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3 years ago

Most of the Earth’s water is found in the rivers and lakes. t or d

Physics

2 answers:

salantis [7]3 years ago

5 0

False most of earths water is found in oceans .

Temka [501]3 years ago

4 0

False the majority of the water on earth is found in the oceans

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A rock is thrown straight upward off the edge of a balcony that is 5 m above the ground. The rock rises 10 m, then falls all the

Dvinal [7]

Displacement = distance and direction from the start-point
to the end-point, regardless of the route followed on the way.

From the throw to the 'plop', the displacement is 5 meters down.

7 0

3 years ago

A slanted vector has a magnitude of 41 N and is at an angle of 23 degrees north of east. What are the magnitude and direction of

jenyasd209 [6]

Answer:

The horizontal component is 37.74 N to east

The vertical component is 16.02 N to North

Explanation:

Any slant vector has two components

→ Horizontal component = R cos Ф

→ Vertical component = R sin Ф

→ R is the magnitude of the vector

→ Ф is the direction of the vector with positive part of the horizontal axis

A slanted vector has a magnitude of 41 N and is at an angle of 23

degrees north of east

23° north of east means the angle between the vector and the

east direction is 23° (east is the positive horizontal direction)

That means R is 41 N and Ф is 23°

→ R = 41 N , Ф = 23°

→ The horizontal component = 41 × cos(23) = 37.74 N east

→ The vertical component = 41 × sin(23) = 16.02 N North

<em>The horizontal component is 37.74 N to east</em>

<em>The vertical component is 16.02 N to North</em>

7 0

3 years ago

Determine the image distance and image height for a 5.00 cm tall object placed 30.0 cm from a double convex lens with a focal le

vlabodo [156]

Answer:

The image distance is 30 cm

image height = - 5 cm

Explanation:

The formula for calculating the image distance is expressed as

1/f = 1/u + 1/v

where

f is the focal length

u is the object distance

v is the image distance

From the information given,

u = 30

f = 15

By substituting these values into the formula,

1/15 = 1/30 + 1/v

1/v = 1/15 - 1/30 = (2 - 1)/30 = 1/30

Taking the reciprocal of both sides,

v = 30

The image distance is 30 cm

magnification = image height/object height = - v/u

Given that object height = 5 cm, then

image height/5 = - 30/30 = - 1

image height = - 5 * 1

image height = - 5 cm

8 0

1 year ago

Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20 g of mass. What

Kobotan [32]

Answer:

Mass of the wooden Block is 20g.

Explanation:

The buoyant force equation will be used here

Buoyant Force= ρ*g*1/2V Here density used is of water

m*g= ρ*g*1/2V

Simplifying the above equation

2m= ρ*V Eq-1

Also we know from the question that

ρ*V = m + 0.020 Eq-2 ( Density = (Mass+20g)/Volume )

Equating Eq-1 & Eq-2 we get

2m = m+0.020

m = 0.020kg

m = 20g

6 0

3 years ago

Z. A force that gives a 8-kg objet an acceleration of 1.6 m/s^2 would give a 2-kg object an

Paladinen [302]

Answer:

$\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}$

Explanation:

We need to find the acceleration of the 2 kilogram object. Let's complete this in 2 steps.

<h3>1. Force of 1st Object </h3>

First, we can find the force of the first, 8 kilogram object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

$F=m \times a$

The mass of the object is 8 kilograms and the acceleration is 1.6 meters per square second.

m= 8 kg
a= 1.6 m/s²

Substitute these values into the formula.

$F= 8 \ kg * 1.6 \ m/s^2$

Multiply.

$F= 12.8 \ kg*m/s^2$

<h3>2. Acceleration of the 2nd Object </h3>

Now, use the force we just calculated to complete the second part of the problem. We use the same formula:

$F= m \times a$

This time, we know the force is 12.8 kilograms meters per square second and the mass is 2 kilograms.

F= 12.8 kg *m/s²
m= 2 kg

Substitute the values into the formula.

$12.8 \ kg*m/s^2= 2 \ kg *a$

Since we are solving for the acceleration, we must isolate the variable (a). It is being multiplied by 2 kg. The inverse of multiplication is division. Divide both sides of the equation by 2 kg.

$\frac {12.8 \ kg*m/s^2}{2 \ kg}= \frac{2\ kg* a}{2 \ kg}$

$\frac {12.8 \ kg*m/s^2}{2 \ kg}=a$

The units of kilograms cancel.

$\frac {12.8}{2}\ m/s^2=a$

$6.4 \ m/s^2=a$

The acceleration is 6.4 meters per square second.

4 0

2 years ago