Explanation for x=73km/h•0.5s= 10.14 pls help how did they get 10.14? What was the work around it

nirvana33 [79]

Linear motion (also called rectilinear motion) is a motion along a straight line, and can therefore be described mathematically using only one spatial dimension.

4 0

3 years ago

Why are chloroplasts essentially to the process of photosynthesis to occurs

lesya [120]

they absorb sunlight and turn it into glucose I think.

8 0

3 years ago

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A horizontal object-spring system oscillates with an amplitude of 2.8 cm. If the spring constant is 275 N/m and object has a mas

Lisa [10]

Answer:

(a) the mechanical energy of the system, U = 0.1078 J

(b) the maximum speed of the object, Vmax = 0.657 m/s

(c) the maximum acceleration of the object, a_max = 15.4 m/s²

Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

(a) the mechanical energy of the system

This is the potential energy of the system, U = ¹/₂KA²

U = ¹/₂ (275)(0.028)²

U = 0.1078 J

(b) the maximum speed of the object

$V_{max} =\omega*A= \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s$

(c) the maximum acceleration of the object

$a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2$

6 0

3 years ago

8. How did the measured angular magnification of the telescope compare with the theoretical prediction?

Genrish500 [490]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The theoretical angular magnification lies within the angular magnification range

Explanation:

From the question we are told that

The focal length of B is $f_{objective } = 43.0 \ cm$

The focal length of A is $f_{eye} = 10.4 \ cm$

The theoretical angular magnification is mathematically represented as

$m = \frac{f_{objective }}{f_{eye}} = \frac{43.0}{10.4}$

$m = \frac{f_{objective }}{f_{eye}} = 4.175$

Form the question the measured angular magnification ranges from 4 -5

So from the value calculated and the value given we can deduce that the theoretical angular magnification lies within the angular magnification range

3 0

3 years ago

Models are particularly useful in relativity and quantum mechanics, where conditions are outside those normally encountered by h

mylen [45]

Answer:

A model is defined as a structure used to represent an object, usually of a different scale.

Explanation:

In quantum mechanics and particle physics, many of the particles are subatomic, meaning that they are smaller than atoms. This is where a model would be useful. A model could help people to visualise what the particle looks like, and in general would make it easier to understand the behaviour of such a particle.

4 0

3 years ago