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kvv77 [185]
3 years ago
15

A ball rolls off a table and it traveling with a horizontal velocity of 2 m/s and 1 point

Physics
2 answers:
morpeh [17]3 years ago
4 0

Answer:

The velocity when the ball hits the ground is obtained using v2. 2 = v1. 2 + 2 g Dy with v1=0 and Dy=h. Thus solving for v2 yields 17.1 m/s v2 = 2 g h =.

21 pages·330 KB

astra-53 [7]3 years ago
4 0

Answer:

I e jofh ndoj eohe nc n very nic e

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if two substances have similar physical properties such as color and texture but different in other physical properties such as
blagie [28]
Since they are made of different chemical make ups, boiling point and density they will most likely be in different groups on the periodic table.
3 0
3 years ago
Car A starts in Sacramento at 11am. It travels along 400 mile route to Los Angeles at 60 mph. Car B starts from Los Angeles at n
damaskus [11]

Answer:

  • 38.89 miles

Explanation:

from the question we have the following:

distance between Sacramento and los angles = 400 miles

speed of car A = 60 mph

start time of car A = 11 am

speed of car B = 75 mph

start time of car B = 12 pm

distance of Fresno from Los Angeles = 150 miles

  • To start off let's allow car A to travel for one hour (from 11 am to 12 pm), during which it would have covered a distance of 60 miles.
  • Now the time would be 12 pm and the distance between the two cars would be 400 - 60 (distance traveled by car A within 11 am to 12 pm) = 340 miles
  • From 12 pm to the time both cars will meet, the distance covered by car A + distance covered by car B would be equal to 340 miles. Therefore
  • Distance covered by car A = speed x time(t) = 60 x t = 60t
  • Distance covered by car B = speed x time(t) = 75 x t = 75t
  • 60t + 75t = 340 miles
  • 135t = 340
  • t = 2.51 hours
  • Recall that at their meeting point, the distance covered by car B = 75t = 75 x 2.62 = 188.89 miles
  • Since Fresno is 150 miles from Los Angeles, car B which is 188.89 miles from Los Angeles at their meeting point would be 188.89 - 150 = 38.89 miles from Fresno
  • 38.89 miles would also be the distance of car A from Fresno since that is their meeting point.

4 0
3 years ago
How much force is required to accelerate 3 kg at 3 m/sec^2
bogdanovich [222]

Answer:

F=?

a=3m/s2

m=3kg

F=ma=[3][3]=9 N

Explanation:

7 0
3 years ago
A race car starting from rest accelerates uniformly at a rate of 3.90 m/s^2 what is the cars speed after it has traveled 200 m.
snow_tiger [21]
The formula for accelerational displacement is at^2/2, so we know that 3.9t^2/2 = 200, or 3.9t^2 = 400. t = \sqrt{400/3.9} \approx 10.12739367, at = v, so 3.9 * 10.12739367 \approx 39.5  m/s
6 0
3 years ago
Read 2 more answers
The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10−31 kg from rest to 3 × 107 m/s within
zaharov [31]

Answer:

Electric field, E = 40608.75 N/C

Explanation:

It is given that,

Mass of electrons, m=9.1\times 10^{-31}\ kg

Initial speed of electron, u = 0

Final speed of electrons, v=3\times 10^7\ m/s

Distance traveled, s = 6.3 cm = 0.063 m

Firstly, we will find the acceleration of the electron using third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(3\times 10^7)^2}{2\times 0.063}

a=7.14\times 10^{15}\ m/s^2

Now we will find the electric field required in the tube as :

ma=qE

E=\dfrac{ma}{q}

E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}

E = 40608.75 N/C

So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.

3 0
3 years ago
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