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3 years ago

What are 2 force pairs when a person is pushing a box across a floor

1 answer:

3 0

Answer: As a person pushes a box across a floor, the energy from the person's moving arm is transferred to the box and floor become warm during this process, energy is conserved

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(II) A 1200-kg car moving on a horizontal surface has speed v = 85 km/h when it strikes a horizontal coiled spring and is brough

Pie

Answer:

k = 138440 N/m

Explanation:

given,

Mass of the car = 1200 Kg

speed of the car = 85 km/h

= 85 x 0.278 = 23.63 m/s

KE of the car

$KE = \dfrac{1}{2} mv^2$

Using Spring Energy Formula

$KE_{spring}=\dfrac{1}{2}kx^2$

Using conservation of energy

$\dfrac{1}{2} mv^2=\dfrac{1}{2}kx^2$

$k = \dfrac{mv^2}{x^2}$

$k = \dfrac{1200\times 23.63^2}{2.2^2}$

k = 138440 N/m

Spring stiffness constant of the spring is equal to k = 138440 N/m

7 0

3 years ago

If a person is standing up in a moving bus that stops suddenly,the person can easily fall forwards.Has a force acted to push the

cupoosta [38]

That is inertia, 1st law of motion, no net force is applied

8 0

4 years ago

Read 2 more answers

A silver cube with an edge length of 2.28 cm2.28 cm and a gold cube with an edge length of 2.66 cm2.66 cm are both heated to 89.

Diano4ka-milaya [45]

Yeah I have no clue what you saw lol yeah

8 0

3 years ago

The amount of friction depends on the weight of the objects pressing together.

natta225 [31]

Answer:

The amount of friction depends on the force pushing the surfaces together. If this force increases, the hills and valleys of the surfaces can come into closer contact. The close contact increases the friction between the surfaces. Objects that weigh less exert less downward force than objects that weigh more.

Explanation:

Friction depends partly on the smoothness of the contacting surfaces, a greater force being needed to move two surfaces past one another if they are rough than if they are smooth.

4 0

3 years ago

A +8.75 μC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 μC point charge by a light, nonc

lisov135 [29]

(a) The tension on the wire when the two charges have opposite signs is 383.5 N.

(b) The tension on the wire if both charges were negative is 3.640.25 N.

The given parameters;

first charge, q₁ = 8.75 μC
second charge, q₂ = -6.5 μC
electric field, E = 1.85 x 10⁸ N/C
distance between the two charges, r = 2.5 cm

(a)

The attractive force between the charges is calculated as follows;

$F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N$

The force on the negative charge due to the electric field is calculated as follows;

$F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N$

The tension on the wire is the resultant of the two forces and it is calculated as follows;

$T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N$

(b) when the two charges are negative

The repulsive force between the two charges is calculated as follows;

$F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N$

The force on the first negative charge due to the electric field is calculated as follows;

$F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N$

The force on the second negative charge due to the electric field is calculated as follows;

$F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N$

The tension on the wire is the resultant of the three forces and it is calculated as follows;

$T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N$

Learn more here:brainly.com/question/19565286

5 0

3 years ago