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3 years ago

Who was the carthaginian general who used elephants to cross the alps in the second punic war

Physics

2 answers:

Sunny_sXe [5.5K]3 years ago

8 0

Hannibal :) hope helped

goldfiish [28.3K]3 years ago

3 0

<span>the answer is Hannibal</span>

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A runner of mass 51.8 kg starts from rest and accelerates with a constant acceleration of 1.31 m/s^2 until she reaches a velocit

Stells [14]

Answer:32.24 s

Explanation:

Given

mass of runner (m)=51.8 kg

Constant acceleration(a)= $1.31 m/s^2$

Final velocity (v)=5.47 m/s

Time taken taken to reach 5.47 m/s

v=u+at

$5.47=0+1.31\times t$

$t=\frac{5.47}{1.31}=4.17 s$

Distance traveled during this time is

$s=ut+\frac{1}{2}at^2$

$s=\frac{1}{2}\times 1.31\times 4.17^2=11.42 m$

So remaining distance left to travel with constant velocity=153.57 m

thus time $=\frac{distance}{speed}$

$t_2=\frac{153.57}{5.47}=28.07 s$

Total time=28.07+4.17=32.24 s

6 0

3 years ago

17. How long does it take a giraffe running at a speed of 33 m/s to run 200 meters !

scZoUnD [109]

Answer:

6600sec

Explanation:

T=sv

T=200m*33m/s

T=6600sec

5 0

2 years ago

25. Which of the following is true for an electromotor? A. It transforms thermal energy to electrical energy. B. It transforms m

Vlad1618 [11]

Answer:

Explanation:

it transforms electrical energy into mechanical energy.

4 0

3 years ago

Two water balloons of mass 0.75 kg collide and bounce off of each other without breaking. Before the collision, one water balloo

bagirrra123 [75]

By the law of momentum conservation:-
=>m¹u¹ + m²u² = m1v1 + m²v² {let East is +ve}
=>u¹ + u² = v¹ + v² {as m1=m2}
=>3.5 - 2.75 = v1-1.5
<span> =>v¹ = 2.25 m/s (East) </span>

5 0

3 years ago

When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a

sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = $\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}$

= $4.59\times 10^{-19} \ J$

or,

= $\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}$

= $2.87 \ eV$

(b)

As we know,

⇒ $Vq=\frac{hc}{\lambda}-\Phi_0$

By substituting the values, we get

⇒ $1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0$

⇒ $\Phi_0=2.3\times 10^{-19} \ J$

or,

⇒ $=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}$

⇒ $=1.4375 \ eV$

5 0

3 years ago