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Kipish [7]
2 years ago
7

two charges are placed at corners A and B of a square of side length. How much work is needed to move a charge from point C to D

Physics
1 answer:
Burka [1]2 years ago
6 0

A study occasionally the effect of anxiety (low vs. high) and stress (low vs. moderate vs. high) on test.

Everyone experiences anxiety occasionally, but persistent anxiety can reduce your quality of life. Though likely best known for altering behavior, worry can have negative effects on our physical health. Anxiety speeds up our heartbeat and breathing, concentrating blood flow to the parts of our brains that need it. You are getting ready for a challenging situation by having this extremely bodily reaction. Test performance may be impacted by anxiety. According to studies, pupils with low levels of test anxiety perform better on multiple-choice question (MCQ) exams than pupils with high levels of anxiety. Studies have occasionally that female students have greater levels of test anxiety than male students.

Learn more about anxiety here:
brainly.com/question/4913240

#SPJ4

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What would happen to mass and accelaration if the force on an object increases?? please help.
kozerog [31]
Wouldn't mass stay the same and acceleration increase or am I mistaken?

6 0
3 years ago
Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur
Alenkasestr [34]

Answer:

the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}

the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder  be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be \phi and \beta respectively.

The velocity of the block to be = v

The equivalent mass of the system = m_e

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

K.E = \frac{1}{2} m_ev^2       --------------- equation (1)

The angular velocity of the cylinder = \omega  :  &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

v = \omega R

\omega =\frac{v}{R}

The kinetic energy of the system in terms of individual mass can be written as:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2

By replacing \omega with \frac{v}{R} ; we have:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2

K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2   ------------------ equation (2)

Equating both equation (1) and (2); we have:

m_e = m_1+m_2+\frac{I}{R^2}

Therefore, the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}    which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by  the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

f_{block }=m_1gsin \beta

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

f_{cylinder} = m_2gsin \phi

Equating the above both equations; we have the equation of motion of the  equivalent system to be

m_e \bar x = f_{cylinder}-f_{block}

which can be written as follows from the previous derivations

(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Finally; the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

8 0
3 years ago
A small bug walks up a wall 2 m. The bug then walks 2 m down the wall. What is the distance?
gizmo_the_mogwai [7]
4m. The big travels 2m twice
5 0
2 years ago
Read 2 more answers
17. A volleyball weighs about 300 grams.
atroni [7]

Answer:

PE = 44.1 J

Explanation:

Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)

We have:

  • 1 kilograms = 1000 grams.

We convert it using a rule of 3, replacing, simplifying units and solving:

  • \boxed{\bold{x=\frac{gr*1\ kg}{1000\ gr}=\frac{300\ gr*1\ kg}{1000\ gr}=\frac{300\ kg}{1000}=\boxed{\bold{0.3\ kg}}}}

==================================================================

Earth's gravity is known to be 9.8 m/s², so we have:

Data:

  • m = 0.3 kg
  • g = 9.8 m/s²
  • h = 15 m
  • PE = ?

Use formula of potencial energy:

  • \boxed{\bold{PE=m*g*h}}

Replace and solve:

  • \boxed{\bold{PE=0.3\ kg*9.8\frac{m}{s^{2}}*15\ m}}
  • \boxed{\boxed{\bold{PE=44.1\ J}}}

Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.

The potential energy of the volleyball is <u>44.1 Joules.</u>

Greetings.

8 0
3 years ago
How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?
Illusion [34]
Power is defined as the rate at which the body is doing work:
P=\frac{W}{t}
Work is defined as displacement done by the force times that displacement:
W=F\cdot h
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
W=62N\cdot10m=620J
of work.
Now we just divide that by 5s to get how much power is required:
P=\frac{620J}{5s}=124W
5 0
3 years ago
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