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notka56 [123]
4 years ago
11

All atoms of the same element must have the same number of

Physics
2 answers:
mamaluj [8]4 years ago
6 0

 the answer protons Explanation:

DIA [1.3K]4 years ago
3 0
Since isotopes of any given element all contain the same number of protons,
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Danny is competing in the high jump. When he is in the air, his body has _______ energy due to its height, and it has _______ en
shusha [124]

Answer:

Gravitational potential energy

kinetic energy

5 0
3 years ago
A ball is thrown vertically downward from the top of a 37.4-m-tall building. The ball passes the top of a window that is 15.4 m
Effectus [21]

Answer:

v= 20.8 m/s

Explanation:

  • Assuming no other forces acting on the ball, from the instant that is thrown vertically downward, it's only accelerated by gravity, in this same direction, with a constant value of -9.8 m/s2  (assuming the ground level as the zero reference level and the upward direction as positive).
  • In order to find the final speed 2.00 s after being thrown, we can apply the definition of acceleration, rearranging terms, as follows:

       v_{f} = v_{o} + a*t  =  v_{o} + g*t (1)

  • We have the value of t, but since the ball was thrown, this means that it had an initial non-zero velocity v₀.
  • Due to we know the value of the vertical displacement also, we can use the following kinematic equation in order to find the initial velocity v₀:

        \Delta y = v_{o} *t + \frac{1}{2} * a* t^{2}  (2)

  • where Δy = yf - y₀ = 15.4 m - 37.4 m = -22 m (3)
  • Replacing by the values of Δy, a and t, we can solve for v₀ as follows:

       v_{o} = \frac{(\Delta y- \frac{1}{2} *a*t^{2})}{t} = \frac{-22m+19.6m}{2.00s} = -1.2 m/s (4)

  • Replacing (4) , and the values of g and t in (1) we can find the value that we are looking for, vf:

       v_{f} = v_{o} + g*t  = -1.2 m/s - (9.8m/s2*2.00s) = -20.8 m/s (5)

  • Therefore, the speed of the ball (the magnitude of the velocity) as it passes the top of the window is 20.8 m/s.
4 0
3 years ago
A robot is on the surface of Mars. The angle of depression from a camera in the robot to a rock on the surface of Mars is 13.69
ra1l [238]

Answer:

The distance between the camera and the rock is 836.6 cm

Explanation:

A right triangle is formed where the hypotenuse (h) is the distance between the rock and the camera. One of the leg (l) is the distance between the camera and the surface. The angle between the hypotenuse and this leg is α = 90° - 13.69° = 76.31°. By definition:

cos α = adjacent/hypotenuse

cos(76.31) = 198.0/h

h = 198.0/cos(76.31)

h = 836.6 cm

3 0
4 years ago
A 1.15-kg grinding wheel 22.0 cm in diameter is spinning counterclockwise at a rate of 20.0 revolutions per second. When the pow
malfutka [58]

Answer:

Angular acceleration = -1.57 Rad/s^2

Number of revolutions = 800 revolutions

Explanation:

Mass = 1.15kg

diameter of the grinding wheel = 22.0cm

angular speed = 20 rev/s

1 Rev/s = 6.283 rad/s

20 Rev/s = 20 * 6.283 = 125.664 rad/s

Angular acceleration = Angular speed/ time taken for for the wheel rotation

Angular accleration = 125.664/80

Angular acceleration = -1.57 rad/s^2 (since the wheel is rotating counterclockwise)

number of revs = 0.5∝t²

number of revs = 0.5 * (1.57/2π) * 80²

number of revs = 800 revolutions

6 0
3 years ago
Read 2 more answers
Parasaurolophus was a dinosaur whose distinguishing feature was a hollow crest on the head. The 1.5-m-long hollow tube in the cr
Inessa05 [86]

Answer:

58.33 Hz

175 Hz

291.67 Hz

Explanation:

L = Length of tube = 1.5 m

v = Speed of sound in air = 350 m/s

The first resonant frequency is given by

f_1=\dfrac{v}{4L}\\\Rightarrow f_1=\dfrac{350}{4\times 1.5}\\\Rightarrow f_1=58.33\ Hz

The first resonant frequency is 58.33 Hz

The second resonant frequency is given by

f_2=3\dfrac{v}{4L}\\\Rightarrow f_2=3\dfrac{350}{4\times 1.5}\\\Rightarrow f_2=175\ Hz

The first resonant frequency is 175 Hz

The third resonant frequency is given by

f_3=5\dfrac{v}{4L}\\\Rightarrow f_3=5\dfrac{350}{4\times 1.5}\\\Rightarrow f_3=291.67\ Hz

The first resonant frequency is 291.67 Hz

8 0
4 years ago
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