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Helga [31]
2 years ago
15

Bianca is standing at x =600m. Firecracker 1, at the origin, and firecracker 2, at x =900m, explode simultaneously. The flash fr

om firecracker 1 reaches Bianca's eye at 4.0 μs . Part A At what time does she see the flash from firecracker 2?
Physics
1 answer:
agasfer [191]2 years ago
4 0

Answer:3 \mu s

Explanation:

Given

Bianca is at x=600 m

i.e. distance between origin and Bianca is 600 m

time taken to reach Bianca eyes is

t=\frac{600}{speed\ of\ light}

t=\frac{600}{3\times 10^8}

t=2\times 10^{-6} s

t=2 \mu s

i.e. Cracker exploded at t=2\mu s because it is observed at t=4\mu s

Time taken by second cracker flash to reach Bianca eyes

t_2=\frac{300}{3\times 10^8}

t_2=10^{-6}

t_2=1 \mu s

Therefore it will be observed at t=3 \mu s

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The ball of dough hits the floor and does not rebound.
tekilochka [14]

Answer:

When the ball goes down its mechanical energy is conserved, ust before touching the ground all the energy is kinetic

When the ball touches the floor, energy has been converted into potential and heat, by the deformation of the ball.

Explanation:

When the ball goes down its mechanical energy is conserved, this is the power energy due to the height it is converted into kinetic energy to medicad that falls, just before touching the ground all the energy is kinetic.

When the ball touches the floor, the kinetic energy is not conserved, but if we define a system formed by the ball and the floor, the amount of movement is conserved, this being an inelastic shock, because the bla and the floor are stuck, so which energy has been converted into potential and  energized and heat by the deformation of the ball.

   Consequently all the mechanical energy that the ball brings before reaching the ground was converted into potential energy and heat during the crash.

8 0
2 years ago
A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 16.0 s before the package strikes the ground, how h
pashok25 [27]
Well we know acceleration from free fall due to gravity is 9.8m/s^2

Lay out

S = displacement is what we need

U

V = 1.5m/s

A = 9.8m/s2

T = 16.0s

Use the equation s=vt-1/2at^2

Where a = acceleration t= time and v= velocity

Sub in the values to get displacement or height from ground

= -1230.4 metres which would be positive as you’re measuring distance (scalar quantity) so it’s 1230.4 metres
8 0
2 years ago
A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
2 years ago
You want to create a spotlight that will shine a bright beam of light with all of the light rays parallel to each other. you hav
Tema [17]
At the focal length of the mirror. The reason why is when parallel light rays hit the mirror, it converges at the focal point. So if we want parallel light rays, it would make sense to put it at the focal point.<span />
5 0
2 years ago
Read 2 more answers
You hang an object with mass m = 0.380 kg from a vertical spring that has negligible mass and force constant k = 60.0 N/m. You p
joja [24]

Answer:

a) 0.500 s

b) greater than 0.500 s

c) greater than 0.500 s

Explanation:

The time period of an oscillating spring-mass system is given by:

T=2\pi \sqrt{\frac{m}{k}}

where, m is the mass and k is the spring constant.

a) As the period of oscillation does not depend on the distance by which the mass is pulled, the period would remain same as 0.500 s for the given system.

b) As the period varies inversely with the square root of spring constant, so with the decrease in the spring constant, the period would increase. So, the new period would be greater than 0.500 s.

c) As the period varies directly with the square root of mass, so with the increase in mass, the period will also increase. The new period will be greater than 0.500 s.

8 0
2 years ago
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