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4 years ago

Consider the following situations:

Physics

1 answer:

Snezhnost [94]4 years ago

4 0

Answer:

Though no question is specified here but let analyzed experiences of each

Explanation:

1. Required a force to stop it (Law of inertial). Constant acceleration

2. will have reduced acceleration (deceleration)

3. Uniform acceleration may set in. it accelerate

4. maximized it speed since no friction affect it motion

5. accelerating. Vo=0 and V gradually increased

6. less or no friction opposes the motion

7. It experience reduce motion based on air friction

8. constant acceleration due to gravity

9. g can be a bit affected but still about equal

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3. A 40-gram ball of clay is dropped from a height, h, above a cup which is attached to a spring of spring force constant, k, of

zheka24 [161]

Answer:

the maximum speed of the ball is 12.65 m/s

Explanation:

Given;

mass of the ball, m = 40 g = 0.04 kg

spring constant, k = 25 N/m

Apply the principle of conservation of energy;

The Elastic potential energy of the spring will be converted into Kinetic of the ball;

$\frac{1}{2} kx^2 = \frac{1}{2} mv^2\\\\ kx^2 = mv^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m}} \\\\v = \sqrt{\frac{(25)(0.506)^2}{0.04}} \\\\v = \sqrt{160.0225} \\\\v = 12.65 \ m/s$

Therefore, the maximum speed of the ball is 12.65 m/s

4 0

3 years ago

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an elect

klio [65]

Complete Question

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?

Answer:

The magnetic field strength is $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

$\Delta V q = \frac{1}{2}mv^2$

Making v the subject

$v = \sqrt{[\frac{2 \Delta V * q }{m}] }$

Where m is the mass of electron

v is the velocity of electron

q charge on electron

$\Delta V$ is the potential difference

Substituting values

$v = \sqrt{\frac{2 * 5.9 *10^3 * 1.60218*10^{-19} }{9.10939 *10^{-31]} }$ f

$= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[ $F = B q v$ ] must be equal to the centripetal force[ $\frac{mv^2}{r}$ ] and this is mathematically represented as

$Bqv = \frac{mv^2}{r}$

making B the subject

$B = \frac{mv}{rq}$

r is the radius with a value = 5.4cm = $= \frac{5.4}{100} = 5.4*10^{-2} m$

Substituting values

$B = \frac{9.1039 *10^{-31} * 4.556 *10^7}{5.4*10^-2 * 1.60218*10^{-19}}$

$= 0.0048 T$

7 0

4 years ago

Show the relationship between the quantity of matter and the mass of objects

IRISSAK [1]

Answer:

actually quantity of matter present in a body is its mass.

3 0

3 years ago

A 50-cm-long spring is suspended from the ceiling. A 330g mass is connected to the end and held at rest with the spring unstretc

Anvisha [2.4K]

Explanation:

Given that,

Length of the spring, l = 50 cm

Mass, m = 330 g = 0.33 kg

(A) The mass is released and falls, stretching the spring by 28 cm before coming to rest at its lowest point. On applying second law of Newton at 14 cm below the lowest point we get :

$kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.14}\\\\k=23.1\ N/m$

(B) The amplitude of the oscillation is half of the total distance covered. So, amplitude is 14 cm.

(C) The frequency of the oscillation is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{23.1}{0.33}} \\\\f=1.33\ Hz$

5 0

4 years ago

Could you please help me on number four

sleet_krkn [62]

What is it asking? It is sort of blurry

3 0

3 years ago