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3 years ago

You don't need to push off with a foot against the ground to start

Physics

1 answer:

Angelina_Jolie [31]3 years ago

4 0

Answer:

No, it does not violate Newton's first law

Explanation:

Newton's first law of motion states that, "A body will continue in its state of rest or uniform motion in a straight unless acted up by a force to make it act otherwise" This means a force is required to initiate movement and also required to halt it.

For a skate positioned to roll down a bank, Here, the force of gravity acting on the skate acting downward will make the skate roll without having to push off the skate with a foot. Because the position of the skate doesn't balance the Gravitational force acting on it. Hence. The Gravitational force is enough to set the skate in motion.

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Now explore friction force. Set the piece of plastic or wood on the table and push it steadily across the tabletop using your fi

nydimaria [60]

Answer: If there is a higher friction, the opposition force is higher so that it can reduce our speed. So, a factor that affects friction is the roughness or smoothness of the surface of the object. In comparison of the table with the fabric, the fabric will have a more opposition force. As the surface of the fabric is usually rougher than the surface of a smooth table. As there is more friction on a fabric, we will feel more opposition force on our finger tip.

Hope it helped! :>

4 0

3 years ago

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

6 0

3 years ago

Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an idea

steposvetlana [31]

Answer

given,

V = 2 L

the left is an ideal gas at P = 100 k Pa and T = 500 K

mass is constant

$m_1 = m_2$

$\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}$

Pressure is same because it's not changing due to process

$\dfrac{V}{500} = \dfrac{2 V}{T_2}$

$T_2 = 1000\ K$

$\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}$

$\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})$

$m = \dfrac{P_1V_1}{RT_1}$

$m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}$

m = 1.39 x 10⁻³ Kg

$\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)$

$\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K$

5 0

3 years ago

for an object that is speeding up in the positive direction, what does the displacement vs. time graph look like?

slava [35]

Answer:

The speeding up is steady, it is a parabola (a=V*t+(at^2)/2), and give it's an equation in connection to time, at that point, it is conceivable to discover the separation recipe by utilizing more substantial amount mathematics(integrals).

Explanation:

7 0

3 years ago

Read 2 more answers

g Two radiation modes (one at the center frequency lIo and the other at lIO+?lI) are excited with 1000 photons each. Determine t

Schach [20]

Answer:

a) P=0.25x10^-7

b) R=B*N2*E

c) N=1.33x10^9 photons

Explanation:

a) the spontaneous emission rate is equal to:

1/tsp=1/3 ms

the stimulated emission rate is equal to:

pst=(N*C*o(v))/V

where

o(v)=((λ^2*A)/(8*π*u^2))g(v)

g(v)=2/(π*deltav)

o(v)=(λ^2)/(4*π*tp*deltav)

Replacing values:

o(v)=0.7^2/(4*π*3*50)=8.3x10^-19 cm^2

the probability is equal to:

P=(1000*3x10^10*8.3x10^-19)/(100)=0.25x10^-7

b) the rate of decay is equal to:

R=B*N2*E, where B is the Einstein´s coefficient and E is the energy system

c) the number of photons is equal to:

N=(1/tsp)*(V/C*o)

Replacing:

N=100/(3*3x10^10*8.3x10^-19)

N=1.33x10^9 photons

7 0

3 years ago