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asambeis [7]
3 years ago
6

Two bowling balls collide. A red 5 kg bowling ball is traveling with a velocity of 1 m/s to the left. A blue 4 kg bowling ball i

s traveling with a velocity of 2 m/s to the right. The 5 kg ball moves 0.8 m/s to the right after the collision. What is the final velocity of the 4kg ball?
Physics
1 answer:
lana66690 [7]3 years ago
6 0

Answer:

idk use google

Explanation:

google

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Ranboo <br><br> ⏚⟒⟒⌿<br><br> :) &lt;3 have a good day
bearhunter [10]
Ranboo oobnar have a good day
4 0
3 years ago
Read 2 more answers
At some instant and location, the electric field associated with an electromagnetic wave in vacuum has the strength 57.5 V/m. Fi
n200080 [17]

Answer:

1.9167\times 10^{-7}\ T

2.9247498849\times 10^{-8}\ J/m^3

8.7742496547\ W/m^2

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

E = Electric field = 57.5 V/m

c = Speed of light = 3\times 10^8\ m/s

Magnetic field is given by

B=\dfrac{E}{c}\\\Rightarrow B=\dfrac{57.5}{3\times 10^8}\\\Rightarrow B=1.9167\times 10^{-7}\ T

The magnetic field strength is 1.9167\times 10^{-7}\ T

Energy density is given by

u=\dfrac{1}{2}\epsilon_0E^2+\dfrac{1}{2\mu_0}B^2\\\Rightarrow u=\dfrac{1}{2}\times 8.85\times 10^{-12}\times 57.5^2+\dfrac{1}{2\times 4\pi \times 10^{-7}}(1.9167\times 10^{-7})^2\\\Rightarrow u=2.9247498849\times 10^{-8}\ J/m^3

The energy density is 2.9247498849\times 10^{-8}\ J/m^3

Power flow per unit area is given by

\dfrac{P}{A}=uc\\\Rightarrow \dfrac{P}{A}=2.9247498849\times 10^{-8}\times 3\times 10^8\\\Rightarrow \dfrac{P}{A}=8.7742496547\ W/m^2

Power flow per unit area is 8.7742496547\ W/m^2

7 0
3 years ago
A woman rides a carnival Ferris wheel at radius 16 m, completing 4.5 turns about its horizontal axis every minute. What are (a)
fredd [130]

Answer:

a)13.33s

b)at highest point, the centripetal acceleration has its direction at downward path towards the center of the circular path, and the radius vector has its direction upward. Then acceleration=3.555m/s^2

c)c)at lowest point the centripetal acceleration has its direction upward towards the center of the circular path, and the radius vector has its direction downward Then acceleration=3.555m/s^2

Explanation

a)number of turns= 4.5

radius= 16m

We know that the period is the time taken by the wheel to complete one turn which can be calculated using below expresion

T= t/n

Where T= period of motion

t= Time taken by the wheel to finish n turns where our n= 4.5

T= (1×60)/4.5= 13.33s

Hence the period is 13.33s

Then the speed of the woman v= 2πr/T

v= (2×π×16)/13.33

v=7.5417m/s

b)at highest point, the centripetal acceleration has its direction at downward path towards the center of the circular path, and the radius vector has its direction upward.

a= v^2/r

Where r= radius

v= speed of the woman= 7.5417m/s

a=(7.5417m/s)^2/16

a=3.555m/s^2

The centripetal acceleration and radius vector are in opposite direction

c)at lowest point the centripetal acceleration has its direction upward towards the center of the circular path, and the radius vector has its direction downward

The magnitude of the acceleration is calculated below

a= v^2/r

Where r= radius

v= speed of the woman= 7.5417m/s

a=(7.5417m/s)^2/16

a=3.555m/s^2

5 0
3 years ago
You overindulged on a delicious dessert, so you plan to work off the extra calories at the gym. To accomplish this, you decide t
iragen [17]

Answer:

17086

Explanation:

We are given that

Weight=5 kg

Distance of elbow from the weight=d=35 cm=\frac{35}{100}=0.35 m

1 m=100 cm

g=9.8m/s^2

Work done  each time on the weight=mgh=5\times 9.8\times 0.35=17.15 J

20%  means 0.20 of food energy=Work done each time on weight

Total work done in each time=\frac{17.15}{0.2}=85.75J

1 food calorie=4186 J

350 food calories=350\times 4186 J

Number of lift needed=\frac{350\times 4186}{85.75}=17086

Hence, you must do 17086 arm raises to work off 350 food calories.

6 0
4 years ago
Why is the escape speed for a spacecraft independent of the spacecraft's mass?
scoray [572]
In order to escape the gravitational pull of our planet, any object must have an escape velocity of 7 km/s  or more, anything lower than that will be slowed down by the pull of gravity, and will eventually returned to the surface of our planet. It is independent of mass, any lighter or heavier object must attain the required escaped velocity to reach space.
7 0
4 years ago
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