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svlad2 [7]
3 years ago
15

A 10 kg object experiences an acceleration of 2 m/s squared. What net force was applied to the object?

Physics
1 answer:
Nastasia [14]3 years ago
4 0

F = ma

= 10kg × 2m/s

= 20N

Answer : 20N

Explanation :

F = Force (N)

m = mass (kg)

a = acceleration (m/s)

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The waves become longer but slower
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For his science project, Austin wants to determine if the pH of standing rainwater changes
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A railroad car of mass M moving at a speed v1 collides and couples with two coupled railroad cars, each of the same mass M and m
marshall27 [118]

Answer:

vf = v₁/3 + 2v₂/3

Explanation:

Using the law of conservation of linear momentum,

momentum before impact = momentum after impact

So, Mv₁ + 2Mv₂ = 3Mv (since the railroad cars combine) where v₁ = initial velocity of first railroad car, v₂ = initial velocity of the other two coupled railroad cars, and vf = final velocity of the three railroad cars after impact.

Mv₁ + 2Mv₂ = 3Mvf

dividing through by 3M, we have

v₁/3 + 2v₂/3 = vf

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6 0
3 years ago
Place the balloon in a bell jar. If available also add some shaving cream and fresh marshmallows. Ask the instructor for help if
irina [24]

Answer:

The balloon will collapse

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5 0
3 years ago
From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up
NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

7 0
3 years ago
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