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3 years ago

1. A ski-plane with a total mass of 1200 kg lands towards the west on a frozen lake at 30.0

Physics

1 answer:

igor_vitrenko [27]3 years ago

6 0

Answer:

d = 229.5 m

Explanation:

It is given that,

Total mass of a ski-plane is 1200 kg

It lands towards the west on a frozen lake at 30.0 m/s.

The coefficient of kinetic friction between the skis and the ice is 0.200.

We need to find the distance covered by the plane before coming to rest. In this case,

$\mu mg=ma\\\\a=\mu g\\\\a=0.2\times 9.8\\\\a=1.96\ m/s^2$

It is decelerating, a = -1.96 m/s²

Now using the third equation of motion to find the distance covered by the plane such that :

$v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(30)^2}{2\times -1.96}\\\\d=229.59\ m$

So, the plane slide a distance of 229.5 m.

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The driver of a car travels at 90 km / h, observes some children playing on the road 50 m away, and applies the brakes, managing

Orlov [11]

Answer:

13,750 N

Yes

Explanation:

Given:

v₀ = 90 km/h = 25 m/s

v = 0 m/s

t = 4 s

Find: a and Δx

a = Δv / Δt

a = (0 m/s − 25 m/s) / (4 s)

a = -6.25 m/s²

F = ma

F = (2200 kg) (-6.25 m/s²)

F = -13,750 N

Δx = ½ (v + v₀) t

Δx = ½ (0 m/s + 25 m/s) (4 s)

Δx = 50 m

6 0

3 years ago

Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

5 0

3 years ago

A disc is thrown through the air for 1.5 min with a power output of 12.5 W. How much work is done when throwing the disc?

klasskru [66]

Answer:

work = 1125 [J]

Explanation:

To solve this problem we must remember the definition of power, which is defined as the relationship between work and time. The power can be calculated using the following equation:

Power = work/time

Power = 12.5 [w]

work = joules [J]

time = 1.5 [min] = 90 [s]

work = 12.5*90

work = 1125 [J]

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What safety feature melts to protect a circuit?

N76 [4]

A fuse melts to protect a circuit.

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scoray [572]

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