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3 years ago

A student is playing with a radio-controlled race car on the balcony of a sixth-floor apartment. An accidental

Physics

1 answer:

pentagon [3]3 years ago

8 0

Answer:

No, it is independent

Explanation:

As we know that car is moving horizontally

so the vertical component of the speed is zero initially

so in order to hit the ground we know that

$\Delta y = v_y t + \frac{1}{2}at^2$

so here we know that

$h = 0 + \frac{1}{2}gt^2$

on solving above equation for time

$t = \sqrt{\frac{2h}{g}}$

so we will say that it will not depends on the initial horizontal speed

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A motion sensor emits sound, and detects an echo 0.0115 s after. A short time later, it again emits a sound, and hears an echo a

Mekhanik [1.2K]

Answer:

1.17 m

Explanation:

From the question,

s₁ = vt₁/2................ Equation 1

Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.

Given: v = 343 m/s, t = 0.0115 s

Substitute into equation 1

s₁ = (343×0.0115)/2

s₁ = 1.97 m.

Similarly,

s₂ = vt₂/2.................. Equation 2

Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo

Given: v = 343 m/s, t₂ = 0.0183 s

Substitute into equation 2

s₂ = (343×0.0183)/2

s₂ = 3.14 m

The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁

s₂-s₁ = (3.14-1.97) m = 1.17 m

7 0

3 years ago

Read 2 more answers

A particle with charge 3.20×10−19 c is placed on the x axis in a region where the electric potential due to other charges increa

lys-0071 [83]

Answer:

-5 V

Explanation:

The charged particle (which is positively charged) moves from point A to B, and its kinetic energy increases: it means that the particle is following the direction of the field, so its potential energy is decreasing (because it's been converted into potential energy), therefore it is moving from a point at higher potential (A) to a point at lower potential (B). This means that the value

vb−va

is negative.

We can calculate the potential difference between the two points by using the law of conservation of energy:

$\Delta K+ \Delta U=0\\\Delta K + q\Delta V=0$