A boat floats because of which force?

Andreyy89

“Once in a blue moon” is a common expression that has been used for a long time, and which means 'not very often,' or 'very rarely. ' It often refers to an extra full moon; however, it has been used to describe the way the moon actually looked, when for different reasons it had turned a blueish color.
www.loc.gov › astronomy › item
What

4 0

3 years ago

A kayaker paddles at 4.0 m/s in a direction 30° south of west. He then turns and paddles at 3.7 m/s in a direction 20° west of s

vichka [17]

<u>Answer:</u>

Magnitude of resultant velocity of kayaker to the nearest tenth = 10 m/s

Direction of resultant velocity of kayaker = 49.32⁰ South of west.

<u>Explanation:</u>

Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.

First kayaker paddles at 4.0 m/s in a direction 30° south of west, kayaker paddles at 4.0 m/s in a direction 210° anticlockwise from positive horizontal axis.

So velocity of kayaker = 4 cos 210 i + 4 sin 210 j = -3.46 i - 2 j

He then turns and paddles at 3.7 m/s in a direction 20° west of south, kayaker paddles at 3.7 m/s in a direction 250° anticlockwise from positive horizontal axis.

So that velocity = -1.27 i - 3.48 j

So resultant velocity of kayaker = -3.46 i - 2 j +(-1.27 i - 3.48 j) = -4.71 i - 5.48 j

Magnitude of resultant velocity of kayaker = $\sqrt{(-4.71)^2+(-5.48)^2} = 7.23 m/s$

Magnitude of resultant velocity of kayaker to the nearest tenth = 10 m/s

Direction of resultant positive horizontal axis, θ = tan⁻¹(-5.48/-4.71) = 229.32⁰ = 49.32⁰ South of west.

6 0

3 years ago

Read 2 more answers

Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.3×109 V and the quantity of charg

Georgia [21]

Answer:

A) ΔU = 3.9 × 10^(10) J

B) v = 8420.75 m/s

Explanation:

We are given;

Potential Difference; V = 1.3 × 10^(9) V

Charge; Q = 30 C

A) Formula for change in energy of transferred charge is given as;

ΔU = QV

Plugging in the relevant values gives;

ΔU = 30 × 1.3 × 10^(9)

ΔU = 3.9 × 10^(10) J

B) We are told that this energy gotten above is used to accelerate a 1100 kg car from rest.

This means that the initial potential energy will be equal to the final kinetic energy since all the potential energy will be converted to kinetic energy.

Thus;

P.E = K.E

ΔU = ½mv²

Where v is final velocity.

Plugging in the relevant values;

3.9 × 10^(10) = ½ × 1100 × v²

v² = [7.8 × 10^(8)]/11

v² = 70909090.9090909

v = √70909090.9090909

v = 8420.75 m/s

4 0

3 years ago

A space probe is coasting through space at a steady speed of 100p feet per second. the booster rocket fires for 1/2 seconds so t

aleksandr82 [10.1K]

Answer:

Acceleration: $9800 ft/s^2$

Explanation:

The acceleration of an object is equal to the rate of change of velocity:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

For the space probe in this problem, we have:

u = 100 ft/s (initial velocity)

v = 5000 ft/s (final velocity)

t = 0.5 s (time taken)

Therefore, the acceleration is

$a=\frac{5000-100}{0.5}=9800 ft/s^2$

6 0

3 years ago

How many newtons does a 5-kg backpack weigh on Earth?

zvonat [6]

A 5kg backpack will weigh 49 newtons on earth

7 0

3 years ago