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Flura [38]
3 years ago
11

When an object threw to the free space to make an angle of 25 degree at an initial speed of 15 m/sec, the ball takes time to rea

ch another position 1.12 sec. Find the travelling path of the object.
Physics
1 answer:
alekssr [168]3 years ago
8 0

Answer:

The horizontal distance traveled by the projectile is 15.23 m.

Explanation:

Given;

angle of projection, θ = 25⁰

initial velocity of the projectile, u = 15 m/s

time of flight, t = 1.12 s

The the travelling path of the object is calculated as the range of the projectile

R = u_x t\\\\R = (15\ Cos \ 25^0) \times 1.12\\\\R = 13.595 \times 1.12\\\\R = 15.23 \ m

Therefore, the horizontal distance traveled by the projectile is 15.23 m.

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galben [10]
Two analogies: 
1. The wires are like hoses full of water. As soon as you turn on the water, water is pushed out the end of the hose. 
2. Although the electrons only move at 1mm/s, the electomagnetic pulse travels through the circuit at near speed of light (varies btw .1c to .9c). It is this pulse that provides the pressure to push the electrons out the end of the wire into the light
5 0
3 years ago
A friend has written to you informing you that she/he giving him/her the name of some place in nepal
sleet_krkn [62]

Answer:

The answer is below

Explanation:

Dear Friend,

I received your letter a few days back. And because I was busy with work, I couldn't write a reply to you immediately. However, here is my well-detailed reply to your request. I hope you find it satisfactory. I am excited to learn that you wish to go on a tour to Nepal. For the advice, here is my suggestion for some places for you to visit.

1. Kathmandu: this is the capital of Nepal. It also doubles as the capital city. The atmosphere there is electric. The streets are wide and there are old monumental buildings there.

2. Bhaktapur: here you find interesting things around. It is the third-largest royal city. Durbar square is a nice place to visit.

3. Pokhara: this place has the largest population and the city is cleaner compared to most other big cities around. It has wonderful places to relax. And if you enjoy hiking this is the place for you.

4. Chitwan National Park: I know you love watching wildlife. So this is a must for you to visit. You get to see amazing animals such as rhinos, Bengal tigers, leopards, etc.

However, due to space and for me not to bore you, here are other honorable mentions

Trekking in the Langtang region, Swayambhunath, known as Monkey Temple. You should also consider the Everest & the Trek to Base Camp.

I hope you will enjoy your stay to the fullest. Don't forget to send me details of your tour after your visit.

I love you, friend.

Yours sincerely,

MyName

4 0
3 years ago
In a motion diagram, the size of the object must be less than the distance moved.
insens350 [35]
Is there  more to this question
4 0
3 years ago
There is a naturally occurring vertical electric field near the Earth’s surface that points toward the ground. In fair weather c
Neko [114]

Answer:

0.00110616446408

Explanation:

r =  Radius of sphere = 11 μm

V = Volume = \dfrac{4}{3}\pi r^3

\rho = Density of pollen = 1000 kg/m³

q = Charge = − 0.550 fC

E = Electric field = 110 N/C

g = Acceleration due to gravity = 9.81 m/s²

Mass is given by

m=\rho V\\\Rightarrow m=1000\times \dfrac{4}{3}\pi (11\times 10^{-6})^3\ kg

Force due to gravity is given by

F_g=mg\\\Rightarrow F_g=1000\times \dfrac{4}{3}\pi (11\times 10^{-6})^3\times 9.81\\\Rightarrow F_g=5.4693494471\times 10^{-11}\ N

Electric force is given by

F_e=qE\\\Rightarrow F_e=-0.55\times 10^{-15}\times 110\\\Rightarrow F_e=-6.05\times 10^{-14}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{6.05\times 10^{-14}}{5.4693494471\times 10^{-11}}\\\Rightarrow \dfrac{F_e}{F_g}=0.00110616446408

The ratio is 0.00110616446408

7 0
3 years ago
As a young boy swings a yo-yo parallel to the ground and above his head, the yo-yo has a centripetal acceleration of 250 m/s2 .
Ahat [919]

Centripetal acceleration ac is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity (tangential speed) v and has the magnitude:

a_c = \dfrac{v^2}{R}

Solving for v:

v = \sqrt{a_c R} =  \sqrt{(250\;m/s^2)(0.50\;m)} = 5\sqrt{5}\;m/s \approx 11.18\;m/s

8 0
2 years ago
Read 2 more answers
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