In classical mechanics there are three base dimensions. length is one of them. what are the other two? view available hint(s)

A ball is thrown upward from the top of a 25.0 m tall building. The ball’s initial speed is 12.0 m/sec. At the same instant, a p

zimovet [89]

<h2>Person must have 8.18 m/s to catch the ball</h2>

Explanation:

Consider the vertical motion of ball

We have equation of motion s = ut + 0.5at²

Initial velocity, u = 12 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = -25 m

Substituting

-25 = 12 x t + 0.5 x -9.81 x t²

4.905 t² -12t - 25 = 0

t = 3.79 sec

Ball hits ground after 3.79 seconds.

So person need to cover 31 m in 3.79 seconds

Consider the horizontal motion of person

We have equation of motion s = ut + 0.5at²

Initial velocity, u = ?

Acceleration, a = 0 m/s²

Displacement, s = 31 m

Time, t = 3.79 seconds

Substituting

31 = u x 3.79 + 0.5 x 0 x 3.71²

u = 8.18 m/s

Person must have 8.18 m/s to catch the ball

6 0

3 years ago

What type of energy does the horse have because of its motion?

Alekssandra [29.7K]

is the horse running/moving? if so then kinetic i believe

7 0

2 years ago

A bullet with a mass of 0.02 kg is fired horizontally into a block of wood hanging on a string. The bullet sricks in the wood an

djyliett [7]

Answer:

u= 20.09 m/s

Explanation:

Given that

m = 0.02 kg

M= 2 kg

h= 0.2 m

Lets take initial speed of bullet = u m/s

The final speed of the system will be zero.

From energy conservation

1/2 m u²+ 0 = 0+ (m+M) g h

m u²=2 (m+M) g h

By putting the values

0.02 x u² = 2 (0.02+2) x 10 x 0.2 ( take g=10 m/s²)

u= 20.09 m/s

7 0

3 years ago

The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi

Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

$\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

B= Magnetic Field

l = length

$\mu_0$ = Vacuum permeability

$\epsilon_0$ = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

$B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}$

Recall that the speed of light is equivalent to

$c^2 = \frac{1}{\mu_0 \epsilon_0}$

Then replacing,

$B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}$

$B = \frac{r}{2C^2} \frac{dE}{dt}$

Our values are given as

$dE = 2150N/C$

$dt = 5s$

$C = 3*10^8m/s$

$D = 0.440m \rightarrow r = 0.220m$

Replacing we have,

$B = \frac{r}{2C^2} \frac{dE}{dt}$

$B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}$

$B =5.25*10^{-16}T$

Therefore the magnetic field around this circular area is $B =5.25*10^{-16}T$

3 0

3 years ago

Take another look at lines 2 and 3. Suppose you use distance and time between any pair of neighboring dots to calculate speed:

Nataly [62]

Answer:

Add the two speeds together.

Then, divide the sum by two. This will give you the average speed for the entire trip. So, if Ben traveled 40 mph for 2 hours, then 60 mph for another 2 hours, his average speed is 50 mph.

8 0

3 years ago

Read 2 more answers