Why do objects accelerate when dropped on Earth?

A biker travels at an average speed of 18 km/hr along a 0.30 km straight segment of a bike path. How much time (in hours) does t

lawyer [7]

Answer: 0.016 h

Explanation:

$\text{Average speed} = \frac{\text {Total Distance}}{\text {total time taken}}$

It is given that, biker has an average speed = 18 km/h

Total distance traveled = 0.30 km

Therefore, time taken by biker to travel this distance:

$\Rightarrow \text{total time taken} = \frac{0.30 km}{18 km/h}=0.016 h$

Thus, the biker takes 0.016 hours to travel the segment of 0.30 km at an average speed of 18 km/h.

7 0

3 years ago

What is the frequency of motion of a 0.50 m long pendulum?

IgorC [24]

Complicated 2-step process.

1. Write down the formula for "frequency of a pendulum" from your textbook. 

f = (1/2pi) * sqrt(g/L) 

2. Plug in g and L.

I hope this can help!

3 0

3 years ago

Read 2 more answers

A laser emitting light with a wavelength of 560 nm is directed at a single slit, producing an interference pattern on a screen t

muminat

Answer:

a) a = 6.72 10⁻⁵ m, b) the slit (a) is smaller, which represents a wider pattern

Explanation:

In is a diffraction experiment since we have a single slit, it is explained by the equation

a sin θ = m λ

where a is the width of the slit

The diffraction pattern is characterized by a very intense central maximum, with a value of 5.0 cm, therefore the distance from the center to the first zero is y = 5.0 / 2 cm = 2.5 10⁻² m

let's use trigonometry to enter the angle

tan θ = y / L

tan θ = sin θ / cos θ = sin θ

sin θ = y / L

substituting into the equation

a y / L = m λ

the first maximum occurs for m = 1

a = λ L / y

let's calculate

a = 560 10⁻⁹ 3.0 / 2.5 10⁻²

a = 6.72 10⁻⁵ m

b) if the width a of the slit (a) is smaller

sin θ = m λ / a

therefore the sinus increases, which implies a greater angle, which represents a wider pattern

c) if the distance to the screen (L) goes away

y = m λL / a

If L increases the width of the pattern they also increase of course the intensity must be less

d) If the wavelength increases

In this case the width of the pattern also increases

e) What happens if the light passes through two slits in this case we have two diffraction patterns one centered in each slit and the resulting pattern is the sum of these patterns, this sum gives the double slit interference that characterizes a series of slits of equal height

4 0

3 years ago

The centripetal force of an object of mass m is given by F(r)=mv2r, where v is the speed of rotation and r is the distance from

mr Goodwill [35]

Answer:

$\frac{dF}{dr} = -4.82$

Explanation:

As we know that the centripetal force is given as

$F = \frac{mv^2}{r}$

now we will have to find the rate of change in force with respect to its radial distance

so we have

$\frac{dF}{dr} = - \frac{mv^2}{r^2}$

so we have

m = 1000 kg

v = 13.89 m/s

r = 200 m

now we have

$\frac{dF}{dr} = -\frac{1000(13.89^2)}{200^2}$

$\frac{dF}{dr} = -4.82$

3 0

4 years ago

A ball is thrown horizontally to the right, from the top of a vertical cliff of height h. A wind blows horizontally to the left,

Akimi4 [234]

Answer:

v = $\sqrt{\frac{y_o \ g}{2} }$

Explanation:

For this exercise we must use the projectile launch ratios, let's start by finding the time it takes to reach the bottom of the cliff, the initial vertical velocity is zero

y = y₀ + $v_{oy}$ t - ½ g t²

at the bottom of the cliff y = 0 and as the body is thrown horizontally the initial vertical velocity is zero

0 = y₀ + 0 - ½ g t²

t = $\sqrt{2y_o/g}$

this time is the same as the horizontal movement.

Let's use Newton's second law to find the acceleration on this x-axis due to the force of the air

F = m aₓ

they tell us that force is equal to the weight of the body

-mg = maₓ

aₓ = -g

the sign indicates that the acceleration is to the left

we write the kinematics equation

x = x₀ + v₀ₓ t + ½ aₓ t²

They indicate that the final position is the foot of the cliff (x = 0), when it leaves the top it is at x₀ = 0 and has a velocity v₀ₓ = v

we substitute

0 = 0 + v t + ½ (-g) t²

v = ½ g t

we use the drop time

v = ½ g $\sqrt{\frac{2yo}{g} }$

v = $\sqrt{\frac{y_o \ g}{2} }$

5 0

3 years ago