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kondor19780726 [428]
3 years ago
7

A train travels 76 kilometers in 2 hours and then 54 kilometers in 5 hours .What is the average speed ?

Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
7 0
V = d ÷ t --> bc d=vt
V = (76+54)÷(2+5) = 130÷7 = 18.57km/hr
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A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the hori
EastWind [94]

Answer:

the child is 1.581 m far from the fence

Explanation:

The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.

From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:

x - x_o = u_xt

\mathtt{x = u_xt  \ \  \ since (x_o = 0)}  ---- (1)

the equation of the motion y is :

\mathtt{y - y_o =u_yt - 0.5 gt^2}

\mathtt{y = u_yt-4.9t^2     \ \ \  since (y_o =0)}

\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2        }

\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}

\mathtt{1 = 5.14t - 4.9t^2}

\mathtt{4.9t^2 - 5.14t +1 = 0}

By using the quadratic formula, we have;

\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}}     }

where;

a = 4.9,   b = -5.14     c = 1

= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}}     }

= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}}     }

= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}}     }

= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8}  \  \ OR \  \  \dfrac{ 5.14- \sqrt{6.8196}}{9.8}}    }

= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8}  \  \ OR \  \  \dfrac{ 5.14- 2.6114}{9.8}}    }

= \mathtt{ \dfrac{ 7.7514}{9.8}  \  \ OR \  \  \dfrac{ 2.5286}{9.8}}    }

= \mathbf{ 0.791 \  \ OR \  \  0.258}    }

In as much as the ball is traveling upward, then we consider t= 0.258sec.

From equation (1)

\mathtt{x = u_x(0.258)}

\mathtt{x = ucos 40^0 (0.258)}

\mathtt{x = 8 \ cos 40^0 (0.258)}

\mathbf{x = 1.581  \ m}

Thus, the child is 1.581 m far from the fence

6 0
3 years ago
A light wave travels through space at a speed of 3 x 108 m/s. If the wavelength of some light wave is 2 x10-3 m, what is the fre
kifflom [539]

Here is your answer

b) \huge 1.5× {10}^{11} Hz

REASON :

We know that

Velocity= Frequency× Wavelength

So,

Frequency= Velocity/wavelength

Here,

V= 3× 10^8 m/s

Wavelength= 2×10^-3 m

Hence,

Frequency= 3×10^8/2×10^-3

= 3/2 × 10^11

= 1.5× 10^11 Hz

HOPE IT IS USEFUL

4 0
3 years ago
A 2-f and a 1-f capacitor are connected in series and a voltage is applied across the combination. the 2-f capacitor has:_______
VARVARA [1.3K]

Half the potential difference of the the1-µF

A circuit must have a capacitance of 2 F across a 1 kV potential difference for an electrical technician. He has access to a sizable number of 1F capacitors, each of which can sustain a potential difference of no more than 400 V. Please suggest a configuration that uses the fewest capacitors possible.

The 2-mu F capacitor has the following characteristics: none of the aforementioned; half the charge of the 1-mu F capacitor; twice the charge of the 1-mu F capacitor; and half the potential difference of the 1-mu F capacitor.

Q = C V, C = Capacitance of the capacitor gives the charge stored by a capacitor with an applied voltage V. V is the applied voltage.

Learn more about capacitor brainly.com/question/21851402

#SPJ4

7 0
1 year ago
Read 2 more answers
If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?
vodomira [7]

<u>Answer:</u>

 Option A is the correct answer.

<u>Explanation:</u>

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

 First walking 1.2 km north,  displacement = 1.2 j km

 Secondly 1.6 km east, displacement = 1.6 i km

 Total displacement = (1.6 i + 1.2 j) km

 Magnitude = \sqrt{1.2^2+1.6^2} = 2 km

 Angle of resultant with positive X - axis = tan^{-1}(1.2/1.6)=36.87^0 = 36.87⁰ east of north.

 

5 0
2 years ago
A hot air balloon has a volume of 65,00 cubic feet of hot air. The balloon is at sea level on a standard day. If the temperature
seraphim [82]
If you heat that air by 100 degrees F, it weighs about 7 grams less. Therefore, each cubic foot of air contained in a hot air balloon can lift about 7 grams. That's not much, and this is why hot air balloons are so huge -- to lift 1,000 pounds, you need about 65,000 cubic feet of hot air.
6 0
3 years ago
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