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3 years ago

Explain how that energy converts itself from POTENTIAL to KINETIC energy.

Physics

1 answer:

Karolina [17]3 years ago

8 0

Answer:

Potential energy can transfer into other forms of energy like kinetic energy. Kinetic energy is energy an object has because of its motion.

Explanation:

The ball was released, as the ball moves faster and faster toward the ground, the force of gravity will transfer the potential energy to kinetic energy.

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Binding of a signaling molecule to which type of receptor leads directly to a change in the distribution of ions on opposite sid

aalyn [17]

Binding of a signaling molecule to ligand-gated ion channel leads directly to a change in the distribution of ions on opposite sides of the membrane

<u>Explanation:</u>

The ligand-gated-ion-channel is trans membrane protein buildings that conduction course through the channel pores in light of the official of a neurotransmitter. They are not quite the same as voltage - gated - ion - channels, which are delicate to film possibilities, and G-protein coupled receptors (GPCR's), which utilize second messengers.

These ion channels, additionally ordinarily referred to as ionotropic receptors, are a gathering of trans membrane -ion-channel proteins which open to permit ions, for example, $N a^{+}, K^{+}$ and potentially $c l^{-}$ to go through the layer in light of the official of a synthetic errand person, for example, a neurotransmitter.

5 0

4 years ago

7) Sound waves travel with a nominal speed of 340m/s.

Savatey [412]

Answer:

The wavelengths of C1 is 10.4m, A6 is 0.193m and B7 is 0.0861m

Explanation:

Using the formula V = f×λ . Then substitute the following values into the formula:

a) v=340m/s

f=32.7 Hz

λ=V ÷ f

= 340 ÷ 32.7

= 10.4m (3s.f)

b) λ=340 ÷ 1760

= 0.193m (3s.f)

c) λ=340÷3951.1

= 0.0861m (3s.f)

(Correct me if I am wrong)

4 0

3 years ago

A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.

mafiozo [28]

Answer:

a. $f=1.22*10^{-15} N$

b. $f=53.6*10^{-17} N$

Explanation:

The force existing between two charges is given as

$f=\frac{kq_{1}q_{2}}{r^{2}}$

where q= charge,

k=constant

r= distance between the two charges

Note: this force can either be repulsive or attractive force depending on the charges involve. it is repulsive if they are similar charge and it is attractive if it is opposite charges.

Also the charge of an electron is

$-1.602*10^{-19}$

A. we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{0.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{0.04}\\f_{21}=1.44*10^{-15}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis

for the -5.50nC the distance between them is 0.600m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.6^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.36}\\f_{23}=-(0.22*10^{-15})N i$

this this force will be repulsive force and it points away from the electron i.e points towards the -ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=1.44*10^{-15}-0.22*10^{-15}\\ f=1.22*10^{-15} N$

b. at distance of x=1.20m, this is shown on the diagram below (attachment 2)

we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{1.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{1.44}\\f_{21}=4.0*10^{-17}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

for the -5.50nC the distance between them is 0.4m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.4^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.16}\\f_{23}=49.6*10^{-17}Ni$