a thought out explanation of the observations done in the nature or that happens in the natural world. They are constructed using several scientific methods by observing many facts and hypothesis.

According to the question, repeating of an experiment a number of times and verifying its result may help in improving a scientific theory as if we get the similar result for each experiments, then it confirms the basis of the hypothesis or the facts that are observe in the nature and conclude a scientific theory.

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Jupiter's gravitational field is stronger than Earth's. On Jupiter, an object's__________ will be greater than on Earth. What on
vesna_86 [32]

Answer: Weight

Explanation:The mass of an object is a measure of the quantity of matter in the object. This quantity remains constant under any circumstance.

However, the same cannot be said about the weight of such object.

The weight of the object is very much dependent on the acceleration due to gravity which is a accelerational pull (by convention-- a pull downwards).

This is why an object tends to fall when it is thrown upwards on the earth for instance.

The statements above consequently infer that since the gravitational field of Jupiter is greater than that of the earth, the acceleration due to gravity on Jupiter is greater than that on earth.

And since the weight of an object(W) is a product of its mass and the acceleration due to gravity at that point.

Consequently, the object's weight on Jupiter would be greater than its weight on earth.

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3 0
3 years ago
⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a)
GREYUIT [131]

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

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