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

When flying in an airplane, you are most likely in which layer of the atmosphere? mesosphere thermosphere stratosphere trosphere

Sergeeva-Olga [200]

Lower stratosphere, this is to avoid turbulence

6 0

3 years ago

Read 2 more answers

12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it

vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

$v_{f} = v_{o}+a\cdot (t-t_{o})$ (1)

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$v_{f}$ - Final velocity, measured in meters per second.

$a$ - Acceleration, measured in meters per square second.

$t_{o}$ - Initial time, measured in seconds.

$t$ - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust ( $v_{o} = 0\,\frac{m}{s}$ , $a = 30\,\frac{m}{s^{2}}$ , $t_{o} = 0\,s$ , $0\,s\le t< 30\,s$ )

$v = 30\cdot t$ (2)

Free fall ( $v_{o} = 900\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s}$ , $t_{o} = 30\,s$ , $30\,s \le t \le 120\,s$ )

$v = 900-9.81\cdot (t-30)$ (3)

Now we created the graph speed-time, which can be seen below.

5 0

3 years ago

A student wants to make an analogy to help her remember the characteristics of the planets. Which analogy makes the most sense?

Gala2k [10]

Answer:

The last one makes the most sense as they combine two like things that are easy to visualize

Explanation:

3 0

3 years ago

Read 2 more answers

Scientists hypothesize that there are no dinosaurs alive today because

Wittaler [7]

Its c after the meteorite hit earth it created a different environment worldwide that animals weren't able to adapt to due to its harsh conditions

8 0

3 years ago

Read 2 more answers

Heat is added to a 2kg piece of ice at a rate of 793kW. How long will it take for ice to melt if it was initially 0?

Ede4ka [16]

Answer:

0.84 s

Explanation:

Step 1

Given information:

Mass of the ice (m) = 2.0 kg

Heat transfer rate (Q/T) = 793.0 kW

Latent heat of fusion of ice (Lf) = 334 kJ/kg

$\frac{Q}{T} = \frac{mLf}{T}$

Substituting the corresponding values we have:

$793.0 kW= \frac{2.0kg(334 kJ/kg)}{T} \\ T = \frac{2.0kg(334 kJ/kg)}{793.0kW} = \frac{668kJ}{793kW} \\ = 0.84s$

8 0

3 years ago