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iogann1982 [59]
3 years ago
12

Identical particles, each with energy E, are incident on the following four potential energy

Physics
1 answer:
amid [387]3 years ago
3 0

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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Pls help i’ll give brainliest if you give a correct answer!!
Anit [1.1K]

Answer:

south

im not sure with this answer

3 0
4 years ago
A 0.140-kg baseball is dropped from rest. It has a speed of 1.20 m/s just before it hits the ground, and it rebounds with an upw
irga5000 [103]

Answer:

22 N upward

Explanation:

From the question,

Applying newton's second law of motion

F = m(v-u)/t....................... Equation 1

Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact

Note: Let upward be negative and downward be positive

Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s

Substitute into equation 1

F = 0.14(-1-1.2)/0.014

F = 0.14(-2.2)/0.014

F = 10(-2.2)

F = -22 N

Note the negative sign shows that the force act upward

6 0
4 years ago
The sum of potential and kinetic energies in the particles of a substance is called
olganol [36]

Internal energy.

Explanation:

In any substance/object, the particles inside it (atoms/molecules) constantly move in random directions and with random speeds (this motion is called Brownian motion). As a result, the particles have some kinetic energy (which is proportional to the temperature of the substance). Moreover, the particles interact with each other due to the presence of electrostatic intermolecular forces, and as a result, the particles also have some potential energy.

The sum of the kinetic energies and potential energies of the particles in a substance is called internal energy.

6 0
4 years ago
Read 2 more answers
An object is placed 96.5 cm from a glass lens (n = 1.51) with one concave surface of radius 24 cm and one convex surface of radi
poizon [28]

Answer:

image is vertical at distance -203.62 cm

magnification is 2.110

Explanation:

given data

n = 1.51

distance u = 96.5 cm

concave radius r1 = 24 cm

convex radius r2 = 19.1 cm

to find out

final image distance and magnification

solution

we will apply here lens formula to find focal length f

1/f = n-1 ( 1/r1 - 1/r2)   .......................1

put here all value

1/f = 1.51 -1 ( -1/24 + 1/19.1)

f = 183.43

so from lens formula

1/f = 1/v + 1/u     .............................2

put here all value and find v

1/183.43 = 1/v + 1/96.5

so

v = −203.62 cm

so here image is vertical at distance -203.62 cm

and

magnification are = -v /u

magnification =  203.62 / 96.5

magnification is 2.110

3 0
4 years ago
A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.
Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

    k x = m g

k = \dfrac{mg}{x}

k = \dfrac{6.4 \times 9.8}{0.28}

      k = 224 N/m

b) f = \dfrac{\omega}{2\pi}

    \omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s

   f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

   f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}

  f =0.94\ Hz

c)  v_b = -v cos \omega t

    v_b = -5.1 \times cos (5.92 \times 0.42)

    v_b = 4.04\ m/s

d)  a_{max} = v \omega

    a_{max} = 4.04 \times 5.92

    a_{max} =23.94\ m/s^2

e)  Y =- A sin (\omega t)

    A = \dfrac{v}{\omega}

    A = \dfrac{4.04}{5.92}

        A = 0.682 m

    Y =- 0.682 \times sin (5.92 \times 0.42)

    Y =- 0.42

Force =m \omega^2 |Y|

          =6.4 \times 5.92^2\times 0.42

F = 94.20 N

4 0
4 years ago
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