1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
babunello [35]
3 years ago
15

The bob (weight) at the end of a pendulum has a mass of 0.3 kilograms. The bob is pulled to position B and allowed to swing. It

goes all the way to position C and swings back.
The potential energy of the bob at position B is >>1.47 joules. If the maximum height of the bob is 0.45 meters when it swings back, ___ joules of energy was transformed to thermal energy.
>>>>>Use g = 9.8 m/s2 and PE = m × g × h.<<<<<

It's NOT 1.323!!!

Physics
1 answer:
Ivahew [28]3 years ago
3 0

Answer:

0.147 J

Explanation:

The total energy that has been transformed into thermal energy is equal to the loss of gravitational potential energy between the initial situation (bob at h=0.5 m above the ground) and the final situation (bob back but at h=0.45 m above the ground).

Therefore, we have

E_{thermal}=\Delta U=mgh_1 - mgh_2 = mg(h_1 -h_2)

where

m = 0.3 kg is the mass of the bob

g = 9.8 m/s^2

h1 = 0.5 m is the initial height

h2 = 0.45 m is the final height

Substituting, we find the thermal energy

E_{thermal}=(0.3 kg)(9.8 m/s^2)(0.5 m-0.45 m)=0.147 J

Therefore, the energy transformed into thermal energy is 0.147 J.

You might be interested in
3 kg of saturated liquid water, contained in a rigid container at p = 1.5 bar, is heated by a 80 W heater. a) What will the phas
Mkey [24]
Is this science also? bc i know that
5 0
3 years ago
A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a
Zielflug [23.3K]

Answer:F_{v} =\mu_{k} mg

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to  start crate moving is F_{s} =\mu_{s} mg.

but once crate starts moving the force of friction is reduced  F_{v} =\mu_{k} mg.

Hence  to keep crate moving at constant velocity we have to reduce the  force pushing crate ie F_{v} =\mu_{k} mg.

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as  forces are balanced.

Magnitude of the force

F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9  N

4 0
4 years ago
A student sets a board to pivot about its center. The student then places a box with mass m at some position r from the pivot po
kvv77 [185]

Answer:

The box should be placed at a distance of \frac{r}{2} from the pivot

Explanation:

In order to be in static equilibrium, both Torques have to be the same magnitude, so:

T_{m} = T_{2m}   Replacing the formula for Torque:

m*r = 2m*X   where X is the distance we need to find.

Solving for X we get:

X = \frac{r}{2}

As we can see, the distance does not depend on the actual value of the mass but on the fact of one being twice as much as the other one.

7 0
4 years ago
Can someone please help?? I don’t understand this material!!!
Debora [2.8K]

Answer:

1)

When the person throws the ball away, the person rolls backward. This is due to the law of conservation of momentum: in fact, the total momentum of the person+ball system must be conserved.

At the beginning,

p_i=0

after throwing the ball, the total momentum is the sum of the momentum of the person and of the ball:

p_f=p_p + p_b

Since momentum is conserved,

p_i = p_f\\0=p_p+p_b

So

p_p = -p_b

Therefore, the person has equal momentum (in magnitude) but opposite direction to the ball, so the person rolls backward.

However, if the person hold to the ball, then they will have same momentum (moving in the same direction). In order to conserve the total momentum (which was zero at the beginning), the only possible solution is that

p_p=p_b=0

which means that both the person and the ball will remain at rest. This is because there are no external forces acting on the system, so the system cannot move.

2)

The change in momentum of an object is given by

\Delta p=m(v-u)

where

m is the mass of the object

v is its final velocity

u is the initial velocity

For the clay ball in this problem, we have:

m = 50 g = 0.050 kg

v = 0 m/s (it sticks on the wall)

u = 1 m/s

So its change in momentum is

\Delta p_c=(0.050)(0-1)=-0.050 kg m/s

For the superball, we have:

m = 50 g = 0.050 kg

v = -0.8 m/s (it bounces back)

u = 1 m/s

So its change in momentum is

\Delta p_s = (0.050)(-0.8-1)=-0.09 kg m/s

So, the superball has a greater change in momentum (in magnitude).

3a)

According to Newton's third law of motion:

"When an object A exerts a force (action force) on an object B, then object B exerts an equal and opposite force (reaction force) on object A".

Here, we have a Hummer and a Beetle colliding head-on: we can identify them as object A and object B. Therefore, according to Newton's third law:

- The action force is the force of impact exerted by the Hummer on the Beetle

- The reaction force is the force of impact exerted by the Beetle on the Hummer

And according to the Law, the two forces are equal in magnitude: so, the two vehicles experience the same force of impact.

3b)

The change in momentum of each vehicle during the collision can be written as

\Delta p = F\Delta t (1)

where

\Delta p is the change in momentum

F is the force experienced by the vehicle

\Delta t is the duration of the collision

in part 3a), we said that the two vehicles experience the same force in the collision.

Moreover, the duration of the collision, \Delta t, is the same for the two vehicles.

As a result, according to formula (1), the two vehicles have same change in momentum (however, the directions would be opposite, since they experience force in opposite directions).

3c)

According to Newton's second law of motion, the acceleration of an object is given by:

a=\frac{F}{m}

where

F is the force experienced by the object

m is its mass

a is its acceleration

In part 3a), we stated that the force experienced by the Beetle and the Hummer is the same. However, the mass of the Beetle is smaller than the mass of the Hummer: from the equation we see that the acceleration is inversely proportional to the mass, therefore the Beetle will experience a greater acceleration.

4a)

The force experienced by the dashboard on the car is given by:

F=\frac{\Delta p}{\Delta t}

Where

\Delta p is the change in momentum

\Delta t is the duration of the collision

In a padded dashboard, the duration of the collision \Delta t is larger than the duration of the collision for a hard dashboard. According to the equation above, the force experienced by the dashboard (and so, the car) is inversely proportional to the duration of the collision: therefore, since the padded dashboard has a larger \Delta t, it will experience a smaller force than the hard dashboard.

4b)

The force experienced by the climber if falling is given by

F=\frac{\Delta p}{\Delta t}

Where

F is the force experienced by the climber

\Delta p is his change in momentum

\Delta t is the duration of fall

Nylon is a very elastic material, so it is able to "soften" the fall by stretching a lot. As a result, the nylon increases the value of \Delta t in the formula. Since the force experienced by the climber is inversely proportional to \Delta t, the climber will feel less force thanks to the nylon.

4c)

This technique is used to exploit the "push" given by the second car of the train to the first car when the brakes are applied.

At first, the engine is started, and the first car starts accelerating, pulling the second car (and the following cars). Then, the brakes are applied on the first car: however, the second car keeps moving by inertia, so then it gives a push forward on the first car. Then, this action is repeated several times, so that this push exerted by the second car is exploited several times.

3 0
3 years ago
Car A m= 2000kg<br> v=20 km/h <br> Car B m= 2000kg <br> v= 20km/h what is the location after impact
Darya [45]

Mass of car, M=2000kg

Velocity, V=72× 5/18 =20ms¹

Apply kinematic equation of motion

v² -u² = as

0 − 20² =2a×20

a=−10ms^-2

Breaking force, F=ma=2000×10=20 kN

Apply first kinematic equation

v = u +at

t= \frac{v - u}{a}  =  \frac{0 - 20}{ - 10}  = 2sec\\

7 0
3 years ago
Other questions:
  • A bear can cover 60 min 4 seconds. What speed can it travel?
    9·2 answers
  • Serena is mixing a material into a beaker filled with a liquid. She notices that the material seems to disappear into the liquid
    9·2 answers
  • 1 question 20points<br> How is frequency related to the sound we here
    13·1 answer
  • At a construction site, the site manager notices that a crane takes 20 seconds to lift a 500kg steel beam up to a height of 15 m
    8·1 answer
  • Which sample has the lowest temperature?
    12·1 answer
  • ** URGENT** The voltage across the primary winding is 350,000 V, and the voltage across the secondary winding is 17,500 V. If th
    15·1 answer
  • When is the force on a current-carrying wire in a magnetic field at its strongest?
    5·2 answers
  • In central Texas there is a large domed hill of pink granite surronund by low hills of more common bedrock of Texas, limestone.
    9·2 answers
  • Potential energy is
    5·1 answer
  • A 72.5-kg hiker starts at an elevation of 1230 m and climbs to the top of a peak 2660 m high.
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!