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2 years ago

In urgent need of help, science is my worse subject!

Physics

1 answer:

taurus [48]2 years ago

5 0

The first reason is that in the past humans didn't necessarily see the connections between the oceans and thus labelled them that way separating them into 5 oceans and the names stuck to this day.

Secondly, while all the oceans of the world constitute one large Ocean, people need some way to identify each part of it since it is easier to think when we have different categories instead of one large category.

Hope that helped!

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A single point charge q is located at the center of both an imaginary cube and an imaginary sphere. How does the electric flux t

saveliy_v [14]

Answer:

Explanation:

Electric flux is defined as the flow of electric field intensity through a given surface.

Mathematically:

$\phi=E.A.cos\ \theta$

where:

E = electric field

A = area

θ = angle between the area vector and the electric field

Electric flux through the surface of a sphere will be uniform throughout the surface area due to a charge at the center of the sphere. The distance of the surface from the center is always at a constant distance of radius of the sphere.

Electric flux through the surface of a cube will be varying as the surface area is at a varying distance from the center of the cube. The distance of the surface from the center is not at a uniform distance from the center of the cube and so the projection of solid angle changes.

7 0

3 years ago

A concrete block is hung from an ideal spring that has a force constant of 100 N/m . The spring stretches 0.129 m .A- What is th

madam [21]

Answer:

1.31498 kg

0.72050 s

Explanation:

m = Mass of block

g = Acceleration due to gravity = 9.81 m/s²

k = Spring constant = 100 N/M

x = Displacement = 0.129 m

The force balance is

$mg=kx\\\Rightarrow m=\dfrac{kx}{g}\\\Rightarrow m=\dfrac{100\times 0.129}{9.81}\\\Rightarrow m=1.31498\ kg$

The mass of the block is 1.31498 kg

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{1.31498}{100}}\\\Rightarrow T=0.72050\ s$

The period of oscillations is 0.72050 s

The time period does not depend on the acceleration due to gravity. It varies with the mass and the spring constant.

Hence, the time period would be the same

8 0

3 years ago

a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi

zubka84 [21]

Answer:

The work done on the car is, W = 75,000 J

The power delivered by the engine, P = 25,000 watts

Explanation:

Given,

The mass of the car, m = 1500 Kg

The initial velocity of the car, u = 0

The final velocity of the car, v = 10 m

The time duration of the travel, t = 3 s

Using the first equation of motion

v = u + at

a = (v - u) / t

Substituting the given values in the above equation

a = (10 - 0) / 3

= 3.33 m/s²

Using the second equations of motion

s = ut + 1/2 at²

= 0 + 0.5 x 3.33 x 3²

= 15 m

The force exerted by the car

F = m x a

= 1500 Kg x 3.33 m/s²

= 5000 N

The work done by the car,

W = F x S

= 5000 N x 15 m

= 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

P = W / t

= 75,000 J / 3 s

= 25,000 watts

The power delivered by the engine, P = 25,000 watts

5 0

3 years ago

Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i

densk [106]

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

6 0

3 years ago

An automobile of mass 2000 kg moving at 30 m/s is braked suddenly with a constant braking force of 10000 N. How far does the car

saveliy_v [14]

Answer:

The car traveled the distance before stopping is 90 m.

Explanation:

Given that,

Mass of automobile = 2000 kg

speed = 30 m/s

Braking force = 10000 N

For, The acceleration is

Using newton's formula

$F = ma$

Where, f = force

m= mass

a = acceleration

Put the value of F and m into the formula

$-10000 =2000\times a$

Negative sing shows the braking force.

It shows the direction of force is opposite of the motion.

$a = -\dfrac{10000}{2000}$

$a=-5\ m/s^2$

For the distance,

Using third equation of motion

$v^2-u^2=2as$

Where, v= final velocity

u = initial velocity

a = acceleration

s = stopping distance of car

Put the value in the equation

$0-30^2=2\times(-5)\times s$

$s = 90\ m$

Hence, The car traveled the distance before stopping is 90 m.

6 0

3 years ago