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Evgen [1.6K]
3 years ago
11

Increased force will increase acceleration true or false.

Physics
2 answers:
zalisa [80]3 years ago
8 0
True. Force is acceleration therefor inverse force is increasing acceleration
Mars2501 [29]3 years ago
7 0
True. For example because force and acceleration are directly related. For example weight is the force of gravity. Weight= mg
So as g increase so does weight. Therefore the answer is true
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A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr
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Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

 a= μ g                          ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g  =ω²X

We know for spring mass system natural frequency given as

\omega=\sqrt{\dfrac{k}{M+m}}

By putting the values

\omega=\sqrt{\dfrac{150}{5.67+1}}

ω = 4.47 rad/s

μ g  =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2

kX^2=(m+M)v^2

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

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3 years ago
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A 2.8-kg physics cart is moving forward with a speed of 45 cm/s. A 1.9-kg brick is dropped from rest and lands on the cart. The
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A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc
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This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so: 

v_{d} =  \frac{J}{n\left | q \right |}

<span>The current I is any motion of charge from one region to another, so this is given by:

</span>I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)

The magnitude of the current density is:

J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})

Being:

A=2mm^{2} = 2x10^{-6}m^{2}
<span>
Finally, for the drift velocity magnitude vd, we find:

</span>v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
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