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CaHeK987 [17]
4 years ago
11

For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this

is simple harmonic motion (in the vertical direction) with amplitude 5 cm and frequency 2 Hz. There is a box on the floor with mass m = 1 kg.
(a) Assuming the box remains in contact with the floor throughout, find the maximum and minimum values of the normal force exerted on it by the floor over an oscillation cycle.
(b) How large would the amplitude of the oscillations have to become for the box to lose contact with the floor, assuming the frequency remains constant? (Hint: what is the value of the normal force at the moment the box loses contact with the floor?)
Physics
1 answer:
ikadub [295]4 years ago
7 0

Answer:

no idea

Explanation:

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Please help my test is going on. Please answer both (i) and (ii). ​
Eduardwww [97]

Answer:

i. 7.5 m

ii. 15000 N

Explanation:

Area under vt graph shows the displacement

so area of triangle ABE= 1/2×15×2

=7.5 m

ii) F=ma

here, m=1000kg

and a=v-u/t

=15-0/1

a=15

F=1000×15=15000N

8 0
3 years ago
A straight segment of wire 35.0 cm long carrying a current of 2.60 A is in a uniform magnetic field. The segment makes an angle
den301095 [7]

Answer:

Magnetic field, B = 0.275 T

Explanation:

Given that,

Length of the wire, L = 35 cm = 0.35 m

Current carried in the wire, I = 2.6 A

The segment makes an angle of 53∘ with the direction of the magnetic field, \theta=53^{\circ}

Magnetic force, F = 0.2 N

To find,

The magnitude of the magnetic field.

Solution,

The magnetic force acting on the wire is given by :

F=ILBsin\theta

\theta is the angle between the length of wire and the magnetic field.

0.2=2.6\times 0.35\times B\times sin(53)

B = 0.275 T

Therefore, the magnitude of the magnetic field is 0.275 T. Hence, this is the required solution.

5 0
3 years ago
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te
schepotkina [342]

Answer:

6.77 minutes

Explanation:

172 degree - 78 degree = (185 degree - 78 degree)e−2 k

=> 94 = 107

e−2 k => 94 ÷ 107

k => ln (94÷107) / 2

147 - 78 = (185 - 78)e ^[ln (94÷107) / 2]

=> 69 = 107 e^ [ln (94÷107) / 2]

e^[ln (94÷107) / 2] =69 ÷ 107

=> t = [ln (69 ÷ 107)] ÷ [ln (94÷107) / 2]

t=> -0.4387 ÷ -0.0648

t => 6.77 minutes.

Therefore, the final answer to the question is 6.77 minutes.

4 0
3 years ago
Question 10 of 10Which statements are true?Check all that apply.O A. If heat is added to a substance, the temperature alwaysincr
inna [77]

The true statements are B and C.

This comes from the following facts:

Even when heat is added to a substance in a change of phase the temperature does not increases since this heat is latent heat.

3 0
1 year ago
The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typically μr=0.002. Suppose a 180
Irina18 [472]

Answer:

t = 0.354 hours

Explanation:

given,

coefficient of rolling friction μr=0.002

mass of locomotive = 180,000 Kg

rolling speed = 25 m/s

The force of friction = μ mg

                                 = (.002) x (180000) x (9.8)

                                 = 3528 N

F = m  a

now,

m a =  3528 N

180000 x a = 3528

a = 0.0196 m/s²

Then apply

v = u + at  

0 = 25 - 0.0196 x t

t = 1275.51 sec

t = 1275.61/3600 hours

t = 0.354 hours

time taken by the locomotive to stop = t = 0.354 hours

7 0
3 years ago
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