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3 years ago

An object has a kinetic energy ke, and potential energy pe. it also has a rest energy e0. what is the object's total energy e?

Physics

1 answer:

Dmitriy789 [7]3 years ago

7 0

Look that one up in you text book PG:678 that is if you got the same book as my friend

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The voltage across a membrane forming a cell wall is 72.7 mV and the membrane is 9.22 nm thick. What is the magnitude of the ele

Blizzard [7]

Answer:

The magnitude of the electric field intensity is $E = 7.89 *10^{6} \ V/m$

Explanation:

From the question we are told that

The voltage is $\epsilon = 72.7 \ mV = 72.7 *10^{-3} V$

The thickness of the membrane is $t = 9.22 \ nm = 9.22 *10^{-9} \ m$

Generally the electric field intensity is mathematically represented as

$E = \frac{\epsilon }{t}$

substituting values

$E = \frac{72.7 *10^{-3} }{9.22 *10^{-9}}$

$E = 7.89 *10^{6} \ V/m$

8 0

3 years ago

A car is travelling with a velocity of 10 m/s and has a mass of 550 kg. The car has energy.

Elenna [48]

Answer: The car has a kinetic energy (because it's in motion) of: $27500J$

Explanation:

$Formula: E=\frac{1}{2}mv^2$

$E=\frac{1}{2}mv^2$

$E=\frac{1}{2}(550kg)(10m/s)^2$

$E=\frac{1}{2}(550kg)(100m^2/s^2)$

$E=\frac{1}{2}(55000J)\\E=27500J$

3 0

3 years ago

A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on

MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

$Ec=\frac{1}{2} *m*v^{2}$

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

$W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}$

$W=\frac{1}{2} *m*(v2^{2}-v1^{2})$

In this case:

W=?
m= 2,145 kg
v2= 12 $\frac{m}{s}$

v1= 25 $\frac{m}{s}$

Replacing:

$W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})$

W= -515,872.5 J

The work required is -515,872.5 J

3 0

3 years ago

In an elastic collision, a 580 kg bumper car collides directly from behind with a second, identical bumper car that is traveling

kozerog [31]

Answer: vl = 2.75 m/s vt = 1.5 m/s

Explanation:

If we assume that no external forces act during the collision, total momentum must be conserved.

If both cars are identical and also the drivers have the same mass, we can write the following:

m (vi1 + vi2) = m (vf1 + vf2) (1)

The sum of the initial speeds must be equal to the sum of the final ones.

If we are told that kinetic energy must be conserved also, simplifying, we can write:

vi1² + vi2² = vf1² + vf2² (2)

The only condition that satisfies (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:

vf1 = vi2 and vf2 = vi1

If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:

vf1 = 2.75 m/s vf2 = 1.5 m/s

8 0

3 years ago

The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the wave shown?

belka [17]

Answer:

B) 0.3Hz

Explanation:

I just took the test i hope i helped and i hope you pass the test

7 0

3 years ago

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