All categories

2 years ago

Which of these is the best example of translational motion?

Physics

2 answers:

Tanya [424]2 years ago

6 0

B, air blowing from across the field is as a bullet fired from a rifle

stich3 [128]2 years ago

5 0

Would definitely be letter b, it makes the most sense

You might be interested in

a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance tra

LenaWriter [7]

Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

$a=\dfrac{\Delta v}{\Delta t}=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.47\,\mathrm s}=11.2\,\dfrac{\mathrm m}{\mathrm s^2}$

Now, we have that

${v_f}^2-{v_0}^2=2a\Delta x$

so we end up with a distance traveled of

$\left(46.1\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(11.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x$

$\implies\Delta x=79.6\,\mathrm m$

6 0

3 years ago

The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far befor

Mekhanik [1.2K]

Answer:

The new height the ball will reach = (1/4) of the initial height it reached.

Explanation:

The energy stored in any spring material is given as (1/2)kx²

This energy is converted to potential energy, mgH, of the ball at its maximum height.

If the initial height reached is H

And the initial compression of the spring = x

So, mgH = (1/2)kx²

H = kx²/2mg

The new compression, x₁ = x/2

New energy of loaded spring = (1/2)kx₁²

And the new potential energy = mgH₁

mgH₁ = (1/2)kx₁²

But x₁ = x/2

mgH₁ = (1/2)k(x/2)² = kx²/8

H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)

6 0

3 years ago

An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in

Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

$T {V}^{\gamma - 1} = \text{constant}$

$\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}$

Given:-

${T}_{1} = 275 \; K$

${T}_{2} = T \left( \text{say} \right)$

${V}_{1} = V$

${V}_{2} = 6V$

$\gamma = \cfrac{5}{3} \;$ (the gas is monoatomic)

$\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}$

$\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}$

T = 275 $\times$ 0.30

T = 83 K.

3 0

3 years ago

Solve the following system of equations by using either substitution or elimination.

strojnjashka [21]

I believe that the answer to the question provided above are the following;

x = 29.8410

y = 16.6794

z = -1.2642
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

5 0

3 years ago

A mass m is attached to an ideal massless spring with spring constant k. In experiment 1 the mass oscillates with amplitude a, a

Trava [24]

Time period remains the same in both the experiment as change in amplitude does not affect time period.

What are the factors on which time period depends in SHM?

Time period is given by:

$T=2\pi \sqrt{\frac{m}{k} }$

where,

T = time period

m = mass

k = spring constant

In a straightforward harmonic motion, we see from the preceding formula that the time period depends only on the object's mass and spring constant (SHM). The time period will adjust to any variations in the object's mass or the spring constant.

What is Spring Constant?

A spring's "spring constant" is a property that quantifies the relationship between the force acting on the spring and the displacement it produces. In other words, it characterises a spring's stiffness and the extent of its range of motion.

Learn more about SHM here:

brainly.com/question/20885248

#SPJ4

6 0

6 months ago