Ask question

Ask question

All categories

3 years ago

7

A 1.0-kg mass (mA) and a 6.0-kg mass (mB) are attached to a lightweight cord that passes over a frictionless pulley. The hanging

masses are free to move. Find the acceleration of the larger mass.

2 answers:

RideAnS [48]3 years ago

8 0

Answer:

Explanation:

Given

$m_1=1\ kg$

$m_2=6\ kg$

Pulley mass String

For $m_1=1\ kg$

$T-m_1g=m_1a$

$T=m_1(g+a)$

for other body $m_2$

$m_2g-T=m_2a$

$T=m_2(g-a)$

Equating value of Tension

$m_2=m_1\times \frac{g+a}{g-a}$

$6=\frac{g+a}{g-a}$

$6(10-a)=10+a$

$50=7a$

$a=\frac{50}{7}$

$a=7.142\ m/s^2$

IrinaVladis [17]3 years ago

8 0

Answer:

Answer:

7 m/s^2

Explanation:

mA = 1 kg

mB = 6 kg

Let a be the acceleration and T be the tension in the string

By use of Newton's second law

T - mA g = mA x a .... (1)

mBg - T = mB x a ..... (2)

Adding both the equations

(mB - mA) g = (mA + mB) x a

(6 - 1) x 9.8 = (6 + 1) x a

7 a = 9.8 x 5

a = 7 m/s^2

Thus, the acceleration in the system is 7 m/s^2.

Explanation:

You might be interested in

suppose you needed to build a raft to cross a fast-moving river. Describe the physical and chemical properties of the raft that

erastovalidia [21]

1.Density of the material building the raft must be lower than the water.
<span>2. Material must not react with water. </span>
<span>3.Material must have high strength. </span>
<span>4.Raft must be wide in-order to avoid drawing in the river.</span>

4 0

3 years ago

Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m

Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

$m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s$

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

$V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

Substituting,

$V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$V_{cm} = 1.034m/s$

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where here $v_f$ is the velocity after the collision.

$v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

$v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$v_f = 1.034m/s$

8 0

3 years ago

When you rub a glass rod with silk. The rod becomes positive charged, and the silk becomes negative.Also when you rub a rubber r

adelina 88 [10]

Answer:

Not sure

Explanation:

7 0

2 years ago

Objects that rest have no forces upon them

Pani-rosa [81]

Answer:

false

Explanation:

every object will always have the force of gravity acting upon it.

6 0

2 years ago

A circular loop ( radius = 0.5 m) carries a current of 3.0 A and has unit normal vector of (2i - j +2k)/3 what is x component of

Roman55 [17]

Answer:

$T=9.42Nm$

Explanation:

From the question we are told that:

Radius $r= 0.5 m$

Current $I= 3.0 A$

Normal vector $n=\frac{(2i - j +2k)}{3}$

Magnetic field $B= (2i-6j) T$

Generally the equation for Area is mathematically given by

$A=\pi r^2$

$A=3.1415 *0.5^2$

$A=0.7853 m^2$

Generally the equation for Torque is mathematically given by

$T=A(i'*B)$

Where

$i'*B= \begin{bmatrix}2&-1&2\\2&-6&0\end{bmatrix}$

$X\ component\ of\ i'*B= [(-1 * 0)-(2*-6)]$

$X\ component\ of\ i'*B=12$

Therefore

$T=0.7853*12$

$T=9.42Nm$

7 0

2 years ago