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Softa [21]
2 years ago
9

What is the kinetic energy of a 100 kg object that is moving with a speed of 12.5m/s

Physics
1 answer:
Doss [256]2 years ago
7 0

The kinetic energy of any moving object is

                           (1/2) (mass) (speed²) .

For the object you described, that's

                            (1/2) (100 kg) (12.5 m/s)²

                         =      (50 kg)  (156.25 m²/s²)

                         =              7,812.5 joules  
______________________________

Your attachment is way out of focus, and impossible to read.

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The blue curve is the plot of the data. The straight orange line is tangent to the blue curve at t = 40 s. A plot has the concen
Sever21 [200]

Answer:

  0.00325 moles/liter/second

Explanation:

The tangent line has a slope of (y2 -y1)/(x2 -x1) = (0.35-0.48)/(40-0) = -0.00325.

The rate of the reaction is about 0.00325 moles/liter/second.

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This is the rate of decrease of the concentration of A.

5 0
2 years ago
Read 2 more answers
A particle with a charge of -1.24 x 10"° C is moving with instantaneous velocity (4.19 X 104 m/s)î + (-3.85 X 104 m/s)j. What is
astra-53 [7]

Answer:

(a) F= 6.68*10¹¹⁴ N (-k)

(b) F =( 6.68*10¹¹⁴ i  + 7.27*10¹¹⁴ j  ) N

Explanation

To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:

F=q* v X B Formula (1 )

q: charge (C)

v: velocity (m/s)

B: magnetic field (T)

vXB : cross product between the velocity vector and the magnetic field vector

Data

q= -1.24 * 10¹¹⁰ C

v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j

B  =(1.40 T)i  

B  =(1.40 T)k

Problem development

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =

            = - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)

1T= 1 N/ C*m/s

We apply the formula (1)

  F= 1.24 * 10¹¹⁰ C*  5.39* 10⁴ m/s* N/ C*m/s (-k)

   F= 6.68*10¹¹⁴ N (-k)

a)  vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =

             =( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T

1T= 1 N/ C*m/s

We apply the formula (1)

F= 1.24 * 10¹¹⁰ C*  (  5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s

F =( 6.68*10¹¹⁴  i  + 7.27*10¹¹⁴  j  ) N

8 0
2 years ago
A 57 kg paratrooper falls through the air. How much force is pulling him down?
just olya [345]

Answer:

Explanation:

Force = Mass * acceleration due to gravity.

Given

Mass of the paratrooper = 57kg

Acceleration due to gravity = 9.81m/s²

Required

Force pulling him down

Substitute unto formula;

F = 57 * 9.81

Force = 559.17 N

Hence the force pulling him down is 559.17N

7 0
2 years ago
In 1990, Dave Campos of the United States rode a special motorcycle called the Easyrider at an average speed of 518 km/h. Suppos
maks197457 [2]

The distance travelled during the given time can be found out by using the equations of motion.

The distance traveled during the time interval is "13810.8 m".

First, we will find the deceleration of the motorcycle by using the first <em>equation of motion</em>:

v_f=v_it+at\\\\

where,

vi = initial velocity = (518 km/h)(\frac{1\ h}{3600\ s})(\frac{1000\ m}{1\ km}) = 143.89 m/s

vf = final veocity = 60 % of 143.89 m/s = (0.6)(143.89 m/s) = 86.33 m/s

a = deceleration = ?

t =time interval = 2 min = 120 s

Therefore,

86.33\ m/s = 143.89\ m/s + a(120\ s)\\\\a = \frac{86.33\ m/s - 143.89\ m/s}{120\ s}

a = -0.48 m/s²

Now, we will use the second <em>equation of motion </em>to find out the distance traveled (s):

s = v_it+\frac{1}{2}at^2\\\\s = (143.89\ m/s)(120\ s)+\frac{1}{2}(-0.48\ m/s^2)(120\ s)^2\\\\s = 17266.8\ m - 3456\ m

<u>s = 13810.8 m = 13.81 km</u>

<u />

Learn more about the equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion.

6 0
2 years ago
A kitten sits in a lightweight basket near the edge of a table. A person accidentally knocks the basket off the table. As the ki
Lesechka [4]

Answer:

the answer is B

Explanation: this was actually an ap exam question a few years back. the reason for answer B is that the only force being applied to the kitten is the force of gravity after being pushed.

4 0
2 years ago
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