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GalinKa [24]
3 years ago
5

Why does someone coming off a slide count as a projectile?

Physics
2 answers:
user100 [1]3 years ago
8 0

Answer:

i do not understand. ve more spacific

Tpy6a [65]3 years ago
6 0

Answer:

oooooooooommmmmmmmmmgggggggg

Explanation:

You might be interested in
What is the strength of electric field EpEp 0.60 mmmm from a proton? Express your answer to two significant figures and include
disa [49]

Answer:

3.99*10^-3N/C

Explanation:

Using

Ep= kq/r²

Where r = 0.6mm = 0.6*10^-3m

K= 8.9*10^9 and q= 1.6*10^-19

So = 8.9*10^9 * 1.6*10^-19/0.6*10^-3)²

= 3.99*10^-3N/C

8 0
3 years ago
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
If a small motor does 520 J of work to move a toy car 260 m, what force does the engine exert on the car?
ohaa [14]

Answer:

<h2>2 N</h2>

Explanation:

The force engine exert on the car can be found by using the formula

f =  \frac{w}{d}  \\

w is the workdone

d is the distance

From the question we have

f =  \frac{520}{260}  = 2 \\

We have the final answer as

<h3>2 N</h3>

Hope this helps you

4 0
3 years ago
A drag racer crosses the finish line doing 212 mi/h and promptly deploys her drag chute. (A.) what force must the drag chute exe
Svet_ta [14]

initial speed of the racer is given as

v_i = 212 mi/h

v_i = 212*\frac{1609}{3600} = 94.75 m/s

after applied force the final speed is given as

v_f = 45 mi/h

v_f = 45 * \frac{1609}{3600} = 20.11 m/s

now during this speed change the racer will cover total distance 185 m

so here we will use kinematics

v_f^2 - v_i^2 = 2 a d

20.11^2 - 94.75^2 = 2*a*185

a = -23.2 m/s^2

now the force that chute will exert on the racer will be given as

F = ma

F = 898* 23.2

F = 2.1* 10^4 N

B) here following is the strategy for solving it

1. first we used kinematics to find the acceleration of the car

2. then we used Newton's II law (F = ma) to find the force

4 0
3 years ago
Different masses are hung on a spring scale calibrated in newtons. The force exerted by gravity on 1.0 kg is shown in the image
Lunna [17]
6.5 x 9.8
=63.7
so the answer is 64N
6 0
3 years ago
Read 2 more answers
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