All categories

1 year ago

Help pls, see picture. Will mark Brainliest

Physics

2 answers:

lilavasa [31]1 year ago

6 0

Answer: The ball is thrown straight UP so when it asked whats the ball's velocity. It'll be graph ii. But.. to me, it seems that graph ii is the only one that provides a true reaction since the ball is thrown straight up. Or it could graph iii assuming the ball is thrown by a regular person. So when it asked for the position, I would pick iii.

Explanation: hope this made sense //Give thanks(and or Brainliest) if helpful (≧▽≦)//

Pani-rosa [81]1 year ago

5 0

Answer:

a ) option 2 is correct

b) -ve acceleration for upward motion ,0 acceleration at top point ,+ve acceleration on downward motion ...

Explanation:

mark me as brainliest ❤️

You might be interested in

(Atwood’s Machine): Two masses, 9 kg and 12 kg, are attached by a lightweight cord and suspended over a frictionless pulley. Whe

Fittoniya [83]

Answer:

Acceleration = 1.428m/s2

Tension = 102.85N

Explanation:

The detailed solution is attached

5 0

2 years ago

Read 2 more answers

Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

7 0

2 years ago

a driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di

daser333 [38]

Answer:

v = 98.75 km/h

Explanation:

Given,

The distance driver travels towards the east, d₁ = 135 km

The time period of the travel, t₁ = 1.5 h

The halting time, tₓ = 46 minutes

The distance driver travels towards the east, d₂ = 215 km

The time period of the travel, t₁ = 2 h

The average speed of the vehicle before stopping

v₁ = d₁/t₁

= 135/1.5

= 90 km/h

The average speed of vehicle after stopping

v₂ = d₂/t₂

= 215/2

= 107.5 km/h

The total average velocity of the driver

v = (v₁ +v₂) /2

= (90 + 107.5)/2

= 98.75 km/h

Hence, the average velocity of the driver, v = 98.75 km/h

7 0

2 years ago

A shopper walks westward 5.4 meters and then eastward 7.8 meters

RoseWind [281]

Answer:

13.2 meters

Explanation:

(5.4) + (7.8)

6 0

2 years ago

In which of the following wave types is the particle motion perpendicular to wave motion?

galben [10]

Answer:

C...................

4 0

2 years ago