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KonstantinChe [14]

2 years ago

13

Ebo throws a ball into the air its velocity at the start is 18m/s at an angle of 37° to the ground. What is the range of the bal

l

1 answer:

NeTakaya2 years ago

6 0

PLS help ASAP I DONT have time to answer this, it also detects if it’s right or wrong.

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Consider a person standing in an elevator that is moving at constant speed upward. The person, of mass m, has two forces acting

larisa [96]

Answer:

The weight of the person has a smaller magnitude.

Explanation:

For an observer in inertial frame of reference for the person in the elevator Newton's Second Law can be written as

Normal reaction acts upwards

Weight acts downwards

$\sum F_{v}=ma_{v}\\\\N-mg=m\times a_{v}\\\\m\times a_{v}> 0\\\\\therefore N-mg> 0\\\\\therefore N> mg$

Here

N is the normal reaction force

mg is the weight of the person

g is acceleration due to gravity

4 0

2 years ago

Read 2 more answers

A ray of light is incident on a plane surface separating two sheets of glass with refractive indexes 1.70 and 1.58. The angel of

dlinn [17]

Answer:

r = 71.8⁰

Explanation:

given,

refractive index of the glass 1 = 1.70

refractive index of glass 2 = 1.58

angle of incidence = 62°

angle of refraction =?

using Snell's law

$\dfrac{sin\ i}{sin\ r} = \dfrac{n_2}{n_1}$

$\dfrac{sin\ 62^0}{sin\ r} = \dfrac{1.58}{1.70}$

1.7 ×sin 62 ^0 = 1.58× sin r

$sinr = \dfrac{1.7\times sin 62^0}{1.58}$

sin r = 0.95

r = sin⁻¹(0.95)

r = 71.8⁰

angle of refraction =r = 71.8⁰

4 0

2 years ago

Read 2 more answers

Which of the following are dwarf planets? Check all that apply. Ceres Namaka Eris Charon Haumea Makemake Pluto

kicyunya [14]

Ceres: Yes!
Namaka: No!
Eris: Yes!
Charon: No. (it's a satellite, and dwarf planet's can't be satellites!)
Haumea: Yes!
Makemake: Yes!
Pluto: Yes!

Glad To Help;)

6 0

1 year ago

Read 2 more answers

a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio

Katena32 [7]

Answer:

a) factor $b=\sqrt{2}$

b) factor $b=\frac{1}{2}$

c) factor $b=1$

d) factor $b=1$

Explanation:

Time period of oscillating spring-mass system is given as:

$T=\frac{1}{f}$

$T={2\pi} \sqrt{\frac{m}{k} }$

where:

$f=$ frequency of oscillation

$m=$ mass of the object attached to the spring

$k=$ stiffness constant of the spring

a) <u>On doubling the mass:</u>

New mass, $m'=2m$

<u>Then the new time period:</u>

$T'=2\pi\sqrt{\frac{m'}{k} }$

$T'=2\pi\sqrt{\frac{2m}{k} }$

$T'=\sqrt{2}\times 2\pi\sqrt{\frac{m}{k} } }$

$T'=\sqrt{2} \times T$

where the factor $b=\sqrt{2}$ as asked in the question.

b) On quadrupling the stiffness constant while other factors are constant:

New stiffness constant, $k'=4k$

<u>Then the new time period:</u>

$T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T$

where the factor $b=\frac{1}{2}$ as asked in the question.

c) On quadrupling the stiffness constant as well as mass:

New stiffness constant, $k'=4k$

New mas, $m'=4m$

<u>Then the new time period:</u>

$T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T$

where factor $b=1$ as asked in the question.

d) On quadrupling the amplitude there will be no effect on the time period because T is independent of amplitude as we can observe in the equation.

so, factor $b=1$

7 0

2 years ago

A spring stretches by 15cm when a mass of 300g hangs down from it,if the spring is then stretched an additional 10cm and release

bearhunter [10]

Answer:

0.1 m

Explanation:

It is given that,

Mass of the object, m = 350 g = 0.35 kg

Spring constant of the spring, k = 5.2 N/m

Amplitude of the oscillation, A = 10 cm = 0.1 m

Frequency of a spring mass system is given by :

Time period:

4 0

2 years ago