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aliya0001 [1]
3 years ago
11

A ball is thrown directly downward with an initial speed of 7.95 m/s, from a height of 29.0 m. After what time interval does it

strike the ground?
Physics
1 answer:
Y_Kistochka [10]3 years ago
4 0

Answer: after 1.75 seconds

Explanation:

The only force acting on the ball is the gravitational force, so the acceleration will be:

a = -9.8 m/s^2

the velocity can be obtained by integrating over time:

v = -9.8m/s^2*t + v0

where v0 is the initial velocity; v0 = -7.95 m/s.

v = -9.8m/s^2*t - 7.95 m/s.

For the position we integrate again:

p = -4.9m/s^2*t^2 - 7.95 m/s*t + p0

where p0 is the initial position: p0 = 29m

p =  -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Now we want to find the time such that the position is equal to zero:

0 = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Then we solve the Bhaskara's equation:

t = \frac{7.95 +- \sqrt{7.95^2 +4*4.9*29} }{-2*4.9} = \frac{7.95 +- 25.1}{9.8}

Then the solutions are:

t = (7.95 + 25.1)/(-9.8) = -3.37s

t = (7.95 - 25.1)/(-9.8) = 1.75s

We need the positive time, then the correct answer is 1.75s

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A 2.4 mm -diameter copper wire carries a 37 A current (uniform across its cross section). Determine the magnetic field at the su
cluponka [151]

Answer:

Explanation:

We shall apply Ampere's circuital law to find out magnetic field . It is given as follows.

∫B.dl = μ₀ I , B is magnetic field , I is current ,  μ₀ is permeability .

Radius of the wire r = 1.2 x 10⁻³ m

magnetic field B will be circular in shape around the wire. If B is uniform

∫B.dl = B x 2πr  

B x 2πr  = μ₀ I

B = μ₀ I / 2πr

= 4π x 10⁻⁷ x 37 /2πx1.2 x 10⁻³

= 10⁻⁷ x 2x37 / 1.2 x 10⁻³

= 61.67 x 10⁻⁴ T

= 62  x 10⁻⁴ T

7 0
2 years ago
Q is granite, an igneous rock. Which of these is NOT a characteristic of igneous rock?
fgiga [73]

Igneous rocks are conglomerates, predominantly composed of rounded

gravel

Explanation:

The statement that igneous rocks are conglomerate predominantly composed of rounded gravel is not correct. Igneous rocks are rocks formed from magma that cools and solidifies.

  • Conglomerates are sedimentary rocks that are made of rounded clasts in a matrix.
  • They differ from igneous rocks that crystallize from molten magma.
  • Sedimentary rocks are usually derived from the process of weathering, erosion, transportation and deposition on the earth surface.
  • Igneous rocks forms from crystallized magma which can be on the surface or deep within the crust.

Learn more:

Sedimentary rocks brainly.com/question/2740663

#learnwithBrainly

8 0
3 years ago
Read 2 more answers
two masses are kept 2 metre apart there is gravitational force of 2 Newton what is the gravitational force when they are kept at
sukhopar [10]

Answer: 0.5N

Explanation:

Gravitational force is calculated using the formula :

F = Gm1m2/r^2

Where G is the gravitational constant (6.67 × 10^-11)

At a distance 'r' of 2metres apart:

Mass of objects are m1 and m2

Gravitational force 'F1' = 2N

Inputting values into the formula :

2 = Gm1m2 / 2^2 - - - - - (1)

At a distance 'r' of 4meters apart:

Mass of objects are m1 and m2

Gravitational force 'F2' = y

Inputting values

F2 = Gm1m2 / 4^2 - - - - - (2)

Dividing equations 1 and 2

2 = Gm1m2 / 2^2 ÷ F2 = Gm1m2 / 4^2

2 / F2 = (Gm1m2 / 4) / (Gm1m2 / 16)

2 / F2 = (Gm1m2 / 4) × (16 / Gm1m2)

2/F2 = 16 / 4

Cross multiply

2 × 4 = 16 × F2

8 = 16F2

F2 = 8/16

F2 = 0.5N

7 0
3 years ago
Use Newton's laws to explain why a falling object dropped from a 57m tower accelerates initially but then reaches constant veloc
snow_lady [41]

Answer:

At the point of dropping the object, by Newton's first law due to gravitational force F_g = m × g, accelerates

By Newton's Second law the object reaches impacts on the air with the gravitational force resulting in changing momentum of m×(Final Velocity - Initial Velocity)

As the velocity increases, the rate of change of momentum becomes equivalent to the gravitational force and by Newton's third law, the action action and reaction are equal and opposite hence they cancel each other out

The body then moves at a constant uniform motion down according to Newton's first law

Explanation:

At the point the object of mass, m, is dropped from the height of the tower, the only force acting on the object is the gravitational force such that the object has an acceleration which is the acceleration due to gravity, g, and the gravitational force is therefore = m × g

As the speed of the object increases while the object is falling with the gravitational acceleration the rate at which the object cuts through layers of air which (by Newton's first law of motion, are at rest ) has some buoyancy effect also increases therefore, the object is constantly increasingly changing the momentum of the air which by Newton's second law results, at an high enough velocity, and by Newton's third law, in a force equal to the applied gravitational force

Therefore, the force of the air drag becomes equal to the gravitational force, cancelling each other out and the object then moves according to Newton;s first law, in uniform motion of a constant speed while still falling down.

5 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
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