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3 years ago

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Mr Smith is working in a muddy garden. When he picks up a paving stone his feet sink deeper into the mud. Explain why his feet s

ink into the mud more when he picks up the paving stone. Write about pressure in your answer

1 answer:

Anni [7]3 years ago

3 0

because he is carrying more mass and as the ground is muddy his feet goes in due to the pull of gravity

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A 12000 kg boat is moving 4.25 m/s. Its engine pushes 9200 N forward, but the current pushes back at 12,500 N. How much times do

Verizon [17]

Answer:

15.5 seconds

Explanation:

Apply Newton's second law:

∑F = ma

-12500 + 9200 = (12000) a

a = -0.275 m/s²

v = at + v₀

0 = (-0.275) t + 4.25

t = 15.5 s

It takes the boat 15.5 seconds to stop.

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3 years ago

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Answer:

0.75 g/cm^3

Explanation:

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$\rho = \frac{m}{V}$

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Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What

vesna_86 [32]

Answer:

$1.02\cdot 10^{-8} N$ , repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have same sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

$q_1=q_2=1.6\cdot 10^{-19}C$ is the magnitude of the two electrons

$r=1.5\cdot 10^{-10} m$ is their separation

Substituting into the formula, we find the electric force between them:

$F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N$

And the force is repulsive, since the two electrons have same sign charge.

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3 years ago

A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$

where

$F_{net}$ is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

$a=3.1 m/s^2$ is its acceleration

Substituting, we find the net force on the bicyclist:

$F_{net}=(60)(3.1)=186 N$

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

$F_{net}=F_a-R$

where:

$F_{net}$ is the net force

$F_a$ is the applied force (forward)

$R$ is the air drag (backward)

In this problem we have:

$F_{net}=186 N$ is the net force (found in part b)

$R=60 N$ is the magnitude of the air drag

Solving for $F_a$ , we find the force produced by the bicyclist while pedaling:

$F_a=F_{net}+R=186+60=246 N$

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3 years ago