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frutty [35]
3 years ago
7

Estimate the weight of a 1000kg car that is accelerating at 3 m/s/s.

Physics
1 answer:
Gekata [30.6K]3 years ago
8 0

Answer:

W = 9800  N

Explanation:

Given that,

Mass of a car, m = 1000 kg

Acceleration of the car, a = 3 m/s²

We need to find the weight of the car. Weight of an object is given by the product of mass and acceleration due to gravity on the Earth.

W = mg

Put all the values,

W = 1000 kg × 9.8 m/s²

= 9800  N

So, the weight of the car is 9800  N.

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In a level parking lot, a man with a mass of 100 kg gives a shove to a boy with a mass of 50 kg who is on roller skates. The boy
natita [175]

Answer:

100N

Explanation:

Newton's third law states that whenever an object exerts a force on a second object, it exerts a force of equal magnitude and direction but in the opposite direction on the first. It is often stated as follows: Each action always opposes an equal but opposite reaction.

The subject 1 of 100kg is making a force F, to move an object from 50Kg to 2m / s ^ 2. This Force the object of 50Kg will reflect it in the opposite direction by Newton's third law.

Once the parameter of the force that both are experiencing is clarified, Newton's second law is applied to their respective calculation.

F = ma = 50kg * 2m / s ^ 2 = 100N

That is the force the boy exert on the man during the shove.

4 0
2 years ago
Could anyone help me out ?
Delicious77 [7]

density = mass/volume = 100kg/10ml = 10kg/ml

voluime = mass/density = 50g/2 g/ml = 25 ml

mass = density x volume = 2x55 = 110 kg

4 0
2 years ago
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Ways in which a teacher plays a role in the literacy development of the learners​
RideAnS [48]

Answer:

encourage all attempts at reading, writing, and speaking

Explanation:

7 0
1 year ago
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

6 0
3 years ago
What player(s) can take a goal kick?
inysia [295]

Answer:

c) goalie

Explanation:

8 0
3 years ago
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