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Jlenok [28]
3 years ago
10

A stone is thrown vertically upward with velocity 9.8ms 1)calculate the time taken to reach maximum height.2)what is the maximum

height reached 3) what is time taken to reach ground 4)what velocity it reaches the ground
Physics
2 answers:
agasfer [191]3 years ago
6 0

Answer:

It would take 6 seconds before hitting the ground

Explanation:

ira [324]3 years ago
5 0

Answer:

1.t=-1.96sec

2.H=4.8m

3.T=1.96sec

4.R=19.2m

Explanation:

u=9.8,t=?,sin theta=1

using formula t=2usintheta/g

t=2x9.8x1=19.6/10

t=1.96seconds

using formula H=u(squared)sin(squared)theta/2g

H=9.8(squared)x1(squared)/2x10

H=96x1/20

H=96/20

H=4.8m

using formula T=2usintheta/g

T=2x9.8x1/10

T=19.6/10

T=1.96sec

using the formula R=u(squared)sin2theta/g

R=9.8(squared)x2/10

R=96x2/10

R=192.08/10

R=19.2

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A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
skelet666 [1.2K]

Answer:

Explanation:

Total length of the wire is 29 m.

Let the length of one piece is d and of another piece is 29 - d.

Let d is used to make a square.

And 29 - d is used to make an equilateral triangle.

(a)

Area of square = d²

Area of equilateral triangle = √3(29 - d)²/4

Total area,

A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}

Differentiate both sides with respect to d.

\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)

For maxima and minima, dA/dt = 0

d = 8.76 m

Differentiate again we get the

\frac{d^{2}A}{dt^{2}}= + ve

(a) So, the area is maximum when the side of square is 29 m

(b) so, the area is minimum when the side of square is 8.76 m

8 0
3 years ago
The velocity of a mass moving in a circular path is..
Alenkasestr [34]
Constantly changing.
5 0
3 years ago
The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cr
tatyana61 [14]

Answer:

The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of  second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of  fourth coil.

Explanation:

The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.

So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of  second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of  fourth coil.

8 0
3 years ago
A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st
Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r

where,

\tau= shear stress at a distance 'r' from the center

T = is the applied torque

I_{p} = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

\tau _{max}=\frac{4}{3}\times \frac{V}{A}

Applying values we get

\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa

3 0
3 years ago
You are moving a wagon with a friend's help you push on the left side of the wagon with 25 of force while your friend pulls from
Pavel [41]

Answer:

10N to the left side towards you

Explanation:

The net force is the resultant force that acts on a body.

Force is a push or pull on a body.    

 Push to left side  = 25N

 Pull to the right  = 15N

Net force  = Push to left side   -  Pull to the right  = 25N  - 15N

 Net force  = 10N to the left side towards you

The net force is therefore 10N to the left side towards you

5 0
3 years ago
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