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3 years ago

Match the term and the definition

Physics

1 answer:

sineoko [7]3 years ago

3 0

Answer:

1. When a cold air mass and a warm air mass meet but are at a standstill, the boundary is called stationary front.

2. If warm or more humid air presses forward and colder or drier air draws back, the boundary is called warm front.

3. If colder or drier air presses forward and warmer or more humid air draws back, the boundary is called cold front.

4. Sometimes a warm air mass can get caught between two cold fronts, which can force the warm air up into a wedge shape is called occluded front.

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A football player runs directly down the field for 45m before turning to the right at an angle of 25 degrees from his original d

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3 years ago

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3 years ago

Calculate the induced electric field in a 54-turn coil with a diameter of 19.5 cm that is placed in a spatially uniform magnetic

storchak [24]

Answer:

E = 3.049 N/C

Explanation:

Induced electric field = e/2 *(pi) *r

Induced electric field = e /(3.14 *19.5) - Eq (1)

e = (pi)r^2*B/t

= 3.14 * (19.5/2*100)^2 * 0.50 T/ 0.1

= 1.867 V

Substituting this value in equation 1, we get –

E = 1.867 V/(3.14 *19.5/100)

E = 3.049 N/C

4 0

3 years ago

Pure water has a hydrogen ion concentration of 1 x 10 (-7) M and a ph of 7. A student measures a basic solution and finds that i

Lena [83]

The student must have made an error in her measurement in the concentration of hydrogen ion in the basic solution given. From her measurement, the concentration is 1 × 10−3 M. If we calculate the pH, we get:

pH = -log ( 1 × 10−3) = 3

A pH of 3 does not signifies a basic solution it is in the acidic range. So, the measurement is wrong.

6 0

3 years ago

A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v

Citrus2011 [14]

Answer:

part (a) $v\ =\ 10.42\ at\ 81.17^o$ towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

velocity of the river due to east = $v_r\ =\ 1.60\ m/s.$
velocity of the boat due to the north = $v_b\ =\ 10.3\ m/s.$

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are $90^o$

Let 'v' be the velocity of the boat relative to the shore.

$\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.$

Let $\theta$ be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore. $\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o$

part (b)

Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river. $\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s$

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river $\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m$

7 0

3 years ago