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kherson [118]
3 years ago
9

3. A particle of charge +7.5 µC is released from rest at the point x = 60 cm on an x-axis. The particle begins to move due to th

e presence of a charge ???? that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved 40 cm if a) ???? = +20 µC and b) ???? = −20 µC?
Physics
1 answer:
Fed [463]3 years ago
3 0

Answer:

HSBC keen vs kg get it yyyyyuuy

Explanation:

hgccccxfcffgbbbbbbbbbbghhyhhhgdghcjyddhhyfdghhhfdgbxbbndgnncvbhcxgnjffccggshgdggjhddh

nnnbvvvvvggfxrugdfutdfjhyfggigftffghhjjhhjyhrdffddfvvvvvvvvvvvbbbbbbbbbvvcxccghhyhhhjjjhjnnnnnnnnnnnnnbhbfgjgfhhccccccvvjjfdbngxvncnccbnxcvbchvxxghfdgvvhhihbvhbbhhvxcgbbbcxzxvbjhcxvvbnnxvnn

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If an object is thrown upward with an initial velocity of 128 ​ft/second, then its height after t seconds is given by the follow
IrinaK [193]

Answer:

The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

Explanation:

Given that,

Initial velocity u= 128 ft/sec

Equation of height

h = 128t-32t^2....(I)

(a). We need to calculate the maximum height

Firstly we need to calculate the time

\dfrac{dh}{dt}=0

From equation (I)

\dfrac{dh}{dt}=128-64t

128-64t=0

t=\dfrac{128}{64}

t=2\ sec

Now, for maximum height

Put the value of t in equation (I)

h =128\times2-32\times4

h=128\ ft

(b). The number of seconds it takes the object to hit the ground.

We know that, when the object reaches ground the height becomes zero

128t-32t^2=0

t(128-32t)=0

128=32t

t=4\ sec

Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

3 0
3 years ago
The amount of fluid displaced by a submerged object depends on its
Daniel [21]
The amount of fluid displaced by a submerged object depends on its volume.
7 0
3 years ago
Read 2 more answers
A force vector F1 points due east and has a magnitude of 200N. A second force F2 is added to F1. The resultant of the two vector
PilotLPTM [1.2K]

Answer:

The second vector \vec{F_2} points due West with a magnitude of 600N

Explanation:

The original vector \vec{F_1} points with a magnitude of 200N due east, the Resultant vector \vec{R} points due west (that's how east/west direction can be interpreted, from east to west) with a magnitude of  400N. If we choose East as the positive direction and West as the negative one, we can write the following vectorial equation:

\vec{F_1}+\vec{F_2}=\vec{R}\implies\vec{F_2}=\vec{R}-\vec{F_1}=-400N-200N=-600N

With the negative sign signifying that the vector points west.

3 0
3 years ago
A Chinook (King) salmon (Genus Oncorynchus) can jump out of water with a speed of 6.75 m / s . If the salmon is in a stream with
Pepsi [2]

Answer:

The maximum height that the fish can jump is 2.19 m.

Explanation:

Hi there!

Please, see the attached figure for a better understanding of the problem.

The motion of the salmon is a parabolic one because when it jumps, it already has a horizontal velocity (see figure).

The position and velocity vectors of the salmon at a time t, can be calculated as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position of the salmon at time t.

x0 = initial horizotal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration of gravity.

Looking at the figure, notice that at the maximum height, the vertical velocity is zero (because the velocity vector is horizontal). Using the equation of the vertical component of the velocity, we can obtain the time at which the salmon is at its maximum height:

vy = v0y + g · t

To find the initial vertical velocity, v0y, let´s look at the figure. Notice that the initial velocity is the hypotenuse of the triangle formed with the horizontal velocity and the vertical velocity. Then:

v0² = v0x² + v0y²

Solving for v0y:

v0y = √(v0² - v0x²)

v0y = √((6.75 m/s)² - (1.65 m/s)²)

v0y = 6.55 m/s

Now, using the equation of the vertical component of the velocity at the maximum height (vy = 0):

vy = v0y + g · t

0 = 6.55 m/s + (-9.8 m/s²) · t

-6.55 m/s / -9.8 m/s² = t

t = 0.67 s

Now, using the equation of the vertical position at t = 0.67 s, we can find the maximum height:

y = y0 + v0y · t + 1/2 · g · t²

y = 0 m + 6.55 m/s · 0.67 s + 1/2 · (-9.8 m/s²) · (0.67 s)²

y = 2.19 m

The maximum height that the fish can jump is 2.19 m.

4 0
3 years ago
Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ
Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of   opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

Fn_{3} = \sqrt{405^{2}+701.5^{2}  }

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0
3 years ago
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