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3 years ago

6

A 74 KG student starting from rest slides down 11.8 m high water slide how fast is it going at the bottom of the slide?

1 answer:

-Dominant- [34]3 years ago

8 0

the answer is going to be 873.2

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PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (40pts)

denpristay [2]

Answer:

False, true

Explanation:

3 0

3 years ago

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. Now assuming Anna's far point was found to be 0.9 m (i.e., her eyes can't focus on any object more than 0.9 m away), what powe

Mama L [17]

Answer:1). Distance of far point x=0.9m

Therefore, since the image is virtual

-f=-x = -0.9m

Power of the concave lenses = 1/f = 1/-0.9

= -1.11D

2 ) near point is 21cm = 0.21m

Power = 4-1/near point

= 4/0.21

= 14.2D.

7 0

4 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

3 years ago

Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the

astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

$B = \frac{\mu_oI}{2\pi R}$

Where $\mu_o$ is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

$B_1 = \frac{\mu_oI_1}{2\pi R}$

$B_2 = \frac{\mu_oI_2}{2\pi R}$

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T = $\frac{\mu_oI_1}{2\pi R}$ - $\frac{\mu_oI_2}{2\pi R}$

or

4.0×10⁻⁶T = $\frac{\mu_o}{2\pi R}\times (I_1-I_2 )$

also

$\frac{I_1}{I_2} = \frac{3}{1}$

⇒ $I_1 = 3\times I_2$

substituting the values in the above equation we get

4.0×10⁻⁶T = $\frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)$

⇒ $I_2 = 1A$

also

$I_1 = 3\times I_2$

⇒ $I_1 = 3\times 1A$

⇒ $I_1 = 3A$

Hence, the larger of the two currents is 3A

3 0

3 years ago

URGENT PLEASE HELP!!!! GIVING BRAINLIEST!! If you answer this correctly ill answer some of your questions you have posted! (29pt

Furkat [3]

Chemical=B

Mechanical=C

Nuclear=A

Electromagnetic=D

Thermal=E

I think these are right but if not I’m sorry

Hope I helped :]

5 0

3 years ago

Read 2 more answers