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Nana76 [90]
3 years ago
9

. What landform do you think this map depicts? a. Valley b. Mountain C. Plain​

Physics
1 answer:
nikklg [1K]3 years ago
6 0
Isn’t that a mountain because it’s going up to the middle
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Free charges do not remain stationary when close together. To illustrate this, calculate the magnitude of the instantaneous acce
ASHA 777 [7]

Answer:

a=2.304×10¹⁶m/s²

Explanation:

Given data

Distance d=2.5 nm=2,5×10⁻⁹m

Mass of proton m=1.6×10⁻²⁷kg

charge of proton q=1.6×10⁻¹⁹C

To find

acceleration a

Solution

Apply the Coulombs Law

F=k\frac{q_{1}q_{2}  }{r^{2} }

Where k is coulombs constant (k=9×10⁹Nm²/C²)

q=q₁=q₂

r=d

So

F=k\frac{|q^{2} |}{d^{2} }\\ as \\F=ma\\ma=k\frac{|q^{2} |}{d^{2} }\\a=\frac{k}{m} \frac{|q^{2} |}{d^{2} }\\a=\frac{(9*10^{9} )*(1.6*10^{-19} )^{2} }{(1.6*10^{-27} )*(2.5*10^{-9} )^{2} }\\ a=2.304*10^{16}m/s^{2}  

4 0
4 years ago
1. A car accelerates uniformly from 12 m/s to 39 m/s in 12 seconds. What is the car's average acceleration
Alinara [238K]

Answer:

(1) 2.25m/s^2

(2) 45.6m

Explanation:

(1) A car accelerates uniformly from 12m/s to 39m/s in 12 seconds

Therefore the average acceleration can be calculated as follows

a = 39-12/12

a = 27/12

a= 2.25m/s^2

(2) A butterfly is flying at 4m/s , it accelerates uniformly at 1.2 m/s for 6 seconds

u= 4

a= 1.2

t= 6

Therefore the distance can be calculated as follows

S= ut + 1/2at^2

= 4×6 + 1/2 × 1.2 × 6^2

= 24 + 1/2 × 1.2 × 36

= 24 + 1/2 × 43.2

= 24 + 21.6

S = 45.6m

Hence the butterfly travels at 45.6m

5 0
3 years ago
Madison applied a force of 150 N in a horizontal direction to a sleigh. Meanwhile the sleigh slid 30.0 m across a level surface
Sergeeva-Olga [200]

Answer:

4500 joules

Explanation:

since work(joules) = force(newtons) x distance(meters),

150N x 30M = 4500 Joules

7 0
4 years ago
Read 2 more answers
28. (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32° ramp at a sp
Ad libitum [116K]

Answer:

a) # buses = 7  

Explanation:

For this exercise we use the kinematic equations, let's find the time it takes to reach the same height

   

     y =v_{oy}  t - ½ g t²

Let's decompose the speed, with trigonometry

      v₀ₓ = v₀ cos θ

      v_{oy} = v₀ sin  θ

      v₀ₓ = 40 cos 32

      v₀ₓ = 33.9 m / s

      v_{oy} = 40 sin32

      v_{oy} = 21.2 m / s

When it arrives it is at the same initial height y = 0

         0 = (v_{oy} - ½ gt) t

That has two solutions

       t = 0                    when it comes out

       t = 2 v_{oy} / g       when it arrives

       t = 2 21.2 /9.8

       t = 4,326 s

We use the horizontal displacement equation

       x = vox t

       x = 33.9   4.326

       x = 146.7 m

To find the number of buses we can use a direct proportions rule

    # buses = 146.7 / 20

    # buses = 7.3

    # buses = 7

The distance of the seven buses is

     L = 20 * 7 = 140 m

b) let's look for the scope for this jump

     R = vo2 sin2T / g

     R = 40 2 without 2 32 /9.8

     R = 146.7 m

As we can see the range and distance needed to pass the seven (7) buses is different there is a margin of error of 6.7 m in favor of the jumper (security)

7 0
3 years ago
Number these from least (1) to most (5) mass ?
alisha [4.7K]

Answer:

1. A feather

2. A baseball

3. A small car

4. A truck

5. A large train

Explanation:

5 0
3 years ago
Read 2 more answers
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