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3 years ago

Drag each tool to the correct category. Identify the common use of each tool used by space scientists and earth scientists

Physics

2 answers:

inysia [295]3 years ago

8 0

<h3>Answer</h3>

Tools that magnify objects Tools that don't magnify objects

magnifying glass rule

telescope binoculars

microscope

<h3>Explanation</h3>

radio dish: is use to receive radio waves from the air.

binoculars: provide a magnified stereoscopic view of distant objects.

streak plate: is a technique which is used to grow bacteria on a growth media surface so that individual bacterial colonies are isolated and sampled.

wind vane: is used to determine wind direction.

ruler: is used for measuring a line.

microscope: is used to magnify the objects which are not visible by naked eyes.

magnifying glass: is used to magnifying tiny objects which are difficult to see yet they are visible.

telescope: use by space scientist to examine astronomical event that may take place in the sky which is too far.

navik [9.2K]3 years ago

4 0

Answer:

Tools that magnify objects:

1. Magnifying Glass

2. Binocular

3. Telescope

4. Microscope

5. Radio Dish

Tools that do not magnify objects:

1. Wind vane

2. Streak Plate

3. Ruler

Explanation:

The tools or devices that can show an object closer and bigger are called magnifying instruments. These tools have lens/mirrors or some other mechanism to make the incident light beams converge at one point and focus. If you notice, Magnifying glass, Binoculars, Telescope, and microscope use convex lens/concave mirrors which converge the light beams at one point. Thus they will magnify. Radio dish also converges the incident radio waves on the receiver. These might not visible to naked eye but magnification do occur in them as well with the help of a concave dish structure.

Rest other tools do not magnify any object as they don't have any such parts in them which can make the light beams converge at one point.

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An 85.0-N crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at 28.0°

irina [24]

Answer:

A 75.1 N and a direction of 152° to the vertical.

B 85.0 N at 0° to the vertical.

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A) The interaction partner of this normal force has what magnitude and direction?

The interaction partner of this normal force is the component of the weight of the crate perpendicular to the ramp. <u>It has a magnitude of 85cos28° = 75.1 N and a direction of 180° - 28° = 152° to the vertical(since it is directed downwards perpendicular to the ramp).</u>

B) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?

Since this force has to balance the weight of the crate, its magnitude is 85.0 N. Its direction has to be vertically opposite to that of the weight.

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3 years ago

A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.

sladkih [1.3K]

Explanation:

Given that,

Charge 1, $q_1=6.75\ nC=6.75 \times 10^{-9}\ C$

Charge 2, $q_2=4.46\ nC=4.46\times 10^{-9}\ C$

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

$F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}$

$F=6.84\times 10^{-8}\ N$

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.

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3 years ago

A billiard ball moving at 5 m/s strikes another ball which is initially at rest. After the collision, the first ball moves at a

ziro4ka [17]

Answer:

The velocity of the second ball is approximately 2.588 m/s

The angle direction of the second ball is 75° counterclockwise from the horizontal

Explanation:

The initial velocity of the first billiard ball = 5 m/s

The initial velocity of the billiard ball the first billiard ball strikes = 0 m/s

The final velocity of the first billiard ball = 4.35 m/s

The final direction of motion of the first billiard ball = 30° below its original motion

For perfectly elastic collision, whereby the target is at rest initially, by conservation of momentum, we have;

m₁ × $\underset{v_1}{\rightarrow}$ = m₁· $\underset{v'_1}{\rightarrow}$ + m₂· $\underset{v'_2}{\rightarrow}$

Which gives;

m₁ × 5·i = m₁·((√3)/2×5·i - 2.5·j) + m₂· $\underset{v'_2}{\rightarrow}$

∴ m₂· $\underset{v'_2}{\rightarrow}$ = m₁ × 5·i - m₁·((√3)/2×5·i - 2.5·j)

m₂· $\underset{v'_2}{\rightarrow}$ = m₁ × 5·(1 - √3/2)·i + m₁·2.5·j = m₁ × 2.5·(2 - √3)·i + m₁·2.5·j

Therefore, given that the mass of both billiard balls are equal, we have, m₁ = m₂, which gives;

m₂· $\underset{v'_2}{\rightarrow}$ = m₁· $\underset{v'_2}{\rightarrow}$ = m₁ × 2.5·(2 - √3)·i + m₁·2.5·j

∴ $\underset{v'_2}{\rightarrow}$ = 2.5·(2 - √3)·i + 2.5·j

The magnitude of the velocity of the second ball is $\underset{v'_2}{\rightarrow}$ = √((2.5·(2 - √3))² + 2.5²) ≈ 2.588 m/s

The direction of the second ball, θ = arctan(2.5/((2.5·(2 - √3))) = 75° counterclockwise from the horizontal.

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3 years ago