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Effectus [21]
3 years ago
10

A firework is launched with a force of 700 N and a momentum of 200 kg-m/s. How much time before is explodes?

Physics
1 answer:
8_murik_8 [283]3 years ago
7 0

Answer:

t = 0.28 seconds

Explanation:

Given that,

Force acting on a firework, F = 700 N

The momentum of the firework, p =200 kg-m/s

We need to find the time before it explodes. Ket the time be t. We know that, the rate of change of momentum is equal to external frce. So,

F=\dfrac{P}{t}\\\\t=\dfrac{P}{F}\\\\t=\dfrac{200}{700}\\t=0.28\ s

So, the required time is equal to 0.28 seconds.

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The rotating blade of a blender turns with constant angular acceleration of 1.51 rad/s^2. How much time does it take to reach an
yawa3891 [41]

time to reach an angular velocity of 36.8 is 24.370 s.

<h3></h3><h3>What is angular acceleration?</h3>

The temporal rate at which angular velocity changes is referred to as angular acceleration. Naturally, there are two forms of angular acceleration, referred to as spin angular acceleration and orbital angular acceleration, just as there are two types of angular velocity, namely spin angular velocity and orbital angular velocity. As opposed to orbital angular acceleration, which is the angular acceleration of a point particle around a fixed origin, spin angular acceleration describes the angular acceleration of a rigid body about its centre of rotation.

w(t) = w(0) + α*t

also w(0) =0

=> time = 36.8/1.51= 24.370 s

to learn more about angular acceleration go to - brainly.com/question/21278452

#SPJ4

6 0
2 years ago
Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circu
inysia [295]

Answer:

v  =  1,582 \ \frac{m}{s}

Explanation:

We know that for circular motion the centripetal acceleration a_c is:

a_c = \frac{v^2}{r}

where v is the speed and r is the radius.

The centripetal acceleration for the astronaut must be the gravitational acceleration due to the gravity, as there are no other force. So

a_c = 1.27 \frac{m}{s^2}.

The radius of the orbit must be the radius of the Moon, plus the 270 km above the surface

r = 1.7 * 10^6 \ m + 270  \ km

r = 1.7 * 10^6 \ m + 270 * 10^3 \ m

r = 1.7 * 10^6 \ m + 0.270 * 10^6 \ m

r = 1.97 * 10^6 \ m

We can obtain the speed as:

v^2  = a_c r

v  = \sqrt{a_c r}

v  = \sqrt{1.27 \frac{m}{s^2} * 1.97 * 10^6 \ m}

v  = \sqrt{ 2.509 \ 10^6 \ \frac{m^2}{s^2}}

v  =  1.582 \ 10^3 \ \frac{m}{s}

v  =  1,582 \ \frac{m}{s}

And this is the orbital speed.

7 0
3 years ago
The A string of a violin is a little too tightly stretched. Beats at 4.00 per second are heard when the string is sounded togeth
Basile [38]

Answer:

T=2.5*10^{-3}s

Explanation:

From the question we are told that:

Beat frequency F_b=4

Frequency F=400Hz

Generally the equation for Frequency of the violin is mathematically given by

 f_v=F_b+F

 f_v=4+400Hz

 f_v=404Hz

Therefore the period of the violin string oscillations is

 T=\frac{1}{f_v}

 T=\frac{1}{404}

 T=2.5*10^{-3}s

3 0
3 years ago
What is the voltage drop across the 10.0 2 resistor?
amid [387]

Answer:

the answer is equal to 2.00v

4 0
2 years ago
Imagine that an electron in an excited state in a nitrogen molecule decays to its ground state, emitting a photon with a frequen
mash [69]
Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.

The energy of the emitted photon is given by:
E=hf
where
h is the Planck constant
f is the photon frequency
Substituting f=8.88 \cdot 10^{14}Hz, we find
E=hf=(6.6 \cdot 10^{-34} Js)(8.88 \cdot 10^{14} Hz)=5.86 \cdot 10^{-19} J

This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
\Delta E= -5.86 \cdot 10^{-19} J
6 0
3 years ago
Read 2 more answers
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