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Novay_Z [31]
3 years ago
10

NEED HELP PLEASE !!!!!!!!!!!!!!!!!!!!!!!!!!

Physics
1 answer:
Anna [14]3 years ago
3 0

Answer:

(1) Exercise should be painful in order for it to be beneficial. FALSE

(2) Proper progression involves a gradual increase in the level of exercise. TRUE

(3) Exercising on a regular, consistent basis provides you with the best results. TRUE

Explanation:

(1) It is a false statement because whenever we do exercise, it should not be painful, it should be comfortable to some extent for a person, if it will be painful then it means one cant get the true and genuine benefit from the exercise.

(2) It is very much true because one must keep on increasing the level of his/her exercise gradually with the passage of time, it will make his/her body synchronized with the exercise properly.

(3) It is also a true statement because, one must use variety of exercises in his/her workout session targeting different muscles and different areas and parts of the body to make exercise really beneficial.

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A vehicle of mass 100kg has a kinetic energy of 5000 J at an instant. The velocity at that instant is​
snow_lady [41]

Answer:

  • \bf\pink{10\:m/s}

Explanation:

<h2><u>Given</u> :-</h2>

  • \sf\red{Mass \ of \ vehicle \ (m) = 100 \ kg}
  • \sf\orange{Kinetic \ energy \ (K.E.) = 5000 \ J}

<h2><u>To Find</u> :-</h2>

  • \sf\green{Velocity\: of\: the \:vehicle \:at \:the \:instant}

<h2><u>Formula to be used</u> :-</h2>

  • \sf\blue{K.E. = \dfrac{1}{2} mv^{2}}

Where,

  • K.E. = Kinetic energy possessed by the body
  • M = Mass of the body
  • V = Velocity of the body

<h2><u>Solution</u> :-</h2>

\to\:\:\sf\red{K.E. = \dfrac{1}{2}mv^{2}}

\to\:\:\sf\orange{5000 = \dfrac{1}{2} \times 100 \times v^{2}}

\to\:\:\sf\green{5000 = 50 \times v^{2}}

\to\:\:\sf\blue{\dfrac{5000}{50} = v^{2}}

\to\:\:\sf\purple{100 = v^{2}}

\to\:\:\sf\red{\sqrt{100} = v}

\to\:\:\sf\orange{ 10 = v}

\to\:\:\bf\pink{v = 10\:m/s}

  • Velocity of the vehicle at the instant is \bf\green{10\:m/s}
7 0
2 years ago
Read 2 more answers
A uniform-density 7 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is
GREYUIT [131]

Answer:

\omega = 22.67 rad/s

Explanation:

Here we can use energy conservation

As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc

Now we have

W = \frac{1}{2}I\omega^2

now we know that work done is product of force and displacement

so here we have

W = F.d

W = (44 N)(0.9 m) = 39.6 J

now for moment of inertia of the disc we will have

I = \frac{1}{2}mR^2

I = \frac{1}{2}(7 kg)(0.21^2)

I = 0.154 kg m^2

now from above equation we will have

39.6 = \frac{1}{2}(0.154)\omega^2

\omega = 22.67 rad/s

5 0
2 years ago
Read 2 more answers
(I will give brainliest whoever helps me !!)
lorasvet [3.4K]
C. Forces have mass and take up space
3 0
2 years ago
A 3,000-N force gives an object an acceleration of 15 m/s^2. The mass of the object is
Bingel [31]
<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

According to the second Newton's Law,

<em>✔ We have : F = m x a ⇔ m = F/a </em>

  • m = 3000/15
  • m = 200 kg

The mass of the object is therefore 200 kg.

7 0
2 years ago
Read 2 more answers
What is the gravitational force on a 70kg that is 6.38x10^6m above the earths surface
NARA [144]

Answer:

171.5 N

Explanation:

The gravitational force on an object due to the Earth is given by

F=mg

where

m is the mass of the object

g is the acceleration due to gravity

The acceleration due to gravity at a certain height h above the Earth is given by

g=\frac{GM}{(R+h)^2}

where:

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

R=6.37\cdot 10^6 m is the Earth's radius

Here,

h=6.38\cdot 10^6  m

So the acceleration due to gravity is

g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})}{(6.37\cdot 10^6 + 6.38\cdot 10^6)^2}=2.45 m/s^2

We know that the mass of the object is

m = 70 kg

So, the gravitational force on it is

F=mg=(70)(2.45)=171.5 N

5 0
3 years ago
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