Ask question

Ask question

All categories

2 years ago

9

Two blocks, 1 and 2, are connected by a rope R1 of negligible mass. A second rope R2, also of negligible mass, is tied to block

2. A force is applied to R2 and the blocks accelerate forward. Find the ratio of the forces exerted on blocks 1 and 2 by the ropes R1 and R2 respectively. Here m1 = 2.11m2.

1 answer:

alekssr [168]2 years ago

3 0

Answer:

Explanation:

Given

Two block are connected by rope $R_1$

$R_2$ rope is attached to block 2

suppose $F_2$ is a force applied to Rope $R_2$

Applied force $F_2$ =Tension in Rope 2

$F_2=(m_1+m_2)a---1$

where a=acceleration of system

Tension in rope $R_1$ is denoted by $F_1$

$F_1=m_1a---2$

divide 1 and 2 we get

$\frac{F_2}{F_1}=\frac{(m_1+m_2)a}{m_1a}$

also $m_1=2.11\cdot m_2$

$\frac{F_2}{F_1}=\frac{2.11m_2+m_2}{2.11m_2}$

$\frac{F_2}{F_1}=\frac{3.11}{2.11}$

$\frac{F_1}{F_2}=\frac{2.11}{3.11}$

You might be interested in

To measure moderately low pressures, oil with a density of 9.0 x 102 kg/m3 is used in place of mercury in a barometer. A change

Yuki888 [10]

Answer:

Δ P = 13.24 Pa

Explanation:

Given that

Density of oil ,ρ₁ = 9 x 10² kg/m³

We know that density for mercury ,ρ₂ = 13.6 x 10³ kg/m³

The change in the height of column ,Δh = 1.5 mm

The pressure given as

P = ρ g h

Change in the pressure

Δ P = ρ₁ g Δh

Now by putting the values

Δ P = 9 x 10² x 9.81 x 1.5 x 10⁻³ Pa

Δ P = 13.24 Pa

Therefor the change in the pressure will be 13.24 Pa.

3 0

2 years ago

Considerable scientific work is currently under way to determine whether weak oscillating magnetic fields such as those found ne

slega [8]

Answer:

$\epsilon = 2.96 \times 10^{-11} \ V$

Explanation:

given,

magnetic field strength = 1.40 ✕ 10⁻³ T

frequency of oscillation = 60 Hz

diameter of RBC = 7.5 μm

EMF = ?

$\epsilon = NBA\omega$

$\epsilon = NB(\pi\ r^2)\ (2\pi f)$

$\epsilon = NB(\pi\ (\dfrac{d}{2})^2)\ (2\pi f)$

$\epsilon = (1)\ 1.4 \times 10^{-3}(\pi\ (\dfrac{7.5 \times 10^{-6}}{2})^2)\ (2\pi\times 60)$

$\epsilon = 2.96 \times 10^{-11} \ V$

maximum emf that can generate around the perimeter of the cell $\epsilon = 2.96 \times 10^{-11} \ V$

5 0

2 years ago

A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra

skelet666 [1.2K]

Answer:

So the acceleration of the child will be $8.05m/sec^2$

Explanation:

We have given angular speed of the child $\omega =1.25rad/sec$

Radius r = 4.65 m

Angular acceleration $\alpha =0.745rad/sec^2$

We know that linear velocity is given by $v=\omega r=1.25\times 4.65=5.815m/sec$

We know that radial acceleration is given by $a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2$

Tangential acceleration is given by

$a_t=\alpha r=0.745\times 4.65=3.464m/sec^$

So total acceleration will be $a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2$

7 0

2 years ago

Which best describes nuclear fusion?

MrMuchimi

Answer:

The statement that best describes nuclear fusion is;

Nuclei combine to form a heavier nucleus, releasing energy

Explanation:

In nuclear fusion, we have the reaction of the nuclei of two or more atoms coming together (combining) to form heavier elements and subatomic particles such as protons and neutrons accompanied by the release or absorption in energy depending on the difference between the mass of the reactants and the products

Some nuclear fusion reaction require an input of energy and such reactions are therefore not spontaneous

The best option is nuclei (two or more nuclei) combine to form a heavier nucleus, releasing energy.

5 0

2 years ago

Read 2 more answers

What is the magnitude of the force a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away?

sweet [91]

The magnitude of the force<span> a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away is 1920 Newtons. The formula used to solve this problem is:

F = kq1q2/r^2

where:
F = Electric force, Newtons
k = Coulomb's constant, 9x10^9 Nm^2/C^2
q1 = point charge 1, C
q2 = point charge 2, C
r = distance between charges, meters

Using direct substitution, the force F is determined to be 1920 Newtons.</span>

7 0

2 years ago