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3 years ago

A mass of 2.50 kg is in a gravitational field of 14.0 N/kg. What force acts on the mass?

2 answers:

7 0

To determine the force that acts on the mass, just multiply the mass by the gravitational field. Using the given data,
F = (2.50 kg)(14 N/kg) = 35 N
Therefore, the force that acts on the mass is equal to 35 N.

Tanzania [10]3 years ago

3 0

Answer:

$F= 2.5 Kg * 14 \frac{N}{Kg}= 35N$

So then the force acting on the mass for this gravitational field is 35N

Explanation:

The gravitational field is defined as the force that acts on an object of mass m. By definition we have that :

$F= ma$ from the second law of Newton.

In our case the value of a represent the acceleration or the gravitational field, so then the expression for the gravitational field is:

$g = \frac{F}{m}$

For this case we know that $g= 14.0 \frac{N}{kg}$ and $m=2.5 kg$

So if we replace and find F we got:

$F= 2.5 Kg * 14 \frac{N}{Kg}= 35N$

So then the force acting on the mass for this gravitational field is 35N

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A 0.950 kg block is attached to a spring with spring constant 16.0 N/m . While the block is sitting at rest, a student hits it w

Bumek [7]

Answer:

Explanation:

Given that,

Mass attached m = 0.95kg

Spring constant k = 16N/m

Instantaneous speed v = 36cm/s = 0.36m/s

Amplitude A=?

When x = 0.7A

Using conservation of energy

∆K.E + ∆P.E = 0

K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0

At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.

Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0

So, the equation becomes

— K.E(initial) + P.E(final) = 0

K.E(initial) = P.E(final)

½mv² = ½kA²

mv² = kA²

0.95 × 0.36² = 16×A²

0.12312 = 16•A²

A² = 0.12312/16

A² = 0.007695

A = √0.007695

A = 0.088 m

A = 8.8cm

B. Speed at x = 0.7A

Using the same principle above

K.E(initial) = P.E(final)

½mv² = ½kA²

Where A = 0.7A = 0.7 × 0.088 = 0.0614m

Then,

½× 0.95 × v² = ½ × 16 × 0.0614²

0.475v² = 0.0310644

v² = 0.0310644/0.475

v² = 0.0635

v = √0.0635

v = 0.252 m/s

v = 25.2 cm/s

8 0

4 years ago

A 24 kg crate is moving at a constant speed because it is being pushed with a force of 53n. What would be the coefficient of kin

____ [38]

Answer:

The coefficient of kinetic friction between the crate and the floor can be calculated using the formula μ = Ff / N, where Ff is the frictional force, N is the normal force, and μ is the coefficient of kinetic friction.

In this case, the normal force is equal to the weight of the crate, which is 24 kg * 9.8 m/s2 = 235.2 N. The frictional force can be calculated using the formula Ff = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.

If we substitute the values for N and Ff into the formula for the coefficient of kinetic friction, we get:μ = 53 N / 235.2 N = 0.225

Therefore, the coefficient of kinetic friction between the crate and the floor is 0.225.

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1 year ago

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3 years ago

What do electric and magnetic fields have in common?

Nastasia [14]

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Similarities between magnetic fields and electric fields: Magnetic fields are associated with two magnetic poles, north and south, although they are also produced by charges (but moving charges). Like pole repel unlike poles attract. Electric field points in the direction of the force experienced by a positive charge.

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3 years ago

a force 2.4E2 N exists between a positive charge of 8E-5 C and a positive charge of 3E-5 C. What distance separates the charges?

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