All categories

2 years ago

A disgruntled autoworker pushes a small foreign import off

Physics

1 answer:

Andrews [41]2 years ago

5 0

Answer:

v = a/√(2h/g) m/s

Explanation:

Lets say the distance away from the cliff is a.

then, a = v t

where v is velocity with which it was thrown and t is time taken to fall.

Using equations of motion, we can also say that

h=1/2gt^2

where h is the height of the cliff

Thus, t^2 = 2h/g and t = √(2h/g)

Thus, v = a/√(2h/g).

the vehicle was pushed off the cliff with the velocity , v = a/√(2h/g). m/s

You might be interested in

The function h(t) = –16t2 + 96t + 6 represents an object projected into the air from a cannon. The maximum height reached by the

Rainbow [258]

Answer:

3 seconds

Explanation:

Since h(t) represents the height and t represents the time, we can set the equation equal to 150 to find t.

-16t^2+96t+6=150

Subtract 150 from both sides to set the equation equal to 0, to find the solutions.

-16t^2+96t-144=0

Factor out -16, because all of the terms are divisible by it.

-16(t^2+6t+9)=0

Now we can focus on the terms inside the parenthesis and factor it again.

t^2-6t+9=0

We need to find two value that can be multiplied to get 9 and added to get -6.

-3 and -3 works.

Thus, we get (x-3)(x-3).

Now solve for 0.

x-3=0

x=3

The object reaches its maximum height after 3 seconds.

6 0

3 years ago

Read 2 more answers

A 5.00 kg crate is on a 21.0 degree hill. Using X-Y axes tilted down the plane, what is the y-component of the normal force?

Rama09 [41]

Answer:

The y-component of the normal force is 45.74 N.

Explanation:

Given that,

Mass of the crate, m = 5 kg

Angle with hill, $\theta=21^{\circ}$

We need to find the y component of the normal force. We know that the y component of the normal force is given by :

$F_y=F\ \cos\theta\\\\F_y=mg\ \cos\theta\\\\F_y=5\times 9.8\ \cos(21)\\\\F_y=45.74\ N$

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.

7 0

2 years ago

An object starts at 4m/s and accelerates to 6m/s in 10 seconds. Find its displacement.

Natalka [10]

Answer:

Explanation:

The 2 equations we need here are, first:

$a=\frac{v_f-v_0}{t}$ and then once we solve for the acceleration here:

$v^2=v_0^2+2a$ Δx

Solving for acceleration:

$a=\frac{6-4}{10}=\frac{1}{5}\frac{m}{s^2}$ and now we will use that in the other equation:

$6^2=4^2+2(\frac{1}{5})$ Δx and

36 = 16 + $\frac{2}{5}$ Δx and

20 = $\frac{2}{5}$ Δx and

$\frac{5}{2}(20)=$ Δx so

Δx = 50 m

6 0

3 years ago

Select the correct answer.

kherson [118]

Answer:

B chemical energy

Explanation:

4 0

2 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

LuckyWell [14K]

For a circular aperture, the first minima (n=1) as an angular separation from the peak of the central maxima given by

Sinθ = 1.22λ / d

Where,

d is the aperture or pupil diameter

d = 4.69 mm = 4.69 × 10^-3m

λ is the wavelength

λ = 545 nm = 545 × 10^-9 m

Then,

Sinθ = 1.22λ / d

Sinθ = 1.22 × 545 × 10^-9 / 4.69 × 10^-3

Sinθ = 1.418 × 10^-4 rad

Then, the head light sources have the same angular separation θ from the eye as the image have inside the eye.

For the headlight

Sinθ ≈ light separation / distantce for the eye

Light separation is give as x = 0.659 m

And let the distance of the eye be D

Then,

Sinθ = x / D

Make D subject of formula

D = x / Sinθ

D = 0.695 / 1.418 × 10^-4

D = 4902.316m

To km, 1km = 1000m

D ≈ 4.9 km

4 0

2 years ago