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garri49 [273]
3 years ago
15

A ball is dropped from a tall bridge of height 48 m. What will the speed of the ball be immediately before it hits the ground, i

n units of m/s?
Physics
2 answers:
soldi70 [24.7K]3 years ago
5 0

Answer:

2700

Explanation:

because calculate the minute1=60×45=2700

Anika [276]3 years ago
4 0

Answer:

v \approx 30.67\,\,\frac{m}{s}

Explanation:

We use the kinematic formula for constantly accelerated motion given below:

v^2=v_0^2+2\,a\Delta x\\v^2=0+2\,g\,(48)\\v^2=940.8\\v \approx 30.67\,\,\frac{m}{s}

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English units of distance include the
Masja [62]

Answer:mile

Explanation: heres a hint think aboyt the distance between your house to school

8 0
3 years ago
A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f = 2.07 Hz. On this tray is an empty cu
Anettt [7]

Answer:

Explanation:

Given

Frequency of SHM is f=2.07\ Hz

Amplitude of SHM is A=3.13\ cm

Cup begins to slip when it overcomes the friction force

Friction force F_s=\mu mg

Applied force F=ma

ma=\mu mg

a=\mu g

and maximum acceleration during SHM is

a=A\omega ^2

a=A(2\pi f)^2

a=3.13\times 10^{-2}\times (2\pi 2.07)^2

a=5.296\ m/s^2

\mu =\frac{a}{g}

\mu =\frac{5.296}{9.8}=0.54

6 0
3 years ago
Read 2 more answers
At the center of the sun, fusion converts hydrogen into
dimaraw [331]
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3 0
3 years ago
If a 0.15 kg ball falls and has a KE of 20 J just before striking the ground, from what height did it fall. A. 1.36m B. 3m C. 13
RUDIKE [14]
According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell. 

Therefore, we can say KE = PE
We know that PE = m·g·h

Which means KE = m·g·h

We can solve for h:

h = KE / m·g
   = 20 / (0.15 · 9.8) 
   = 13.6m

The correct answer is: the ball has fallen from a height of 13.6m.

5 0
3 years ago
Froghopper insects have a typical mass of around 11.3 mg and can jump to a height of 58.8 cm. The takeoff velocity is achieved a
allochka39001 [22]

Answer:

2874.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s

Now H-h = 0.588 - 0.002 = 0.586 m

The final velocity will be the initial velocity

v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.586\\\Rightarrow a=-\frac{0.586\times -9.81}{0.002}\\\Rightarrow a=2874.33\ m/s^2

Acceleration of the frog is 2874.33 m/s²

6 0
3 years ago
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