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3241004551 [841]
2 years ago
15

Tim is pushing a box with a 7 N force. The weight of the box is 5 N. A frictional force

Physics
1 answer:
Vesnalui [34]2 years ago
5 0

Friction and the applied force are not balanced. The 7 N push is countered by the frictional force of 3 N to push the box with a net force of 4 N.

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If it takes 150 n of force to accelerate an object at 30 m/s what is the mass of the object
iris [78.8K]
F  =  ma.         

Note Force should  be = 150 N.  Acceleration = 30 m/s2   ( Am presuming for your question you meant to write 30m/s2  and not 30m/s as you wrote)

150 = m ( 30)

150/30  = m.

5 = m

Mass of the object = 5 Kg.
3 0
3 years ago
What best describes the energy in a closed system at the beginning of a day and the energy at the end of the same day? Check all
jeka94
The second, fourth, and seventh answers apply. Energy in a closed system is conserved, but it can change form
6 0
3 years ago
Read 2 more answers
A ball is thrown upward from an initial height of 1.5m the ball reaches a height of 5m then falls to the ground . What Is the di
jeka94

Answer:

The distance traveled by the ball is 8.5 m

Explanation:

Initial height of the ball, h₁ = 1.5 m above the ground

final height of the ball, h₂ = 5m

Upward distance = distance traveled by the ball from a height of 1.5m to 5m = 5m - 1.5m = 3.5 m

Downward distance = distance traveled by the ball from 5m height to the ground =5m - 0 = 5m

Total distance traveled = upward distance + downward distance

Total distance traveled = 3.5 m + 5m = 8.5 m

Therefore, the distance traveled by the ball is 8.5 m

4 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
Points a and b lie in a region where the y-component of the electric field is Ey=α+β/y2. The constants in this expression have t
Drupady [299]

Answer:

V_{a} - V_{b} = 89.3

Explanation:

The electric potential is defined by

         V_{b} - V_{a} = - ∫ E .ds

In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.

         V_{b} - V_{a} = - ∫ E ds

We substitute

         V_{b} - V_{a} = - ∫ (α + β/ y²) dy

We integrate

          V_{b} - V_{a} = - α y + β / y

We evaluate between the lower limit A  2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m

           V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)

            V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33

            V_{b} - V_{a} = - 89.3 V

As they ask us the reverse case

             V_{b} - V_{a} = - V_{b} - V_{a}

             V_{a} - V_{b} = 89.3

3 0
3 years ago
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