An object weighing 150 N and is suspended from the ceiling by a wire. What is the tension in the cord?​

Answer:

2 hrs

Explanation:

time = distance ÷ speed

I would suggest an easier route. So for example, I would research light energy, where it comes from, and what causes it. I won't be able to do the work for you though.

Answer:

when x=0 kinetic energy is maximum and potential energy is high

when it is in mean position the potential energy is maximum and vise versa

Explanation:

And x=0 when the pendulum is at highest point when moving like velocity become zero when ball reached highest point

mean position is the point when pendulum comes back to original position  

I hope u understand

Answer:

2 m/s²

30 m/s

Explanation:

1) F = ma

1000 N = (500 kg) a

a = 2 m/s²

2) Given:

v₀ = 0 m/s

a = 2 m/s²

t = 15 s

Find: v

v = at + v₀

v = (2 m/s²) (15 s) + 0 m/s

v = 30 m/s

Answer:

210 m

Explanation:

The speed of train 2 relative to train 1 is 15 m/s + 20 m/s = 35 m/s.

It takes 6 seconds for the train to pass, so the length of the train is:

(35 m/s) (6 s) = 210 m

Answer:

The correct option is (a) "Changing".

Explanation:

If a coil is placed in the magnetic field, due to the change in its magnetic flux an emf will induce in it. EMF is like voltage. So, due to the induced emf and current is induced in it. Hence, we can say that changing the magnetic field can create an electric current.

Explanation:

Are the compounds formed by the ionic bonding or electronic bonding. They are formed by transferring the electron form one element's valance shell to other element's shell.

i hope it helps...

Explanation:

potential energy ===> kinetic energy

when an object is stretched there will be a stored energy in it enough to react to the upcoming change

that is the potential energy

similar to a stretched elastic band

kinetic energy is the energy on a moving object

*If a kinetic energy takes place there must gave been a potential energy before*

A catapult is used to fire an object into the air, the potential energy transfers as the elastic potential energy and to the kinetic energy when the catapult is released and stretched.

The potential energy is associated with bodies that are above a reference point, at that reference point the value of potential energy is taken as zero. For example, if an object with a mass M is in a tower that is 140 meters high, its potential energy is defined as the product of the mass of the object by gravity and by the height at which the object is measured from the reference point.

Ep =mgh

When talking about elastic objects such as springs or elastic fibers, it is usually understood that there is an elastic potential energy associated with the constant of the spring. This can be calculated using the following equation:

U ela = 1/2×k×(Dx)²

where:

k = elastic constant of the elastic material [N/m]

Dx = distance stretched or compressed of the elastic material [m]

U = elastic energy [Joules]

So, there are difference between the elastic potential energy and the potential energy.

In a catapult, Initially, energy is stored in the form of potential energy which is later converted into kinetic energy. Potential energy can be stored in the form of elastic potential energy in the case of elastic. It suddenly releases the energy to propel the object into the air.

Learn more about catapult,

https://brainly.com/question/20899342

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Answer:

Explanation:

Given,

Force ( f ) = 30 N

Acceleration(a) = 7.6 m/s

Now, Let's find the mass of the ball

Using the Newton's second law of motion:

We get:

[tex]force \: = mass \: \times acceleration[/tex]

plug the value

[tex]30 \: = m \: \times 7.6[/tex]

Use the commutative property to reorder the terms

[tex] 30 = 7.6 \: m[/tex]

Swap the sides of the equation

[tex]7.6m = 30[/tex]

Divide both sides of the equation by 7.6

[tex] \frac{7.6 \: m}{7.6} = \frac{30}{7.6} [/tex]

Calculate

[tex]m = 3.94 \: kg[/tex]

Hope this helps..

Best regards!!

Answer:

[tex]\displaystyle \boxed{\mathrm{3.95 \: kg }}[/tex]

Explanation:

[tex]\mathrm{force \: (N) = mass \: (kg) \times acceleration \: (m/s^2)}[/tex]

[tex]\mathrm{force = 30N}[/tex]

[tex]\mathrm{acceleration = 7.6 \: m/s^2 }[/tex]

[tex]\mathrm{Find \: the \: mass.}[/tex]

[tex]\mathrm{30 = m \times 7.6}[/tex]

[tex]\displaystyle \mathrm{m =\frac{30}{7.6} }[/tex]

[tex]\displaystyle \mathrm{m = 3.947... }[/tex]

Answer and Explanation:

For computing the displacement at point B and D we need to determine the following calculations

[tex]P_Net = P_C + P_E + P_B[/tex]

= 250 + 350 - 50

= 550 N

Now the deflection for bar AB is

[tex]\delta_{AB} = \frac{PL_{AB}}{AE} \\\\ = \frac{550 \times 500}{6,000 \times 200 \times 10^{3}}[/tex]

Now for bar BC it is

[tex]\delta_{BC} = \frac{PL_{BC}}{AE} \\\\ = \frac{(550 + 50) \times 250}{5,000 \times 200 \times 10^{3}} \\\\ = 1.5 \times 10^{-04} mm[/tex]

And for bar CD it is

[tex]\delta_{CD} = \frac{PL_{CD}}{AE} \\\\ = \frac{(550 -250 + 50) \times 250}{5,000 \times 200 \times 10^{3}} \\\\ = 0.875 \times 10^{-4} mm[/tex]

Now the displacement is as follows

For B

2.292 × 10^{-4} mm

For D, it is

[tex]= 2.292 \times 10^{-4} + 1.5 \times 10^{-4} + 0.875 \times 10^{-4} mm \\\\ = 4.667 \times 10^{-4} mm[/tex]

We simply applied the above formulas for determining the  displacements at points B, D and the same is to be considered  

Answer:

Explanation:

This problem bothers on the simple harmonic motion of a loaded spring

Given data

mass attached, m= 0-.675 kg

spring constant, k= 72.4 N/m

the period of oscillation can be solved for using the formula bellow

[tex]T= 2\pi \sqrt{\frac{m}{k} }[/tex]

Substituting the given data into the expression above we have

[tex]T= 2*3.142\sqrt{\frac{0.675}{72.4} }\\T= 6.284*\sqrt{0.0093 }\\T= 0.6[/tex]

T= 0.6 sec

Answer:

0.607

Explanation:

Trust me

Answer:

Q = 1.404 × 10^(5) KJ

Explanation:

We are given:

Mass;m = 500 g = 0.5kg

Temperature 1;T1 = 28 °C

Temperature:T2 = 150 °C

Specific heat capacity;c_p = 4183 J/Kg °C

Latent heat of vaporization;L = 2.26 × 10^(6) J/Kg.

The heat energy needed is given by;

Q = sensible heat energy + Latent heat

Formula for sensible heat is;

Sensible heat energy = mc(t2 - t1)

Formula for Latent heat is ;

Latent heat = mL

Thus:

Q = mc(t2 - t1) + mL

Q = m[c(t2 - t1) + L]

Q = 0.5((4183(159 - 28) + (2.26 × 10^(6)))

Q = 1.404 × 10^(8) J = 1.404 × 10^(5) KJ

Answer:

Sphere is a geometrical object in dimensional space  that surface of a circle and ball.

Explanation:

Sphere is that the circular objects in the two dimensional space (1) circle

(2) disk. Two dimensional space is a set of points and the distance of that point,The two points of Sphere that length and center.

Sphere can constructed as the named of surface form circle about  any diameter. circle is the special type of the revolution replacing the circle,

sphere is the distance r is the radius of the ball and circle  is the center of mathematical ball,as the center and the radius of the sphere is to respectively.

The ball and sphere has not be maintained mathematical references  as a solid references. A sphere of any radius is centered at the number of zero.

Explanation:

Power = work / time

Power = force × distance / time

P = (650 kg) (10 m/s²) (20 m) / (15 s)

P = 8667 W

Answer:

2980 N

Explanation:

The fundamental frequency of a string is:

f₁ = √(T/µ) / (2L)

78 Hz = √(T / (0.03 kg/m)) / (2 (2.02 m))

315.12 m/s = √(T / (0.03 kg/m))

99,300 m²/s² = T / (0.03 kg/m)

T = 2980 N

Answer:

The tension in string P is 25 N, while that of Q is 85 N.

Explanation:

Considering the conditions for equilibrium,

i. Total upward force = Total downward force

                     [tex]T_{P}[/tex] + [tex]T_{Q}[/tex] = 110 N

ii. Taking moment about P,

clockwise moment = anticlockwise moment

110 × (2.5 - 0.8) = [tex]T_{Q}[/tex] × (3 - 0.8)

110 × 1.7 = [tex]T_{Q}[/tex] × 2.2

187 = 2.2[tex]T_{Q}[/tex]

[tex]T_{Q}[/tex] = [tex]\frac{187}{2.2}[/tex]

[tex]T_{Q}[/tex] = 85 N

From the first condition,

[tex]T_{P}[/tex] + [tex]T_{Q}[/tex] = 110 N

[tex]T_{P}[/tex] + 85 N = 110 N

[tex]T_{P}[/tex] = 110 - 85

[tex]T_{P}[/tex] 25 N

Therefore, the tension in string P is 25 N while that of Q is 85 N.

Answer:

The answer to your question is given below.

Explanation:

Mechanical advantage (MA) = Load (L)/Effort (E)

MA = L/E

Velocity ratio (VR) = Distance moved by load (l) / Distance moved by effort (e)

VR = l/e

Efficiency = work done by machine (Wd) /work put into the machine (Wp) x 100

Efficiency = Wd/Wp x100

Recall:

Work = Force x distance

Therefore,

Work done by machine (wd) = load (L) x distance (l)

Wd = L x l

Work put into the machine (Wp) = effort (E) x distance (e)

Wp = E x e

Note: the load and effort are measured in Newton (N), while the distance is measured in metre (m)

Efficiency = Wd/Wp x100

Efficiency = (L x l) / (E x e) x 100

Rearrange

Efficiency = L/E ÷ l/e x 100

But:

MA = L/E

VR = l/e

Therefore,

Efficiency = L/E ÷ l/e x 100

Efficiency = MA ÷ VR x 100

Efficiency = MA / VR x 100

Answer:

we have to use formula of volume to find volume of a cuboid. ( i.e v = l ×b ×h)

Explanation:

here, let your length of cuboid be x cm, breadth be y cm and height be z cm .

now, formula to find volume of cuboid = length ×

breadth × height.

so, v( volume)= l (length)× b (breadth)× h (height)

or, v= x cm × y cm × z cm

therefore, volume is xyz cm^3..... answer.

hope it helps..

Answer: length times width times height

Explanation:

Answer: 5.075Ns

Explanation:

Given the following :

Mass of ball = 145g

Initial Speed of ball = 15m/s

Final speed of ball when hit by the batter = - 20m/s ( Opposite direction)

The impulse of a body is represented using the relation:

Force(f) * time(t) = mass (m) * (final Velocity(V) - initial velocity(u))

Therefore, using:

m(v - u) = impulse

Mass of ball = 145 / 1000 = 0.145kg

Impulse = 0.145(- 20 - 15)

Impulse = 0.145(-35)

Impulse = 5.075Ns

Answer:D

Explanation:

Apex

Explanation:

It is given that,

Object distance from the mirror, u = -30 cm

Focal length of the mirror, f = +15 cm

Size of the object, h = 7.5 cm

We need to find the image distance and the size of the image.

Mirror's formula, [tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]

v is image distance

[tex]\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(15)}-\dfrac{1}{(-30)}\\\\v=10\ cm[/tex]

Let h' is the size of the image. So,

[tex]\dfrac{h'}{h}=\dfrac{-v}{u}\\\\h'=\dfrac{-vh}{u}\\\\h'=\dfrac{-10\times 7.5}{-30}\\\\h'=2.5\ cm[/tex]

So, the image is located at a distance of 10 cm and the size of the image is 2.5 cm.

Answer:

t = 6

Explanation:

Displacement is equal to the area under a velocity vs time graph.

In this case, the area is a triangle.  At time t, the base of the triangle is t.  The height of the triangle can be found with similar triangles:

h / t = 8 / 12

h = ⅔ t

So the distance traveled at time t is:

d = ½ (t) (⅔ t)

d = ⅓ t²

The distance traveled at time 12 is:

D = ½ (12) (8)

D = 48

We want to find when d = D/4.

d = D/4

⅓ t² = 48/4

⅓ t² = 12

t² = 36

t = 6

Alternatively, since the acceleration is constant here, we could use a constant acceleration equation.

Δx = v₀ t + ½ at²

Given v₀ = 0 m/s and a = ⅔ m/s²:

Δx = (0) t + ½ (⅔) t²

Δx = ⅓ t²

When t = 12, Δx = 48.

⅓ t² = 48/4

t = 6

Answer:

The relative velocity of an object A with respect to another object B.

Explanation:

The relative velocity of an object A with respect to another object B is the velocity that object A would appear to have to an observer situated on object B moving along with it.

Answer:

(a) The projectile takes approximately 4.420 seconds to reach the water, (b) The horizontal scope of the projectile is 1591.2 meters, (c) The remaining height to descend after 2 seconds of being launched is 63.624 meters.

Explanation:

The projectile experiments a parabolic motion, where horizontal speed remains constant and accelerates vertically due to the gravity effect. Let consider that drag can be neglected, so that kinematic equation are described below:

[tex]x = x_{o}+v_{o,x} \cdot t[/tex]

[tex]y = y_{o} + v_{o,y}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}[/tex]

Where:

[tex]x_{o}[/tex], [tex]y_{o}[/tex] - Initial horizontal and vertical position of the projectile, measured in meters.

[tex]v_{o,x}[/tex], [tex]v_{o,y}[/tex] - Initial horizontal and vertical speed of the projectile, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]x[/tex], [tex]y[/tex] - Current horizontal and vertical position of the projectile, measured in meters.

Given that [tex]x_{o} = 0\,m[/tex], [tex]y_{o} = 80\,m[/tex], [tex]v_{o,x} = 360\,\frac{m}{s}[/tex], [tex]v_{o,y} = 0\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the kinematic equations are, respectively:

[tex]x = 360\cdot t[/tex]

[tex]y = 80-4.094\cdot t^{2}[/tex]

(a) If [tex]y = 0\,m[/tex], the time taken for the projectile to reach the water is:

[tex]80 - 4.094\cdot t^{2} = 0[/tex]

[tex]t = \sqrt{\frac{80}{4.094} }\,s[/tex]

[tex]t \approx 4.420\,s[/tex]

The projectile takes approximately 4.420 seconds to reach the water.

(b) The horizontal scope is the horizontal distance done by the projectile before reaching the water. If [tex]t \approx 4.420\,s[/tex], the horizontal scope of the projectile is:

[tex]x = 360\cdot (4.420)[/tex]

[tex]x = 1591.2\,m[/tex]

The horizontal scope of the projectile is 1591.2 meters.

(c) If [tex]t = 2\,s[/tex], the height that remains to descend is:

[tex]y = 80-4.094\cdot (2)^{2}[/tex]

[tex]y = 63.624\,m[/tex]

The remaining height to descend after 2 seconds of being launched is 63.624 meters.

Answer: Upthrust = Weight - Apparent weight

= 0.250 kgf - 0. 150 kgf

= 0.100 kgf

Density = mass / volume

volume = mass / density

= 0.100 kg / (1000 kg / m³)

= 0.0001 m³

density of object  = mass / volume

= 0.250 kg / 0.0001 m³

= 2500 kg / m³

upthrust of oil = Weight - Apparent weight

= 0.250 kgf - 0.125 kgf

= 0.125 kgf

density = mass / volume

= 0.125 kg / 0.0001 m³

= 1250 kg/m³

density of the object = 2500 kg / m³

density of oil = 1250 kg / m³

Answer:

Janaki temple and Pashupatinath Temple are found in Nepal. These structures are religious and they increase the pride of the country through tourism. These temples are important in the country and are mostly visited by tourists.

This also helps to bring in income for the country and also helps by attracting many visitors and a corresponding development of its tourism industry.

Answer:

5702.88 J or 5.7mJ

Explanation:

Given that :

C 1 = 6.0-μF

C 2 = 4.0-μF

V 1 = 50V

V 2 = 34V

Note that : Q = CV

Q 1 = C1 * V1

Q 1 = 50×6 = 300μC

Q 2 = 34×4 = 136μC

Parallel connection = C 1 + C 2

= 6+4 = 10μC

V = Qt/C

Where Qt = Q1+Q2

V = Q1+Q2/C

V = 300+136/10

V = 437/10

V = 43.6volts

Uc1 = 1/2×C1V^2

= 1/2 × 6μF × 43.6^2

= 1/2 × 6μF × 1900.96

= 3μF × 1900.96volts

= 5702.88J

= 5702.88J/1000

= 5.7mJ

Answer:

The unit of current is defined as the flow of 1 coulomb of charge in one second

Answer:

Explanation:

Reddening of sun's rays at sunset and sunrise is due to scattering of light . The white light consisting of seven colours coming from the sun are scattered in different directions when they fall on the air particles present in atmosphere . Red coloured light scatters least and it travels straight forward to the viewer on the earth . On the other hand other colours scatter most and therefore they go out of area of vision for the viewer on the earth . Since only red colour reaches the eye of the viewer , sun's ray appear red . This happens during sunrise and sunset . It is so because during this period , sun rays travel far greater distance through  atmosphere , so scattering is most pronounced .