An arrow is fired horizontally at a target that is 40 m away (horizontal d). The arrow drops 0.5 m before striking the target. With what velocity is the arrow fired (what is the horizontal vi)?

Answers

Answer 1

Answer:

Answer:

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Explanation:

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Displacement is the change in velocity of an object.

True or false

Answers

Words less true are seldom if ever spoken.

False, it is the change in distance. Chance in velocity would be Δv

Why was the idea of plate tectonics difficult for many scientists to accept for many years after it was first introduced?SCIENCE QUESTION

Answers

It's kind of funny really. The guy who invented the idea in the first place had actually been made fun of because the thought of that happening was improbable in their mind. Also, Wegener ( the guy who invented the idea) was not even a geologist but a meteorologist. But later it had been backed up due to earthquakes being record consecutively in one area, and the graphs turning out differently.

Which would not increase voltage and generate an electrical current?A. a car battery

B. a thick wire

C. a solar cell

D. a generator

Answers

Answer: The correct answer is a thick wire.

Explanation:

In the given options of the problem, a car battery is needed to generate an electrical current. It forces the electrons to flow. It makes the current flow in the circuit.

Solar cell is a device which converts the solar energy into the electrical energy. It can increase the voltage and generate an electrical current.

Generator is a device which converts the mechanical energy into the electrical energy. It can increase the voltage and generate an electrical current.

A thick wire is not used to increase voltage and generate an electrical current. The thick wire has low resistance in comparison to the thin wire.

Therefore, the correct option is (B).

B. A thick wire
I think this is right, hope this helps

How do the units of work and power compare? a) The unit for work is a watt. The unit for power is a joule, which is a watt-second. b)The unit for work is a watt. The unit for power is a joule, which is a watt per second. c)The unit for work is a joule. The unit for power is a watt, which is a joule-second.
d)The unit for work is a joule. The unit for power is a watt, which is a joule per second.

Answers

Answer:

Option (d) is correct.

Explanation:

Work done is given by :

W = Fd, where F is force and d is displacement

Unit of work done :

The SI unit of force is Newton (N) and that of displacement is meter (m). So, the unit of work done is N-m. It is call Joule. It means that the unit of work done is Joule.

Power is given by rate at which the work is done. It is given by :

P = W/t, W is work done and t is time

Unit of power:

Unit of work is Joule (J) and that of time is second (s). It means that the unit of power is Watt and it is equal to Joule/second

Hence, the correct option is (d) "The unit for work is a joule. The unit for power is a watt, which is a joule per second".

Answer: The unit for work is a joule. The unit for power is a watt, which is a joule per second.

Explanation:

You're in an airplane that flies horizontally with speed 1100 km/h (300 m/s ) when an engine falls off. Neglecting air resistance, assume it takes 26 s for the engine to hit the ground.Find the height of airplane.

Answers

Answer:

The height of the airplane is 3312.4 meters.

Explanation:

Given that,

Speed of the airplane, v = 1100 km/h = 300 m/s

It takes 26 s for the engine to hit the ground. We need to find the height of the airplane. Using second equation of motion to find it as :

$h=ut+(1)/(2)at^2$

Here, u = 0 and a = g

$h=(1)/(2)gt^2\n\nh=(1)/(2)* 9.8* (26)^2\n\nh=3312.4\ m$

The height of the airplane is 3312.4 meters.

A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates.(a) A voltmeter reads 45.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of this material?
(b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

Answers

(A) 3.9

When a dielectric is inserted between the plates of a capacitor, the capacitance of the capacitor increases according to the equation:

$C' = k C$ (1)

where

C' is the final capacitance

k is the dielectric constant

C is the original capacitance

The capacitance is inversely proportional to the to voltage across the plates:

$C=(Q)/(V)$ (2a)

where Q is the charge stored and V the potential difference across the plates. We can rewrite C' (the capacitance of the capacitor filled with dielectric) as

$C'=(Q)/(V')$ (2b)

Substituting (2a) and (2b) into (1), we find

$V'=(V)/(k)$ (3)

where

V = 45.0 V is the original voltage across the capacitor

V' = 11.5 V is the voltage across the capacitor filled with dielectric

Solving for k,

$k=(V)/(V')=(45.0 V)/(11.5 V)=3.9$

(B) 22.8 V

When the dielectric is partially pulled away, the system can be assimilated to a system of 2 capacitors in parallel, of which one of them is filled with dielectric and the other one is not.

Keeping in mind that the capacitance of a parallel-plate capacitor is proportional to the area of the plates:

$C \propto A$

and in this case, the area of the capacitor filled with dielectric is just 1/3 of the total, we can write:

$C_1 = (2)/(3)C\nC_2 = (1)/(3)kC$

where C1 is the capacitance of the part non-filled with dielectric, and C2 is the capacitance of the part filled with dielectric. The total capacitance of the system in parallel is

$C'=C_1 + C_2 = (2)/(3)C+(1)/(3)kC=((2)/(3)+(1)/(3)k)C$

Substituting,

$C'=((2)/(3)+(1)/(3)(3.9))C=1.97 C$

This is equivalent to a capacitor completely filled with a dielectric with dielectric constant k=1.97. Therefore, using again eq.(3), we find the new voltage:

$V'=(V)/(k)=(45.0 V)/(1.97)=22.8 V$

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