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3 years ago

Help please It’s kinda urgent

Physics

1 answer:

user100 [1]3 years ago

4 0

Answer:

a = - 50 [m/s²]

Explanation:

To solve this problem we simply have to replace the values supplied in the given equation.

Vf = final velocity = 0.5 [m/s]

Vi = initial velocity = 10 [m/s]

s = distance = 100 [m]

a = acceleration [m/s²]

Now replacing we have:

$(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]$

The negative sign of acceleration means that the ship slows down its velocity in order to land.

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A block of mass 0.490 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T

Dennis_Churaev [7]

Answer:

The value of x is $x = 0.1955\ m$

The value of the velocity at the top is $v_{top} = 2.25 \ m/s$

Explanation:

The diagram of this process is on the first uploaded image

From the question we are told that

The mass of the block is $m_b = 0.490kg$

The distance of compression is $x$

The force constant is $k = 450 N/m$

The radius of the circular track is $R =1.00m$

The speed at the bottom of the track is $v_A = 13.4 \ m/s$

The frictional force experienced is $F_f = 7.00 \ N$

Now looking at this process we see that the potential energy of the spring is been transformed into the kinetic energy of the block . So,

$PE \ of \ spring = KE \ of \ block$

mathematically i.e

$\frac{1}{2}k x^2 = \frac{1}{2} m_bv^2_A$

$0.5 * 450 * x^2 = 0.5 * 0.490 * 13.4^2$

Making x the subject of the formula

$x = \sqrt{\frac{0.5 * 0.490 * 13.4^2}{ 0.5 * 450} }$

$= 0.1955\ m$

The average workdone by friction is $W_f = F_f * \pi$

Here $\pi is \ the\ net\ displacement$

$W_f = 7 * 3.142$

$= 21.99 J$

The kinetic energy at the bottom is

$KE = \frac{1}{2} * 0.490 * 13.4^2$

$= 43.99 \ N$

The potential energy gained at the top of the circle is

$PE = m_bgh$

Here h is the height which is equal to d(diameter) = 2r = 2 × 1 = 2 m and

g is acceleration due to gravity $= 9.8m/s^2$

Now substituting values

$PE = 0.490 * 9.8 *2$

$=9.604 J$

Since energy is can not be created nor destroyed but transformed according to first law of thermodynamics

$KE_{at \bottom} = PE _ {\ at \ top } + W_f + KE_{\ at \ top}$

$43.99 = 9.604 +21.99 + [\frac{1}{2} m_b * v_{top} ^2]$

$12.369= [2.45* v_{top} ^2]$

$v_{top} =\sqrt{\frac{12.369}{2.45} }$

$= 2.25 \ m/s$

8 0

3 years ago

In Fraunhofer diffraction wave front used is __________. A. Spherical B. Circular C. Plane D. Conical

Zepler [3.9K]

I am pretty sure the answer is C.

4 0

3 years ago

A cylindrical rod of mass M. length L and radius R has two cords wound around it whose ends are a

Rudik [331]

Answer:

T = mg/6

Explanation:

Draw a free body diagram (see attached). There are two tension forces acting upward at the edge of the cylinder, and weight at the center acting downwards.

The center rotates about the point where the cords touch the edge. Sum the torques about that point:

∑τ = Iα

mgr = (1/2 mr² + mr²) α

mgr = 3/2 mr² α

g = 3/2 r α

α = 2g / (3r)

(Notice that you have to use parallel axis theorem to find the moment of inertia of the cylinder about the point on its edge rather than its center.)

Now, sum of the forces in the y direction:

∑F = ma

2T − mg = m (-a)

2T − mg = -ma

Since a = αr:

2T − mg = -mαr

Substituting expression for α:

2T − mg = -m (2g / (3r)) r

2T − mg = -2/3 mg

2T = 1/3 mg

T = 1/6 mg

The tension in each cord is mg/6.

7 0

3 years ago

The conventional relatively small unit for work(ignoring time) such as raising one pound one foot is the foot-pound(ft. lb.). Si

Lelechka [254]

Answer:

1 joule = 0.737 foot-pound

Joule is the unit of work.

1 J = 1 N·m

In SI units

1 J = 1 kg· m/s²

0.737 foot-pound is the amount of work to raise 0.737 pounds one foot or raising one pound to 0.737 ft.

6 0

3 years ago

Sergio [31]

see

below

Explanation:

refractive index = speed of light in vacuum / speed of light in medium

light travels at a speed of 3.0 x 10^8 m/s in vacuum

refractive index = 3.0 x 10^8 / 2.0 x 10^8

refractive index = 1.5

hope this helps, please mark it

4 0

3 years ago