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3 years ago

3 simple machines that make up a can opener

1 answer:

6 0

Hello There

Answer: The wheel and axle, the lever, and the wedge

Reason: The two long arms that clamp onto the can are levers. The handle which is used to rotate the can is the wheel and axle. Finally the blade is the wedge

I hope I helped
-Chris

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A car starts from rest with an acceleration of 5 ft/s. What is its velocity after it has gone 600 ft?

Soloha48 [4]

Answer:

First, let’s correct the question. Acceleration is the rate of change in velocity. Its unit therefore is ft/sec/sec. If S is the distance traveled for a given duration, S = Vot + (1/2)at^2 where Vo is the initial velocity, a is the acceleration and t is the time. For Vo = 0, a = 6m/sec/sec and t = 3 sec. The distance traveled is S = 0 + (1/2) x 6 x 3^2 = 27 meters

3 0

3 years ago

An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop-the-loop maneuver. the acceleration

Yanka [14]

The acceleration due to gravity serves as the centripetal acceleration of the objects that orbits the Earth. The centripetal acceleration due to gravity is calculated through the equation,

a = v²/r

where v is the speed and r is the radius. Substituting the known values to the equation,

9.8 m/s² = (420 m/s)² / r

The value of r from the equation is 18000 m or equal to 18 km.

Answer: 18 km

6 0

3 years ago

If a car moves with a uniform speed of 2m/s in a circle of radius 0.4m, what is its angular speed?

REY [17]

Answer:

The car's angular speed is $\frac{rad}{s}$ .

Explanation:

Angular velocity is usually measured with $\frac{radians}{sec}$ , so I'm going to use that to answer your question.

The relationship between tangential velocity and angular velocity (ω) is given by:

$V = \omega R$

Using the values from the question, we get:

$2 \frac{m}{s} = \omega (0.4m)$

$\omega = 5 \frac{rad}{s}$

Therefore, the car's angular speed is $5 \frac{rad}{s}$ .

Hope this helped!

3 0

3 years ago

If the absolute pressure of a gas is 550.280 kPa, its gage pressure is A. 101.325 kPa. B. 651.605 kPa. C. 448.955 kPa. D. 277.28

guajiro [1.7K]

Answer:

Option C is the correct answer.

Explanation:

Absolute pressure is sum of gauge pressure and atmospheric pressure.

That is

$P_{abs}=P_{gauge}+P_{atm}$

We have

$P_{abs}=550.280 kPa\\\\P_{atm}=1atm=101325Pa=101.325kPa$

Substituting

$P_{abs}=P_{gauge}+P_{atm}\\\\550.280=P_{gauge}+101.325\\\\P_{gauge}=448.955kPa$

Option C is the correct answer.

7 0

3 years ago

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is . Find the time i

a_sh-v [17]

Answer:

a) It took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

b) The minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensitie. That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies:

Gamma rays

X-rays

Ultraviolet rays

Visible region

Infrared

Microwave

Radio waves.

Any radiation that belongs to electromagnetic spectrum has a speed in vacuum of $3x10^{8}m/s$ .

a) Find the time it took for his voice to reach the Earth via radio waves.

To know the time that took for Neil Armstrong's voice to reach the Earth via radio waves, the following equation can be used:

$c = \frac{d}{t}$ (1)

Where v is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

$t = \frac{d}{c}$ (2)

The distance from the Earth to the Moon is $3.85x10^{8} m$ , therefore.

$t = \frac{3.85x10^{8} m}{3x10^{8}m/s}$

$t = 1.28s$

Hence, it took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

b) Determine the minimum time that will be required for a message from Mars to reach the Earth via radio waves.

The distance from the Earth to the Mars at its closest approach is $5.76x10^{10}m$ , therefore.

$t = \frac{5.76x10^{10}m}{3x10^{8}m/s}$

$t = 192s$

Hence, the minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

3 0

3 years ago