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inessss [21]
3 years ago
14

If the speed of an object is doubled, its kinetic energy will

Physics
2 answers:
otez555 [7]3 years ago
7 0
3. remain the same would be the asnwer
Kamila [148]3 years ago
7 0
I would say 1 or 3
I think this website would help you : )
https://quizlet.com/143048348/year-10-physics-forces-motion-and-energy-flash-cards/

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While in a car at 4.47 meters per second a passenger drops a ball from a height of 0.70 meters above the top of a bucket how far
viktelen [127]

Answer:

1.7 m

Explanation:

v_x = Velocity of ball in x direction = 4.47 m/s

u_y = Velocity of ball in y direction = 0

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

t = Time taken

s_y = Vertical displacement = 0.7 m

s_y=u_yt+\dfrac{1}{2}gt^2\\\Rightarrow 0.7=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{0.7\times 2}{9.81}}\\\Rightarrow t=0.38\ \text{s}

Horizontal displacement is given by

s_x=v_xt\\\Rightarrow s_x=4.47\times 0.38\\\Rightarrow s_x=1.7\ \text{m}

The passenger should throw the ball 1.7 m in front of the bucket.

5 0
3 years ago
An​ airplane, flying horizontally at an altitude of 3 miles​, passes directly over an observer. If the constant speed of the air
natima [27]
I’m not sure if this helped you so here you go

6 0
3 years ago
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc
julia-pushkina [17]

Answer:

E = 2.84 * 10^5 N/C

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

v^2 = u^2 + 2as

(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C

8 0
3 years ago
Find the radius using the centripetal acceleration formula: acceleration = 6.6 x 10^6 m/s^2 velocity = 2,000 rev/s
aksik [14]

the radius is 231

Explanation:

6 0
3 years ago
Hii please help i’ll give brainliest if you give a correct answer please
Diano4ka-milaya [45]

Answer:

A

Explanation:

6 0
2 years ago
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