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ahrayia [7]
3 years ago
6

How does a steel ship float,when a steel marble sinks?

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
4 0
I think it probably because the steel ship has more surface area then the the marble 
hope i helped
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6. A 4 kg object hangs below a 6 kg object by a string of negligible mass. If the 6 kg object is pulled upward by a force of 440
MrRissso [65]

Answer:

T =176 N

Explanation:

from diagram

F -(m_1+m_2_g) = (m_1+m_2_g)a

440 - (6+4)g = (6+4)a

a =\frac{440-10*9.8}{10}

a =34.2 m/s^2

frrom free body diagram of mass m2 = 4kg

T -m_2g =m_2a

T = m_2(g +a)

T = 4(9.81+34.2)

T =176 N

7 0
4 years ago
Give two examples of vernier calliper
omeli [17]

Answer:

Example 1, if a vernier caliper output a measurement reading of 2.13 cm, this means that: The main scale contributes the main number(s) and one decimal place to the reading

E.g. 2. 1 cm, whereby 2 is the main number and 0.1 is the one decimal place number

Explanation:

plz mark as brainliest and hope it helps you

4 0
3 years ago
A positive charge of 8.0 × 10-4 C is in an electric field that exerts a force of 3.5 × 10-4 N on it. What is the strength of the
Gennadij [26K]

Answer:

E = 0.437 N/C

Explanation:

Given that,

Charge, q=8\times 10^{-4}\ C

Electric force, F=3.5\times 10^{-4}\ N

Let the strength of the electric field is E. We know that, the electric force is given by :

F = qE

Where

E is the electric field strength

E=\dfrac{F}{q}\\\\E=\dfrac{3.5\times 10^{-4}}{8\times 10^{-4}}\\E=0.437\ N/C

So, the strength of the electric field is equal to 0.437 N/C.

6 0
3 years ago
A playground merry-go-round has a mass of 50 kg and a diameter of 4.0 m. There are 4 children who want to ride on it. They have
mixer [17]

Answer:

B) I1 = 1680 kg.m^2          I2 = 1120 kg.m^2

C) V = 0.84m/s      T = 29.92s

D) ω2 = 0.315 rad/s

Explanation:

The moment of inertia when they are standing on the edge:

I1 = 1/2*M*R^2 + (m1+m2+m3+m4)*R^2   where M is the mass of the merry-go-round.

I1 = 1680 kg.m^2

The moment of inertia when they are standing half way to the center:

I2 = 1/2*M*R^2 + (m1+m2+m3+m4)*(R/2)^2

I2 = 1120 kg.m^2

The tangencial velocity is given by:

V = ω1*R = 0.84m/s

Period of rotation:

T = 2π / ω1 = 29.92s

Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:

I1*ω1 = I2*ω2    Solving for ω2:

ω2 = I1*ω1 / I2 = 0.315 rad/s

5 0
3 years ago
A force vector points due east and has a magnitude of 150 newtons. A second force is added to . The resultant of the two vectors
balandron [24]

Answer:

a. 240 N due east

b. 540 N due west

Explanation:

Let east be the reference direction

(a) if the resultant force has a magnitude of 390 N and points east, and the 1st force is 150N due East, then the additional force would also due east and has a magnitude of

390 - 150 = 240 N

(b) if the resultant force has a magnitude of 390 N and points west, it would be -390N is eastern reference, and the 1st force is 150N due East, then the additional force would also due east and has a magnitude of

-390 - 150 = -540 N

This force would point west

6 0
3 years ago
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