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4 years ago

Using Newton’s second law, why do you think a cotton ball may not be used as a baseball in a baseball game.

Physics

1 answer:

stich3 [128]4 years ago

3 0

Answer:

as its mass and velocity will less so its momentum will be less than that of baseball

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Two forces,

serg [7]

First compute the resultant force F:

$\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N$

$\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N$

$\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N$

Then use Newton's second law to determine the acceleration vector $\mathbf a$ for the particle:

$\mathbf F=m\mathbf a$

$(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a$

$\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}$

Let $\mathbf x(t)$ and $\mathbf v(t)$ denote the particle's position and velocity vectors, respectively.

(a) Use the fundamental theorem of calculus. The particle starts at rest, so $\mathbf v(0)=0$ . Then the particle's velocity vector at t = 10.4 s is

$\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du$

$\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}$

If you don't know calculus, then just use the formula,

$v_f=v_i+at$

So, for instance, the velocity vector at t = 10.4 s has x-component

$v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}$

(b) Compute the angle $\theta$ for $\mathbf v(10.4\,\mathrm s)$ :

$\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ$

so that the particle is moving at an angle of about 313º counterclockwise from the positive x axis.

(c) We can find the velocity at any time t by generalizing the integral in part (a):

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

Then using the fundamental theorem of calculus again, we have

$\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du$

where $\mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m$ is the particle's initial position. So we get

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du$

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m$

So over the first 10.4 s, the particle is displaced by the vector

$\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m$

or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).

(d) See part (c).

3 0

3 years ago

A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch

Evgen [1.6K]

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

8 0

3 years ago

a. What is the peak wavelength for AM0? What temperature T corresponds to this peak wavelength for a blackbody source? Assuming

atroni [7]

Answer:

A) T = 5510 K , B) I = 5,226 10⁷ W / m² , C) I₂ = 1128 W / m²

Explanation:

A) The mass of air is defined as the ratio between the shortest path that sunlight must pass to reach the planet's surface and the length of the beam, is called AM

Am = 1 / cos θ

The AM0 value corresponds to solar radiation in the outer part of the Earth's atmosphere.

The peak of this emission is the peak that emitted from the sun

λ = 526 nm

To find the temperature that corresponds to this emission we use the Wien displacement law

λ T = 2,898 10⁻³

T = 2,898 10⁻³ / 526 10⁻⁹

T = 5510 K

i) The radiance on the surface of the sun is

I = P / A

We can calculate the potency by Stefan's law, for a black body

P = σ A e T⁴

P / A = σ e T⁴

The σ constant is value 5,670 10⁻⁸ W / m²K⁴, we will assume that the Sun emits as a black body, so e = 1

I = sig T⁴

I = 5,670 10⁻⁸ 5510⁴

I = 5,226 10⁷ W / m²

ii) the irradiation at a distance of 1 ua (1,496 1011 m)

Let's use the relationship

P = I A

I₁ A₁ = I₂ A₂

I₂ = I₁ A₁ / A₂

The area of a sphere is

A = 4π R²

Let's replace

I₂ = I₁ (r₁ / r₂)²

Index 1 corresponds to the sun and the index to Earth that is an astronomical unit

r₁ = 6.96 10⁸m (Sun radius)

r₂ = 1,498 1011 m (Earth-Sun distance)

Calculous

I₂ = 5,226 10⁷ (6.96 10⁸ / 1,498 10¹¹)²

I₂ = 1.1281 10³ W / m²

I₂ = 1128 W / m²

8 0

3 years ago

What are dangers of synthetic drugs

Julli [10]

These are some potential dangers of synthetic drug abuse

7 0

3 years ago

A car accelerates at 2 m/s2. Assuming the car starts from rest, how much time does it need to accelerate to a speed of 20 m/s? A

Ludmilka [50]

The answer would be
C) 20 seconds.

8 0

3 years ago

Read 2 more answers