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
baherus [9]
2 years ago
10

PLEASE HELP AS SOON AS POSSIBLE!!

Physics
1 answer:
Naily [24]2 years ago
5 0

Answer:

Saved Energy ?

Explanation:

You might be interested in
Two boxes are connected to each other by a string as shown in the figure. The 10-n box slides without friction on the horizontal
FromTheMoon [43]

Answer:

T=7.4 N hence T<30 N

Explanation:

The figure is likely to be similar to the one attached. Writing the equation for forces we have

F-T=Fa/g where F is the force, T is tension, a is acceleration and g is acceleration due to gravity. Substituting the figures we have the first equation as

30 N - T = (30/9.81)a

Also, we know that T=F*a/g and substituting  10N for F we obtain the second equation as

T = (10/9.81)a

Adding the first and second equations we obtain

30 = 4.077471967

a Hence

a=\frac {30}{4.077471967}=7.3575 m/s^{2}

and T=a hence

T is approximately 7.4 N

5 0
3 years ago
Read 2 more answers
Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge
xxMikexx [17]

Answer:

new force is 6 times of the initial force.

Explanation:

Let the charges on two objects is q₁ and q₂. The electric force between charges is given by :

F=\dfrac{kq_1q_2}{r^2}

Objects 1 and 2 attract each other with a electrostatic force of 18.0 units

If the charge  of Object 1 is doubled and the charge of object 2 is tripled, it means, q_1'=2q_1 and q_2'=3q_2. New force is given by :

F'=\dfrac{kq_1'q_2'}{r^2}\\\\F'=\dfrac{k(2q_1)(3q_2)}{r^2}\\\\F'=6\dfrac{kq_1q_2}{r^2}\\\\F'=6F

So, the new electrostatic force between objects will become 6 times of the initial force.

5 0
3 years ago
The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p
leva [86]

1) 1.86\cdot 10^6 rad/s^2

2) 2418 rad/s

3) 27000 m/s^2

4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

\theta=\omega_i t +\frac{1}{2}\alpha t^2

where:

\theta is the angular displacement

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

\theta=90^{\circ} = \frac{\pi}{2}rad is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

Solving for \alpha, we find:

\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

\omega_f = \omega_i + \alpha t

where

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

Therefore, the final angular speed is:

\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s

3)

The tangential acceleration is related to the angular acceleration by the following formula:

a_t = \alpha r

where

a_t is the tangential acceleration

\alpha is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

v=u+at

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

a=27900 m/s^2 (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

v=0+(27900)(0.0013)=36.3 m/s

5 0
3 years ago
30 POINTS!
brilliants [131]

Answer:

d

Explanation:

5 0
3 years ago
Read 2 more answers
The spring to launch a pinball in a pinball machine is compressed 25 cm and has a spring constant of 140 N/m.
mote1985 [20]

Answer:

I think it is 5.6. This is my answer

8 0
3 years ago
Other questions:
  • Please help me with this homework
    10·2 answers
  • A cell is _______ if it lacks a nucleus and membrane-bound organelles.
    5·2 answers
  • An object is placed so that the image formed is a real image of the same size as the object. What is the position of the object?
    7·1 answer
  • When you throw a pebble straight up with initial speed V, it reaches a maximum height H with no air resistance. At what speed sh
    7·1 answer
  • How much heat in joules is required to melt an ice cube with a mass of 18.6 g at 0 °C? The Lf for water is 333 J/g.
    11·1 answer
  • Why is their displacement the same when the distance each traveled was different?
    7·1 answer
  • Help I need to know this answer!!
    11·1 answer
  • I really need help with this, please make sure the answer is correct.
    9·1 answer
  • If my candle has a brightness of 15 candela from a distance of 2 meter. What would the brightness be at a distance of 6 meters?
    13·1 answer
  • David is investigating the properties of soil using the sample shown.
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!