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3 years ago

A 3.0 kg block is pushed by a 14 N force. If µ = 0.6, will the block move?

Physics

1 answer:

Anna71 [15]3 years ago

4 0

Answer:

The block will not move.

Explanation:

We'll begin by calculating the frictional force. This can be obtained as follow:

Coefficient of friction (µ) = 0.6

Mass of block (m) = 3 Kg

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (R) = mg = 3 × 10 = 30 N

Frictional force (Fբ) =?

Fբ = µR

Fբ = 0.6 × 30

Fբ = 18 N

From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.

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If a lever is 10m long and the effort is applied 4m from the fulcrum, what is the MA of the lever?

Marizza181 [45]

We Know: MA = d₁ / d₂
Here, d₁ = 4
d₂ = 10-4 = 6

Substitute their values into the expression:
MA = 4/6
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So, Your Answer would be 2/3

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3 years ago

Read 2 more answers

Assume the average value of the vertical component of Earth's magnetic field is 42 μT (downward) in some region that has an area

Oliga [24]

Answer:

The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

Explanation:

Given that,

Magnetic field = 42 μT

Area $A=3.71\times10^{5}\ km^2$

$A=3.71\times10^{11}\ m^2$

We need to calculate the flux per unit area

$flux\ per\ unit\ area=\dfrac{42\times10^{-6}}{3.71\times10^{11}}$

$flux\ per\ unit\ area=1.132\times10^{-16}\ T/m^2$

We need to calculate the total earth's surface area

$A'=4\pi r^2$

$A'=4\times\pi\times(6.3781\times10^{6})^2$

$A'=5.1120\times10^{14}\ m^2$

We need to calculate the rest of earth's area

$A''=A-A'$

Put the value into the formula

$A''=5.1120\times10^{14}-3.71\times10^{11}$

$A''=5.10829\times10^{14}\ m^2$

We need to calculate the net magnetic flux through the rest of Earth's surface

$B'=5.10829\times10^{14}\times1.132\times10^{-16}$

$B'=5.782\times10^{-2}\ T$

Hence, The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

3 0

4 years ago

One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor

zhannawk [14.2K]

Answer:

Part a)

v = 16.52 m/s

Part b)

v = 7.47 m/s

Explanation:

Part a)

(a) when the large-mass object is the one moving initially

So here we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v$

since this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

$v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}$

$v = \frac{(8.4\times 24 + 3.8\times 0)}{3.8 + 8.4}$

$v = 16.52 m/s$

Part b)

(b) when the small-mass object is the one moving initially

here also we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v$

Again this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

$v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}$

$v = \frac{(8.4\times 0 + 3.8\times 24)}{3.8 + 8.4}$

$v = 7.47 m/s$

4 0

3 years ago