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3 years ago

___________ force can be thought of as either a push or a pull, as long as it makes the object move in a circular path.

Physics

2 answers:

gogolik [260]3 years ago

7 0

Explanation:

Centripetal force is a force required by an object to move in a circular path. Mathematically, it is given by :

$F_c=\dfrac{mv^2}{r}$

m is the mass of an object

v is the velocity

r is the radius of circular path

The centripetal force acts towards the center of the circle. The speed of object remains constant while its velocity keeps on changing always. In the absence of the centripetal force, the object moves in straight line.

fomenos3 years ago

5 0

I believe it would be the Centripetal force.

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You have three objects of varying shapes and sizes: Object 1 is a rectangular block of tin. Object 2 is a cube of aluminum. Obje

Over [174]

Answer: a. m = 7.7 kg

b. V = 435.52 in³

c. m = 1927 kg

d. V = 335.37 cm³

e. m = 3 kg

Explanation: <u>Density</u> is the ratio of mass per volume, i.e., it's the measure of an object's compactness. Its representation is the greek letter ρ.

The formula for density is

$\rho=\frac{m}{V}$

Density's unit in SI is kg/m³, but it can assume lots of other units.

Some unit transformations necessary for the resolution of the question:

1 L = 1 dm³ = 1000 cm³

1 in³ = 16.3871 cm³

1 g = 0.001 kg

a. V = 1.34 L = 1340 cm³

$\rho=\frac{m}{V}$

$m=\rho.V$

m = 5.75 * 1340

m = 7705 g => 7.705 kg

Mass of object 1 with volume 1.34L is 7.7 kg.

b. A cube's volume is calculated as V = side³

V = 7.58³

V = 435.52 in³

Volume of object 2 is 435.52 in³.

c. Using 1 in³ = 16.3871 cm³ to change units:

V = 435.52 * 16.3871

V = 713689.4 cm³

Then, mass will be

$m=\rho.V$

m = 2.7 * 713689.4

m = 1926961.4 g => 1927 kg

Mass of object 2 is 1927 kg.

d. Volume of a sphere is calculated as $V=\frac{4}{3}.\pi.r^{3}$

Diameter is twice the radius, then r = 4.31 cm.

Volume is

$V=\frac{4}{3}.\pi.(4.31)^{3}$

V = 335.37 cm³

Volume of object 3 is 335.37 cm³.

e. $m=\rho.V$

m = 8.96 * 335.37

m = 3004.91 g => 3 kg

Mass of object 3 is 3 kg.

4 0

3 years ago

Consider the possible scenarios and determine which of them are true for positive feedbacks or negative feedbacks.A. Increased p

avanturin [10]

Answer:

Scenario A, B and E is True.

Explanation:

Scenario A) True. Removing carbon dioxide from atmosphere decreases greenhouse effect of atmosphere. Thus, temperature rise decreases.

Scenario B) True. The more evaporation creates the more greenhouse effect. Therefore, temperature rise increases.

Scenario C) False. Removing carbon dioxide from atmosphere decreases greenhouse effect of atmosphere. Thus, temperature rise decreases.

Scenario D) False. The more evaporation creates the more greenhouse effect. Therefore, temperature rise increases.

Scenario E) True. If reflected radiation increases from Earth, temperature rise of the Earth will decrease. Ice cover increases reflectivity which leads temperature level decrease.

Scenario F) False. If reflected radiation increases from Earth, temperature rise of the Earth will decrease. Ice cover increases reflectivity which leads temperature level decrease.

5 0

3 years ago

Imagine a glass box that is completely sealed and in that box is a cactus in a small pot which statement is most accurate?

Dennis_Churaev [7]

I THINK C BECAUSE IF IT IS A GLASS BOX HOW DID A CACTUS GET IN AND NOTHING CAN GET IN OR OUT OF THE BOX SO THERE IS NO CACTUS IN THE BOX

6 0

3 years ago

Read 2 more answers

After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T

oksian1 [2.3K]

Answer:

required distance is 233.35 m

Explanation:

Given the data in the question;

Sound intensity $I$ = 1.62 × 10⁻⁶ W/m²

distance r = 165 m

at what distance from the explosion is the sound intensity half this value?

we know that;

Sound intensity $I$ is proportional to 1/(distance)²

i.e

$I$ ∝ 1/r²

Now, let r² be the distance where sound intensity is half, i.e $I$ ₂ = $I$ ₁/2

Hence,

$I$ ₂/ $I$ ₁ = r₁²/r₂²

1/2 = (165)²/ r₂²

r₂² = 2 × (165)²

r₂² = 2 × 27225

r₂² = 54450

r₂ = √54450

r₂ = 233.35 m

Therefore, required distance is 233.35 m

6 0

3 years ago

The gold has a density of 19300 kg/m3 calculate the mass of one gold bar 1= 2.54cm

icang [17]

Fair enough, but you'll have to tell us the volume of the bar first.

5 0

3 years ago