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3 years ago

Julian tests four springs and records the results in the table

Physics

2 answers:

Tju [1.3M]3 years ago

8 0

Answer:

list of spring in order of increasing spring constant is given by,

W, Y, X, Z

Explanation:

Spring W:

Magnitude of Force applied on the spring is given by,

F = k x

Where, F = force applied

k = spring constant

x displacement

k = F/x

k = 20 / 0.5

k = 40 N/m

Spring X:

Magnitude of Force applied on the spring is given by,

F = k x

Where, F = force applied

k = spring constant

x displacement

k = F/x

k = 60 / 0.3

k = 200 N/m

Spring Y:

Magnitude of Force applied on the spring is given by,

F = k x

Where, F = force applied

k = spring constant

x displacement

k = F/x

k = 40 / 0.4

k = 100 N/m

Spring Z:

Magnitude of Force applied on the spring is given by,

F = k x

Where, F = force applied

k = spring constant

x displacement

k = F/x

k = 50 / 0.1

k = 500 N/m

Thus, list of spring in order of increasing spring constant is given by,

W, Y, X, Z

Cerrena [4.2K]3 years ago

4 0

the correct answer is a. w,y,x,z.

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Tutorial Exercise An unstable atomic nucleus of mass 1.83 10-26 kg initially at rest disintegrates into three particles. One of

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Answer:

A) v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) K_total = 373.08 × 10^(-15) J

Explanation:

We are given;

Mass of unstable atomic nucleus; M = 1.83 × 10^(-26) kg

Mass of first particle; m1 = 5.03 × 10^(-27) kg

Speed of first particle in y-direction; v1 = (6 × 10^(6) m/s) j^

Mass of second particle; m2 = 8.47 × 10^(-27) kg

Speed of second particle in x - direction; v2 = (4 × 10^(6) m/s) i^

Now, we don't have the mass of the third particle but since we are told the unstable atomic nucleus disintegrates into 3 particles, thus;

M = m1 + m2 + m3

1.83 × 10^(-26) = (5.03 × 10^(-27)) + (8.47 × 10^(-27)) + m3

m3 = (1.83 × 10^(-26)) - (13.5 × 10^(-27))

m3 = 4.8 × 10^(-27) kg

A) Applying law of conservation of momentum, we have;

MV = (m1 × v1) + (m2 × v2) + (m3 × v3)

Now, the unstable atomic nucleus was at rest before disintegration, thus V = 0 m/s.

Thus, we now have;

0 = (m1 × v1) + (m2 × v2) + (m3 × v3)

We want to find the velocity of the third particle v3. Let's make it the subject of the formula;

v3 = [(m1 × v1) + (m2 × v2)]/(-m3)

Plugging in the relevant values, we have;

v3 = [(5.03 × 10^(-27) × 6 × 10^(6))j^ + (8.47 × 10^(-27) × 4 × 10^(6))i^]/(-4.8 × 10^(-27))

v3 = [(30.18 × 10^(-21))j^ + (33.88 × 10^(-21))i^]/(-4.8 × 10^(-27))

v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) Formula for kinetic energy is;

K = ½mv²

Now,total kinetic energy is;

K_total = K1 + K2 + K3

K1 = ½ × 5.03 × 10^(-27) × (6 × 10^(6))²

K1 = 90.54 × 10^(-15) J

K2 = ½ × 8.47 × 10^(-27) × (4 × 10^(6))²

K2 = 67.76 × 10^(-15)

To find K3, let's first find the magnitude of v3 because it's still in vector form.

Thus;

v3 = √[(-6.29 × 10^(6))² + (-7.06 × 10^(6))²]

v3 = 9.46 × 10^(6) m/s

K3 = ½ × 4.8 × 10^(-27) × (9.46 × 10^(6))²

K3 = 214.78 × 10^(-15) J

K_total = (90.54 × 10^(-15)) + (67.76 × 10^(-15)) + (214.78 × 10^(-15))

K_total = 373.08 × 10^(-15) J

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Answer:

Magnetic force, $F = 3.52\times 10^{-13}\ N$

Explanation:

Given that,

A beryllium-9 ion has a positive charge that is double the charge of a proton, $q=2\times 1.6\times 10^{-19}\ C=3.2\times 10^{-19}\ C$

Speed of the ion in the magnetic field, $v=5\times 10^6\ m/s$

Its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location.

The magnitude of the field is 0.220 T.

We need to find the magnitude of the magnetic force on the ion. It is given by :

$F=qvB\\\\F=3.2\times 10^{-19}\times 5\times 10^6\times 0.22\\\\F=3.52\times 10^{-13}\ N$

So, the magnitude of magnetic force on the ion is $3.52\times 10^{-13}\ N$ .

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3 years ago

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