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galben [10]
3 years ago
13

The chances of being involved in a fatal crash is higher at night then during the day

Physics
1 answer:
Ierofanga [76]3 years ago
3 0
Indeed because some leave headlights off and ignore that fact since there are street lights around.
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Prove the identity <br>Trigonometry grade 10​
g100num [7]

Answer:

and is in photo given.I didn't get time to type.

4 0
3 years ago
Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the p
PSYCHO15rus [73]

Answer:

Energy density will be 14.73 J/m^3

Explanation:

We have given capacitance C=225\mu F=225\times 10^{-6}F

Potential difference between the plates = 365 V

Plate separation d = 0.200 mm 0.2\times 10^{-3}m

We know that there is relation between electric field and potential

E=\frac{V}{d}, here E is electric field, V is potential and d is separation between the plates

So E=\frac{V}{d}=\frac{365}{0.2\times 10^{-3}}=1825000N/C

Energy density is given by E=\frac{1}{2}\varepsilon _0E^2=\frac{1}{2}\times 8.85\times 10^{-12}\times (1.825\times 10^6)^2=14.73J/m^3

5 0
3 years ago
Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke
umka2103 [35]

Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

        Thrust = v_{e}  \frac{dM}{dt}

        v_{f}-v₀ = v_{e} ln ( \frac{M_{o} }{M_{f}} )

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

       M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

     

The final mass is the mass of the engines + the mass of the rocket

      M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

       thrust = 5.26 N

       t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

          thrust = v_{e} \frac{M_{f} - M_{o}  }{t_{f} - t_{o}  }

          v_{e} = thrust  \frac{\Delta t}{\Delta M}

          v_{e} = 5.26 \frac{1.90}{0.080 -0.0927}

          v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

            v_{f}-0 = v_{e} ln ( \frac{M_{o} }{M_{f} } )

we calculate

            v_{f} = 786.93 ln (0.0927 / 0.080)

            v_{f} = 115.95 m / s

5 0
3 years ago
I will mark brainlist
tensa zangetsu [6.8K]

Answer:

False

Explanation:

A wave is a disturbance that transfers energy from one place to another without transferring matter.

6 0
2 years ago
Read 2 more answers
Calcular la aceleración que produce una fuerza de 40 N sobre un cuerpo con 88 Kg de masa. Expresar el resultado en m⁄s^2 *
tigry1 [53]

Answer:

a = 0.45 m/s²

Explanation:

The given question is ''Calculate the acceleration that produces a force of 40 N on a body with 88 kg of mass".

Given that,

Force, F = 40 N

Mass of the body, m = 88 kg

The net force acting on the body is given by :

F = ma

Where

a is the acceleration of the body

a=\dfrac{F}{m}\\\\a=\dfrac{40\ N}{88\ kg}\\\\a=0.45\ m/s^2

So, the required acceleration is 0.45 m/s².

4 0
3 years ago
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