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3 years ago

Please help me! Science is not my best subject

Physics

1 answer:

Paul [167]3 years ago

8 0

1)solid
2)contact
3)conduction
4)Fluid
5)Less
6)More
7)Convection
8)Waves
9)Radiation
10)emit
11)absorb

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What is the best way to start a summary paragraph

pochemuha

Start with what the paragraph is about and put it basically in your own words

8 0

4 years ago

Read 2 more answers

Three capacitors having capacitances of 8.40, 8.40, and 4.20 μFμF, respectively, are connected in series across a 36.0-V potenti

son4ous [18]

Answer:

a)Q=71.4 μ C

b)ΔV' = 10.2 V

Explanation:

Given that

C ₁= 8.7 μF

C₂ = 8.2 μF

C₃ = 4.1 μF

The potential difference of the battery, ΔV= 34 V

When connected in series

1/C = 1/C ₁ + 1/C₂ + 1/C₃

1/ C= 1/8.4 +1 / 8.4 + 1/4.2

C=2.1 μF

As we know that when capacitor are connected in series then they have same charge,Q

Q= C ΔV

Q= 2.1 x 34 μ C

Q=71.4 μ C

As we know that when capacitor are connected in parallel then they have same voltage difference.

Q'= C' ΔV'

C'= C ₁+C₂+C₃ (For parallel connection)

C'= 8.4 + 8.4 + 4.2 μF

C'=21 μF

Q'= C' ΔV'

Q'=3 Q

3 x 71.4= 21 ΔV'

ΔV' = 10.2 V

3 0

3 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

4 years ago

a man hikes 6.6 km north along a straight path with an average velocity of 4.2 km/h to the north. he rest at a bench for 15 min.

SSSSS [86.1K]

Answer:

2.6h

Explanation:

I attached the image below of the work hope you can see it. Hope this helps!

6 0

3 years ago

Two particles A and B start simultaneously from a Point P with velocities 20 m/s and 30 m/s respectively. A and B move with acce

zysi [14]

Answer:

20 m/s

Explanation:

Given

u(A) = 20 m/s
u(B) = 30 m/s
acceleration equal in magnitude but opposite in direction

Solving

Velocity of A at Q = 30 m/s
From, P to Q, Δv(A) = 30 - 20 = +10 m/s
Therefore, velocity of B at Q will be decreased by 10 as it is equal in magnitude but opposite in direction to A
Δv(B) = v(B at Q) - u(B at P)
-10 m/s = v(B at Q) - 30 m/s
v(B at Q) = 30 - 10 = 20 m/s

6 0

3 years ago