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Rama09 [41]
3 years ago
15

Assertion(A):The distance moved by an object in unit time is called its speed. Reason (R):Faster vehicles have higher speeds. i)

A is correct while R is incorrect. ii) R is correct while A is incorrect. iii) Both A and R are correct iv) Both A and R are incorrect.
Physics
1 answer:
ikadub [295]3 years ago
8 0

Answer:

it's ii) R is correct while A is incorrect

Explanation:

cuz, both statements are correct but the reason is not the correct reason for the assertion,

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What does this mathematical expression mean? D 1/T
san4es73 [151]
The question is incomplete. If D=1/T then the answer is D, but if instead of =, there is a proportional sign, then the answer is C.
7 0
3 years ago
Read 2 more answers
Two 20.0 g ice cubes at − 20.0 ∘ C are placed into 285 g of water at 25.0 ∘ C. Assuming no energy is transferred to or from the
Lelechka [254]

Answer:

Ft = 17.48°C

Explanation:

Ft is the final temperature. However, ice absorbs heat during two process of melting and cooling and as such, there is no loss of heat to or from the surrounding hence by conservation of energy.

Therefore,

Heat absorbed by water of 20g = heat rejected by water of 265g.

So; M(ice)[C(ice) [(ΔT) + LH(ice) + C(water)(ΔT)] = C(water) M(water) (ΔT)

So, 20[(2.108) [0 - (-20)] + 333.5 + 4.187(Ft - 0)]] = (285)(4.187) (25 - Ft)

To get;

7513 + 83.74 Ft = 29832.4 - 1193.3 Ft

So factorizing, we get;

83.74 Ft + 1193.3 Ft = 29832.4 - 7513

So; 1277.04 Ft = 22319.4

So; Ft = 22319.4/1277.04 = 17.48°C

7 0
2 years ago
On the moon, the acceleration due to gravity is one-sixth that of earth. That is gmoon = gearth /6 = (9.8 m/s2 )/6 = 1.63 m/s2 .
MAVERICK [17]
The pendulum would differ from 300 inches
8 0
2 years ago
Please help. I don’t understand this
skad [1K]

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>

and under constant acceleration,

<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2

According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so

∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2

∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2

∆<em>x</em> = 20.00 m

5 0
2 years ago
Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission
bija089 [108]

Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law                     P = σ A e T⁴

Wien displacement law   λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

       P₁ = σ A T₁⁴

T = 12000K

       P₂ = σ A T₂⁴

       P₂ / P₁ = T₂⁴ / T₁⁴

       P₂ / P₁ = (12000/3000)⁴

       P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

       λ T = 2,898 10-3

       λ₁ = 2.89810-3 / T

       λ₁ = 2,898 10-3 / 3000

       λ₁ = 0.966 10-6 m

      λ₁ = 966 nm

T = 12000K

      λ₂ = 2,898 10-3 / 12000

      λ₂ = 0.2415 10-6 m

      λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

5 0
3 years ago
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