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3 years ago

If a biker starts at rest and accelerates to 15 m/s in 7.5 seconds, what is his acceleration? *

Physics

1 answer:

sergeinik [125]3 years ago

6 0

Answer:

2 m/s^2

Explanation:

from the question

v=15 m/s

t=7.5

a=?

from the first equation of motion

v=u+at

where,

v=final velocity

u=initial velocity

a=acceleration

t=time

from the question (u) will be zero because the body started at rest

v=u+at

15=(0)+a×7.5

15=7.5a

a=15/7.5

a=2 m/s^2

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Consider a box with two gases separated by an impermeable membrane. The membrane can move back and forth, but the gases cannot p

padilas [110]

Answer:

The answer is " $0.3336\ m^3$ "

Explanation:

Using the Promideal gas law:

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$=\frac{1+2(4\times 10^{-4})}{3}\\\\=0.3336\ m^3$

8 0

3 years ago

A solid circular plate has a mass of 0.25 kg and a radius of 0.30 m. It starts rolling from rest at the top of a hill 10 m long

KATRIN_1 [288]

To solve the problem it is necessary to apply the equations related to the conservation of both <em>kinetic of rolling objects</em> and potential energy and the moment of inertia.

The net height from the point where it begins to roll with an inclination of 30 degrees would be

$h=Lsin30$

$h=10sin30$

$h=5m$

In the case of Inertia would be given by

$I = \frac{mR^2}{2}$

In general, given an object of mass m, an effective radius k can be defined for an axis through its center of mass, with such a value that its moment of inertia is

$I = mk^2$

$\frac{mR^2}{2}= mk^2$

$\frac{k^2}{R^2}=\frac{1}{2}$

Replacing in Energy conservation Equation we have that

Potential Energy = Kinetic Energy of Rolling Object

$mgh = \frac{1}{2}mv^2(1+\frac{k^2}{r^2})$

$9.8*5=\frac{1}{2}v^2(1+\frac{1}{2})$

$v^2 (1.5) = 98$

$v=8.0829m/s$

Therefore the correct answer is C.

3 0

3 years ago

A 44 kg block lies on a horizontal frictionless surface. A horizontal force of 110 N is applied to the block. (a) Determine the

Oksana_A [137]

44k jkjk know the answer

8 0

3 years ago

A thick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.33 m, and is initially unc

Vesnalui [34]

A) $E=\frac{4.50\cdot 10^{10}}{r^2} V/m$

r < a

We can find the magnitude of the electric field by using Gauss theorem. Taking a Gaussian spherical surface of radius r centered in the centre of the sphere, the electric flux through the surface of the sphere is equal to the ratio between the charge contained in the sphere and the vacuum permittivity:

$E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$

For r < a, the charge contained in the gaussian sphere is the point charge:

$q=5.00 C$

So the electric field in this region is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{5.00 C}{4\pi (8.85\cdot 10^{-12} F/m)}\frac{1}{r^2}=\frac{4.50\cdot 10^{10}}{r^2} V/m$

B) E = 0

a < r < b

The region a < r < b is the region between the inner and the outer surface of the shell. We have to keep in mind that the presence of the single point charge +q = 5.00 C at the center of the sphere induces an opposite charge -q on the inner surface (r=a), and a charge of +q at the outer surface (r=b).

Using again Gauss theorem

$E\cdot 4 \pi r^2 = \frac{q'}{\epsilon_0}$

this time we have that the gaussian sphere contains both the single point charge +q and the negative charge -q induced at r=a, so the net charge contained in the sphere is

q' = +q - q = 0

And so, the electric field in this region is zero.

C) $E=\frac{4.50\cdot 10^{10}}{r^2} V/m$

r > b

Here we are outside of the sphere. Using Gauss theorem again

$E\cdot 4 \pi r^2 = \frac{q'}{\epsilon_0}$

this time we have that the gaussian sphere contains the single point charge +q, the negative charge -q induced at r=a, and the positive charge +q induced at r=b, so the net charge contained in the sphere is

q' = +q - q +q = q

And so the electric field is identical to the one inside the sphere:

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{5.00 C}{4\pi (8.85\cdot 10^{-12} F/m)}\frac{1}{r^2}=\frac{4.50\cdot 10^{10}}{r^2} V/m$

D) -12.29 C/m^2

We said that the charge induced at the inner surface r=a is

-q = -5.00 C

The induced charge density is

$\sigma = \frac{-q}{A}$

where A is the area of the inner surface of radius r = a = 0.18 m, so it is

$A=4\pi a^2 = 4 \pi (0.18 m)^2=0.407 m^2$

So the induced charge density is

$\sigma = \frac{-5.00 C}{0.407 m^2}=-12.29 C/m^2$

E) +3.65 C/m^2

We said that the charge induced at the outer surface r=b is

+q = +5.00 C

The induced charge density is

$\sigma = \frac{+q}{A}$

where A is the area of the outer surface of radius r = b = 0.33 m, so it is

$A=4\pi b^2 = 4 \pi (0.33 m)^2=1.368 m^2$

So the induced charge density is

$\sigma = \frac{+5.00 C}{1.368 m^2}=+3.65 C/m^2$

7 0

3 years ago

You throw a baseball at an angle of 30.0 degrees above the horizontal. It reaches the highest point of its trajectory 1.05 s lat

vfiekz [6]

Answer:

The ball leaves your hand at 20.6 m/s

Explanation:

Hi there!

The equation of the vertical velocity of the ball is the following:

vy = v0 · sin α + g · t

Where:

vy = vertical component of the velocity vector at time t.

v0 = initial velocity.

α = throwing angle.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

t = time.

When the ball reaches the highest point of its trajectory, its vertical velocity is zero. So, using the equation f vertical velocity we can solve it for the initial velocity:

vy = v0 · sin α + g · t

0 m/s = v0 · sin(30°) - 9.8 m/s² · 1.05 s

9.8 m/s² · 1.05 s / sin(30°) = v0

v0 = 20.6 m/s

The ball leaves your hand at 20.6 m/s

4 0

3 years ago

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