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yulyashka [42]
3 years ago
6

IF YOU PLAY METAL GEAR RISING COMMENT ON THIS QUESTION

Physics
1 answer:
ella [17]3 years ago
5 0

Answer:

Hey baby, I love you. Yes i do play metal gear with snake

Explanation:

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The length of a bicycle pedal arm is 0.152 m, and a downward force of 118 N is applied to the pedal by the rider's foot. What is
Veseljchak [2.6K]

Answer:8.968 N-m

Explanation:

Given

Length of arm=0.152 m

Downward force=118 N

angle made by arm with vertical=30^{\circ}

Force can be divided into two components

It's sin component will contribute towards torque while cos component will not contibute

Torque =Fsin30\times L

T=118\times sin30\times 0.152

T=8.968 N-m

5 0
3 years ago
A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of
koban [17]

<u>Answer:</u>

 Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of truck = 25 m/s north = 25 j m/s

  Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

                          = (1.43 i + 1.00 j) m/s

    Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

    Magnitude of velocity = 26.04 m/s

    Angle from positive horizontal axis = 86.85⁰

 So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0
3 years ago
A long line of charge with uniform linear charge density λ1 is located on the x-axis and another long line of charge with unifor
Dahasolnce [82]

Answer:

A.The positive z-direction

Explanation:

We are given that

Linear charge density of long line which is  located on the x-axis=\lambda_1

Linear charge density of another long line which is  located on the y-axis=\lambda_2

We have to find the direction of electric field at z=a on the positive z-axis if \lambda_1 and \lambda_2 are positive.

The direction of electric field  at z=a on the positive z-axis  is positive z-direction .

Because \lambda_1 and \lambda_2 are positive and the electric field is  applied away from the positive charge.

Hence, option A is true.

A.The positive z-direction

6 0
3 years ago
he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?
Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}

With the following condition :

  • \sf{\omega} = angular frequency (rad/s)
  • \sf{\theta} = change of angle value (rad)
  • t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

  • \sf{\theta} = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
  • t = interval of the time = 54.9 s.

What was asked :

  • \sf{\omega} = angular frequency = ... rad/s

Step by step :

\sf{\omega = \frac{\theta}{t}}

\sf{\omega = \frac{2,000 \pi}{54.9}}

\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0
2 years ago
Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r
Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

v=u-gt

u=gt

t=\dfrac{u}{g}

t=\dfrac{13.7}{9.8}

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

h=ut+\dfrac{1}{2}at^2

Here, a = -g

h=ut-\dfrac{1}{2}gt^2

h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0
3 years ago
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