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1 month ago

If the position of an object is zero at one instant, what is true about the velocity of that object?

1 answer:

6 0

If the position of an object is zero at a particular moment, this does not provide any indication about its velocity. It might simply be moving through that point, and you observed it exactly when it was at zero.

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A student solving a physics problem for the range of a projectile has obtained the expression r= v20sin(2θ)g where v0=37.2meter/

ValentinkaMS [3465]

The formula for range is:

$R = \frac{v_o^2 sin2\theta}{g}$

Given values are:

$v_0=37.2m/s$

where θ equals 14.1 degrees

$g=9.80m/s^2$

Using the equation above,

$R = \frac{37.2^2 sin2*14.1}{9.80}$

The calculated range is 66.7 meters.

Therefore, the range is approximately 66.1 meters.

5 0

2 months ago

Read 2 more answers

Incoming solar radiation (light energy) is absorbed by the Earth's surface and converted to outgoing infrared radiation (heat en

Softa [3030]

An increase in temperature (Global warming) is observed. The solar radiation is transformed into heat energy absorbed by Earth's surface. In line with the law of conservation of energy, energy can only transition forms rather than disappear. If an increasing quantity of energy accumulates on Earth with minimal release, this imbalance in energy demand leads to a rise in temperature due to excessive heat absorption, largely a result of pollution from fossil fuel combustion releasing CO2 and other harmful emissions. Ordinarily, the residual solar energy would escape back into space, but CO2 and similar contaminants trap this heat, thus elevating Earth's temperature.

8 0

1 month ago

A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold

Maru [3345]

Answer:

(a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

$Q=\dfrac{V}{t}$

Insert the values into the formula

$Q=\dfrac{405}{90}$

$Q=4.5\ cm^3/s$

(a). We will compute the coefficient of permeability

Applying the formula for permeability

$Q=kiA$

$k=\dfrac{Q}{iA}$

$k=\dfrac{Ql}{Ah}$

Where, Q=discharge

l = length

A = cross-sectional area

h=constant head that causes flow

Plugging the value into the formula

$k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}$

$k=8.6\times10^{-3}\ cm/s$

The coefficient of permeability measures as $8.6\times10^{-3}\ cm/s$ .

(c). To ascertain the discharge velocity during the testing phase

Utilizing the discharge velocity formula

$v=ki$

$v=\dfrac{kh}{l}$

Substituting the values into the equation

$v=\dfrac{8.6\times10^{-3}\times38}{17.5}$

$v=0.0187\ cm/s$

The discharge velocity during the test measures 0.0187 cm/s.

Thus, (a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity computes at 0.0330 cm/s.

(c). The observed discharge velocity during the test equals 0.0187 cm/s.

8 0

1 month ago

Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fish

Sav [3153]

Response:

In this scenario, the refractive index of seawater is 1.33, while the index for air is 1. Because of this, the refraction angle is smaller than the angle of incidence, making the fish appear closer

Conversely, when viewed from the fish's perspective looking at a person's face, the angle has increased, making the person appear further away

Clarification:

This discussion is analyzed by applying the refraction law, which states that when a light ray crosses from one medium to another, its path bends due to differing indices of refraction,[ [TAG_20]]

n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the refractive indices and θ represents the angles for each medium.

Here, with seawater being 1.33 and air at 1, the refraction angle remains lesser than the angle of incidence, leading to the fish appearing nearer

1 sin θ₁ = 1.33 sin θ₂

θ₂ = sin⁻¹ ( 1/1.33 sin θ₁)

When the fish gazes at the human face, the angle's reason increases, hence making the face seem more distant

4 0

23 days ago

A scared elephant has a mass of 7000 kg. The mouse that frightened the elephant is 0.02 kg. The distance between the elephant an

Maru [3345]

a. $9.34\cdot 10^{-9} N$

The equation for the gravitational force acting between two masses is given by:

$F=G\frac{Mm}{r^2}$

where

G represents the gravitational constant

M and m refer to the masses of the two objects

r indicates the distance separating the objects

In this scenario, we have:

M = 7000 kg, the elephant's mass

m = 0.02 kg, the mouse's mass

r = 1 m, the distance between them

Plugging these values into the equation will allow us to compute the gravitational force that the elephant applies on the mouse:

$F=(6.67\cdot 10^{-11}) \frac{(7000)(0.02)}{1^2}=9.34\cdot 10^{-9} N$

b. $-9.34\cdot 10^{-9}N$

The next part can be tackled by recalling Newton’s third law, which asserts that:

"When object A applies a force (action) on object B, object B simultaneously applies an equivalent and opposite force (reaction) on object A". In this illustration, the elephant is object A and the mouse is object B: thus, the gravitational force from the elephant on the mouse is the action, while the force from the mouse on the elephant acts as the reaction. Both forces match in magnitude yet differ in direction: therefore, the gravitational force the mouse exerts on the elephant can be expressed as

$F=-9.34\cdot 10^{-9}N$

where the negative signifies an opposite direction.

6 0

1 month ago