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VikaD [51]
3 years ago
12

Give a specific example of a vector quantity. why this type of quantity would be used to describe this event ?

Physics
1 answer:
Anarel [89]3 years ago
3 0

example force. because you can say" I applied  3 newtons downward to the floor".

force has magnitude and direcion

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Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal ener
AVprozaik [17]

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

=  1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

8 0
2 years ago
In an HVAC system, the purpose of pneumatics is to serve as A. a controller B. conditional air flow C. an energy source D. a sou
Naya [18.7K]

Answer: <u><em>A</em></u>

Explanation:

A pneumatic control system uses compressed air as a method of control for HVAC systems. ... Each senor responds to changes in temperature, humidity, and static pressure as examples, to provide feedback in a control loop to open or close the actuator to meet the control set point.

3 0
3 years ago
Homer Agin leads the Varsity team in home runs. In a recent game, Homer hit a 34.5 m/s sinking curve ball head on, sending it of
Aneli [31]

Answer:

–77867 m/s/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration = (final velocity – Initial velocity) /time

a = (v – u) / t

With the above formula, we can obtain acceleration of the ball as follow:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

a = (v – u) / t

a = (–23.9 – 34.5) / 0.00075

a = –58.4 / 0.00075

a = –77867 m/s/s

Thus, the acceleration of the ball is –77867 m/s/s.

3 0
3 years ago
A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.
NemiM [27]

Answer:

\dfrac{K_t}{K_r}=\dfrac{5}{2}

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, \omega=2.8\ rad/s

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

K_t=\dfrac{1}{2}mv^2

K_t=\dfrac{1}{2}m(r\omega)^2

K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2

The rotational kinetic energy of the ball is :

K_r=\dfrac{1}{2}I \omega^2

K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2

K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2

Ratio of translational to the rotational kinetic energy as :

\dfrac{K_t}{K_r}=\dfrac{5}{2}

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

4 0
3 years ago
A gun shoots a bullet with a velocity of 500 m/s. The gun is aimed horizontally and fired from a height of 1.5 m. How far does t
MrMuchimi

The bullet travels a horizontal distance of 276.5 m

The bullet is shot forward with a horizontal velocity u_x. It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>

The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

y=u_yt+\frac{1}{2} gt^2

Substitute 0 m/s for u_y, 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.

y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s

The horizontal distance traveled by the bullet is given by,

x=u_xt

Substitute 500 m/s for u_x and 0.5530s for t.

x=u_xt\\ =(500m/s)(0.5530s)\\ =276.5m

The bullet travels a distance of 276.5 m.


5 0
3 years ago
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