Ask question

Ask question

All categories

Makovka662 [10]

2 years ago

7

Two objects (42.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h

angs from the ceiling. Find (a) the acceleration of the objects and (b) the tension in the string.

1 answer:

Kazeer [188]2 years ago

8 0

Answer:

a = 3.27 m/s²

T = 275 N

Explanation:

Given that:

Mass m₁ = 42.p0 kg

Mass m₂ = 21.0 kg

Consider both masses to be in a whole system, then:

The acceleration can be determined as:

$(m_1+m_2)a = g(m_1-m_2)$

Making acceleration the subject in the above formula;

$a =\dfrac{g(m_1-m_2)}{(m_1+m_2)}$

$a =\dfrac{9.8(42.0-21.0)}{(42.0+21.0)}$

$a =\dfrac{9.8(21.0)}{(63.0)}$

$a =\dfrac{205.8}{(63.0)}$

a = 3.27 m/s²

in the string, the tension is calculated using the formula:

$T = \dfrac{2m_1m_2g}{(m_1+m_2)}$

$T = \dfrac{2(42)(21)(9.81)}{(42+21)}$

$T = \dfrac{17304.84}{63}$

T = 274.68 N

T ≅ 275 N

You might be interested in

A man, a distance d=3~\text{m}d=3 m from a target, throws a ball at an angle \theta= 70^\circθ=70 ∘ above the horizontal. If

lbvjy [14]

Answer:

The ball doesn't strike the building because it strikes the ground at d=1.62 meters.

Explanation:

V= 5 m/s < 70º

Vx= 1.71 m/s

Vy= 4.69 m/s

h= Vy * t - g * t²/2

clearing t for the flying time of the ball:

t= 0.95 s

d= Vx * t

d= 1.62 m

4 0

3 years ago

b. A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kg  m2. If the string is pulled with a force

Oduvanchick [21]

Answer:

f = 8 N

Explanation:

Data provided in the question

Radius of the pulley = r = 0.05 m

Moment of inertia = (I) = 0.2 kg.m^{2}

Angular acceleration = ∝ = 2 rad/sec

Based on the above information

As we know that

Torque is

$= force \times radius$

$= f \times r$

And,

Torque is also

$= moment\ of\ inertia \times angular\ acceleration$

$= I \times \alpha$

So,

We can say that

$f \times r = I \times \alpha$

$f \times 0.05 = 0.2 \times 2$

0.05f = 0.4

f = 8 N

We simply applied the above formulas

8 0

3 years ago

Montano1993 [528]

I think the answer is a

4 0

3 years ago

Atoms of elements at the top of a group on the periodic table are smaller than the atoms of elements at the bottom of the group.

serg [7]

Well formation of metallic bond depends on free electrons.smaal sized atoms hold their electrons more firmly as compared to large size atoms ,this z due to distance of outer shell electrons by nucleus..in this way no of free electrons affect strength of metallic bond..smaal sized atoms release less free electrons..

8 0

3 years ago

Read 2 more answers

Complete the equation???

soldi70 [24.7K]

Here in nuclear reaction we can say that sum of neutrons and protons in reactant side and product side will be same always

Here mass number on the product side is given to us

so sum of mass number is given as

$A_1 + A_2 = 265 + 1 = 266$

now on the reactant side also the number must be same

$A_1' + A_2' = 58 + x$

now we will have

$58 + x = 266$

$x = 208$

Now number of protons on product side is given as

$P_1 + P_2 = 108 + 0$

now we also know that atomic number of Fe is 26

so now we will have

$P_1' + P_2' = 108$

$26 + P_2' = 108$

$P_2' = 82$

now the equation is given as

$_{26}^{58}Fe + _{82}^{208}Pb = _{108}^{266}Hs + _0^1X$

8 0

2 years ago