A train moves at constant velocity of 90km/h. How far will it go in 0.25h.

2.

likoan [24]

Answer:

The Storm Surge of a hurricane

Explanation:

A hurricane can ultimately cause a tsunami and is just as deadly.

4 0

3 years ago

At a certain altitude above the Earth's surface, the electric field has a magnitude of 135 V/m. How much energy is stored in 1.0

Paul [167]

Answer:

Explanation:

Given that,

Electric field E=135V/m

Energy stored in 1m³of air=?

The energy stored in an electric field is given as

u = ½ εo E²

Where

U is the energy stored

εo is permissivity and it value is 8.85×10^-12C²/N..m²

And E is the electric field

Then,

U=½×8.85×10^-12×135²

U=8.06×10^-8J/m³

Then, the energy stored in 1m³ of air is 8.06×10^-8 J/m³

4 0

3 years ago

Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 48.

galina1969 [7]

Given:

The force of attraction is F = 48.1 N

The separation between the charges is

$\begin{gathered} r=\text{ 60.9 cm} \\ =\text{ 60.9}\times10^{-2}\text{ m} \end{gathered}$

Also, the magnitude of charge q1 = q2 = q.

To find the magnitude of charge.

Explanation:

The magnitude of charge can be calculated by the formula

$\begin{gathered} F=\frac{k(2q)}{r^2} \\ q=\frac{Fr^2}{2k} \end{gathered}$

Here, k is the Coulomb's constant whose value is

$k\text{ = 9}\times10^9\text{ N m}^2\text{ /C}^2$

On substituting the values, the magnitude of charge will be

$\begin{gathered} q=\frac{48.1\times(60.9^\times10^{-2})^2}{2\times9\times10^{^9}} \\ =9.91\times10^{-10}\text{ C} \\ =9.91\times10^{-4}\text{ }\mu C \end{gathered}$

Thus, the magnitude of each charge is 9.91 x 10^(-4) micro Coulombs.

6 0

2 years ago

A block is given a very brief push up a 20.0 degree frictionless incline to give it an initial speed of 12.0 m/s.(a) How far alo

Orlov [11]

Explanation:

(a) Net force acting on the block is as follows.

$F_{net} = -mg Sin (\theta)$

or, ma = -mg Sin (\theta)[/tex]

a = $-g Sin (\theta)$

= $-9.8 \times Sin (20^{o})$

= -3.35 $m/s^{2}$

According to the kinematic equation of motion,

$v^{2} - v^{2}_{o} = 2as$

Distance traveled by the block before stopping is as follows.

s = $\frac{v^{2} - v^{2}_{o}}{2a}$

= $\frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}$

= 21.5 m

According to the kinematic equation of motion,

v = $v_{o} + at$

0 = $12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t$

$t_{1}$ = 7.16 sec

Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.

(b) When the block is moving down the inline then net force acting on the block is as follows.

$F_{net} = -mg Sin (\theta)$

ma = $mg Sin (\theta)$

a = $g Sin (\theta)$

= $9.8 m/s^{2} \times Sin (20^{o})$

= 3.35 $m/s^{2}$

Kinematics equation of the motion is as follows.

s = $v_{o}t + \frac{1}{2}at^{2}$

21.5 m = $0 + \frac{1}{2} \times 3.35 m/s^{2} \times t^{2}$

$t_{2}$ = $\sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}$

= 3.58 sec

Hence, total time taken by the block to return to its starting position is as follows.

t = $t_{1} + t_{2}$

= 7.16 sec + 3.58 sec

= 10.7 sec

Thus, we can conclude that 10.7 sec time it take to return to its starting position.

3 0

4 years ago

1. How much heat must be absorbed by 375 grams of water to raise its

antiseptic1488 [7]

Answer:

39225J

Explanation:

Given parameters:

Mass of water = 375grams of water

Change in temperature = 25°C

Specific heat capacity of water = 4.184J/g°C

Unknown:

Amount of heat absorbed = ?

Solution:

To solve this problem, we use the expression below:

H = m c Ф

H is the heat absorbed

m is the mass

c is the specific heat capacity

Ф is the change in temperature

Insert the parameters and solve;

H = 375 x 4.184 x (25) = 39225J

5 0

3 years ago