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2 years ago

A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?

1 answer:

6 0

<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>

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The diagram shows monochromatic light passing through two openings.

rusak2 [61]

Answer:

constructive interference in which waves strengthen each other

Explanation:

Some definitions:

- Costructive interference occurs when two (or more) waves meet each other in phase, so with same displacement at the same point. In such situation, the two waves strengthen each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves

- Destructive interference occurs when two waves meet each other in anti-phase, so with opposite displacement at the same point. In such situation, the two waves cancel each other out, and the amplitude of the resultant wave is the difference of the amplitudes of the individual waves (which means zero if the two waves are identical)

For light waves interfering with each other, 'white' means costructive interference, while 'black' means destructive interference (because black is absence of colors, so this means that the waves cancel each other out). In this problem, we see that point X, Y and X are white, therefore they are point of constructive interference, where the waves strengthen each other.

5 0

2 years ago

Read 2 more answers

A turntable reaches an angular speed of "45 rev/min" in "4.10 s" after being turned on. What is its angular acceleration?

Marina86 [1]

Answer:

1.15 rad/s²

Explanation:

given,

angular speed of turntable = 45 rpm

= $45\times \dfrac{2\pi}{60}$

= $4.71\ rad/s$

time, t = 4.10 s

initial angular speed = 0 rad/s

angular acceleration.

$\alpha = \dfrac{\omega_f-\omega_0}{t}$

$\alpha = \dfrac{4.71-0}{4.10}$

$\alpha = 1.15\ rad/s^2$

Hence, the angular acceleration of the turntable is 1.15 rad/s²

3 0

2 years ago

A ball is tossed from an upper-story window of a building. the ball is given an initial velocity of 8.00 m/s at an angle of 20.0

otez555 [7]

A) The motion of the ball consists of two indipendent motions on the horizontal (x) and vertical (y) axis. The laws of motion in the two directions are:
$x(t)=v_0 \cos \alpha t$
$y(t)=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
where
- the horizontal motion is a uniform motion, with constant speed $v_0 \cos \alpha$ , where $v_0 = 8.00 m/s$ and $\alpha=20.0^{\circ}$
- the vertical motion is an uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ , initial position h (the height of the building) and initial vertical velocity $v_0 \sin \alpha$ (with a negative sign, since it points downwards)

The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
$x(3.00 s)=v_0 \cos \alpha t=(8 m/s)(\cos 20^{\circ})(3.0 s)=22.6 m$

b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
$0=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
which becomes
$h=(8 m/s)(\sin 20^{\circ})(3.0 s)+ \frac{1}{2}(9.81 m/s^2)(3.0 s)^2=52.3 m$

c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
$y(t)=h-10$
If we substitute this into the equation of y(t), we have
$h-10 = h-v_0 \sin \alpha t- \frac{1}{2}gt^2$
$\frac{1}{2}gt^2+v_0 \sin \alpha t -10 =0$
$4.9 t^2 +2.74 t-10 =0$
whose solution is $t=1.18 s$ (the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.

4 0

3 years ago

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.

trapecia [35]

Answer

given,

diameter of planet = 1.8 x 10⁷ m

radius of planet = 0.9 x 10⁷ m

time period = 22.3 hours

the planet orbits 2.2 x 10¹¹ m period of 402 earth days.

acceleration= 12.2 m/s²

we know

$g = \dfrac{GM_{p}}{r^2}$

$M_{p} = \dfrac{gr^2}{G}$

$M_{p} = \dfrac{12.2 \times (0.9 \times 10^7)^2}{6.67 \times 10^{-11}}$

M_p = 1.48 x 10²⁵ Kg

b) Formula to calculate the mass of star

$M_{s} = \dfrac{4\pi^2}{GT_s^2}(R^3)$

$M_{s} = \dfrac{4\pi^2}{6.67 \times 10^{-11} \times (402 \times 86400)^2}((2.2 \times 10^{11})^3)$

M_s = 5.22 x 10³³ Kg

3 0

2 years ago

Two infinite planes of charge lie parallel to each other and to the yz plane. One is at x = -5 m and has a surface charge densit

skad [1K]

Answer:

a) -180.7 kN/C

b) -474.3 kN/C

c) 180.7 kN/C

Explanation:

For infinite planes the electric field is constant on each side, and has a value of:

E = σ / (2 * e0) (on each side of the plate the field points in a different direction, the fields point towards positive charges and away from negative charges)

The plate at -5 m produces a field of:

E1 = 2.6*10^-6 / (2 * 8.85*10^-12) = 146.8 kN/C into the plate

The plate at 3 m:

E2 = 5.8*10^-6 / (2 * 8.85*10^-12) = 327.5 kN/C away from the plate

At x < -5 m the point is at the left of both fields

The field would be E = 146.8 - 327.5 = -180.7 kN/C

At -5 m < x < 3 m, the point is between the plates

E = -146.8 - 327.5 = -474.3 kN/C

At x > 3 m, the point is at the right of both plates

E = -146.8 + 327.5 = 180.7 kN/C

3 0

2 years ago