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3 years ago

A fundamental particle cannot be: Accelerated Divided Observed Charged

Physics

1 answer:

tangare [24]3 years ago

7 0

A fundamental article cannot be Accelerated

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Sally travels by car from one city to another. She drives for 26.0 min at 83.0 km/h, 52.0 min at 41.0 km/h, and 45.0 min at 60.0

Anna007 [38]

The average speed is determined by the following formula:

average speed = [sum of (speed * time for which that speed was traveled)] / total time

average speed = [(83 * 26 + 41 * 52 + 60 * 45 + 0 * 15) / 60] / [(26 + 52 + 45 + 15) / 60]
*note: The division by 60 is to convert minutes to hours. We see that the 60 cancels from the top and bottom of the division

average speed = 50.65 km/hr

The total distance traveled is equivalent to the numerator of the fraction we used in the first part. This is:
Distance = (83 * 26 + 41 * 52 + 60 * 45 + 0 * 15) / 60

Distance = 116.5 kilometers

6 0

4 years ago

Is bacteria used in photosynthesis

madreJ [45]

Yes it is because the teacher told me in science lol ☺☺

6 0

3 years ago

When pieces of a plant fall to the ground take root and become a new plant what type of reproduction has occurred

Thepotemich [5.8K]

Vegetative propagation

5 0

4 years ago

A motorboat traveling from one shore to the other at a rate of 5m/s east encounters a current flowing at a rate of 3.5m/s north

nikdorinn [45]

Answer:

Resultant velocity will be equal to 6.10 m/sec

Explanation:

We have given a motorbike is traveling with 5 m/sec in east

And a current is flowing at a rate of 3.5 m /sec in north

We know that east and north is perpendicular to each other

So resultant velocity will be vector sum of both velocity

So resultant velocity $v=\sqrt{5^2+3.5^2}=6.10m/sec$

So resultant velocity will be equal to 6.10 m/sec

6 0

4 years ago

The air is less dense at higher elevations, so skydivers reach a high terminal speed. The highest recorded speed for a skydiver

swat32

Answer:

the terminal velocity v_t= 202.96 m/s≅203 m/s

Explanation:

The expression for the terminal velocity

$v_t= \sqrt{\frac{2mg}{\rho AC_d} }$

here, C_d is the drag coefficient for the cylinder is 1.15

The surface density of the air at 20°C is

ρ_surface = 1.2041 kg/m^3

the density of air at an altitude of 39000 m

ρ= 4.3/100×39000 = 0.05177 kg/m^3

now substitute these values in equation above

we get

$v_t= \sqrt{\frac{2\times90\times9.81}{0.05177\times0.72\times1.15} }$

v_t= 202.96 m/s≅203 m/s

the terminal velocity v_t= 202.96 m/s≅203 m/s

5 0

4 years ago