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Luba_88 [7]
3 years ago
10

1.Which of the following is evidence that the Earth is round?

Physics
2 answers:
Tju [1.3M]3 years ago
7 0
1. B (the ship slowly appears on the horizon as it approaches)
2.B (false)
PtichkaEL [24]3 years ago
5 0

Answer:

1\D. Both A and C

2\A true

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An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming
Vlad [161]

Answer:

Explanation:

Given

mass of archer m=0.3\ kg

Average force F_{avg}=201\ N

extension in arrow x=1.3\ m

Work done to stretch the bow with arrow

W=F\cdot x

W=201\times 1.3=261.3\ m

This work done is converted into kinetic Energy of arrow

W=\frac{1}{2}mv^2

where v= velocity of arrow

261.3=\frac{1}{2}\times 0.3\times v^2

v=\sqrt{1742}

v=41.73\ m/s

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

W=mgh

261.3=0.3\times 9.8\times h

h=\frac{261.3}{0.3\times 9.8}

h=88.87\ m

   

4 0
3 years ago
A printed circuit board (PCB) is supported by a chassis that is attached to a vibrating motor. The board is 1.6 mm thick, 200 mm
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5 0
3 years ago
Which of the following describes the charge of an atom before any electrons are transferred?
Nadya [2.5K]
I am pretty sure that<span> the following which describes the charge of an atom before any electrons are transferred is </span>neutral charge. According to the fact that <span> atom is always like that before ionization, this answer is definitely correct. Hope it helps!</span>
7 0
3 years ago
A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0
3 years ago
the amount of time for one particle of the medium to make one comlpete vibration cycle is a good description of
-Dominant- [34]
That's the definition of the PERIOD of the vibration.
It's exactly the reciprocal of the vibration's frequency.
7 0
3 years ago
Read 2 more answers
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