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Elis [28]
3 years ago
12

Explain the characteristics of the Gypsum Hills regionIf we have a sample of silicon (Si) atoms that has 14 protons, 14 electron

s, and 18 neutrons
What is the name of this specific silicon isotope?

silicon-14
silicon-32
silicon-46
silicon-153
Physics
1 answer:
amid [387]3 years ago
8 0

Answer:

second one

Explanation:

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1 kg ball rolls off a 33 m high cliff, and lands 23 m from the base of the cliff. Express the displacement and the gravitational
enyata [817]

Answer:

d = <23, 33, 0> m ,    F_W = <0, -9.8, 0> ,   W = -323.4 J

Explanation:

We can solve this exercise using projectile launch ratios, for the x-axis the displacement is

         x = vox t

Y Axis  

         y = v_{oy} t - ½ g t²

It's displacement is

      d = x i ^ + y j ^ + z k ^

Substituting

      d = (23 i ^ + 33 j ^ + 0) m

Using your notation

   d = <23, 33, 0> m

The force of gravity is the weight of the body

         W = m g

        W = 1  9.8 = 9.8 N

In vector notation, in general the upward direction is positive

         W = (0 i ^  - 9.8 j ^ + 0K ^) N

         W = <0, -9.8, 0>

Work is defined

           W = F. dy

             W = F dy cos θ

In this case the force of gravity points downwards and the displacement points upwards, so the angle between the two is 180º

          Cos 180 = -1

           W = -F y

           W = - 9.8 (33-0)

           W = -323.4 J

6 0
2 years ago
Charlotte is driving at 66.5 mi/h and receives a text message. She looks down at her phone and takes her eyes off the road for 3
Alborosie

Answer:

the distance traveled by Charlotte in feet is 338.44 ft

Explanation:

Given;

speed of Charlotte, u = 66.5 mi/h

time of motion, t = 3.47 s

The distance traveled by Charlotte in feet is calculated as;

Distance = Speed \ \times \ time \\\\D = ut\\\\D = (\frac{66.5 \ mi}{h} \times  \frac{5280 \ ft}{1 \ mi} \times \frac{1 \ h}{3600 \ s} )(3.47 \ s)\\\\D = 338.44 \ ft

Therefore, the distance traveled by Charlotte in feet is 338.44 ft

7 0
2 years ago
A car is traveling at a 20.0 m/s for 7.00 s and then suddenly comes to a stop over a 3 s period.
Delicious77 [7]

Answer: A) Deceleration of the car is -6.6667 m/s² while it came to stop.

B) The total distance the car travels is 200 meter during the 10 s period.

Explanation:

Given Data

Initial velocity of the car ($$v_{i}$$) =   20.0 m/s

Final velocity of the car (v_{f}) = 0 m/s

Time (in motion) =7.00 s

Time (in rest) =3 s

To find - A) car's deceleration while it came to a stop

              B) the total distance the car travels in 10 s

A) The formula to find the deceleration is

Deceleration = (( final velocity - initial velocity ) ÷ Time)        (m/s²)

Deceleration = ((v_{f}) - ($$v_{i}$$)) ÷ time     (m/s²)

Deceleration =  ( 0.0 - 20 ) ÷ 3      (m/s²)

Deceleration =   (- 20) ÷ 3  (m/s²)

Deceleration   =  - 6.6667 m/s²

(NOTE : Deceleration is the opposite of acceleration so the final result must have the negative sign)

The car's deceleration is  - 6.6667 m/s² while it came to a stop

B) The formula to find the distance traveled by the car is  

Distance traveled by the car is equals to the product of the speed and time

Distance = Speed × Time  (meter)

Distance = 20.0 × 10

Distance = 200 meters

The total distance the car travels during the period of 10 s is 200 meters

7 0
3 years ago
The elements from this section of the periodic table all belong to the same
Lady_Fox [76]

Answer:

The answer is:

B) period

3 0
3 years ago
How does the amount of friction affect the sum of forces
iren [92.7K]

Answer:

<h3>Newton's 2nd law states acceleration is proportional to the net force acting on an object. The net force is the vector sum of all the forces applied to the object. ... In this case the acceleration (slowing down) of the puck is proportional to the amount of friction.</h3>

Explanation:

<h3>mark as brainliast</h3>

6 0
2 years ago
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