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Len [333]
2 years ago
11

A 25 kg lamp is hanging from a rope. What is the tension force being supplied by the rope?

Physics
1 answer:
Elan Coil [88]2 years ago
7 0

The tension force being supplied by the rope is 245 N.

<h3>What is tension force?</h3>
  • Tension force is the force exerted on a rope or cord due to the weight of an object suspended from it.

The tension force on the given rope due to the weight of the lamp hanging from the rope is calculated by applying Newton's second law of motion as shown below;

T = mg

where;

  • m is the mass = 25 kg
  • g is acceleration due to gravity = 9.8 m/s²

T = 25 x 9.8

T = 245 N

Thus, the tension force being supplied by the rope is 245 N.

Learn more about tension force here: brainly.com/question/12797227

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A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.
LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

W = \frac{1}{2}(50)(5.61^2) = 786.8025 J

Now, we can define work:

\large\boxed{W = Fdcos\theta}}

Now, plug in the values:

786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta

Solve for theta:

cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}

4 0
2 years ago
28. Sound can be heard around a corner because of
Lesechka [4]

Answer:

Diffraction of sound wavelengths.

Explanation:

Diffraction-A wave is able to bend around a corner due to the effects of diffraction. sound aves are capable of bending around corners in the same magnitude as it's wavelength making it possible to hear sounds around corners.

5 0
3 years ago
Green light (λ = 518 nm) strikes a single slit at normal incidence. What width slit will produce a central maximum that is 3.00
MariettaO [177]

Answer:

6.9066 × 10⁻⁵ m

Explanation:

For constructive interference, the expression is:

d\times sin\theta=m\times \lambda

Where, m = 1, 2, .....

d is the distance between the slits.

The formula can be written as:

sin\theta=\frac {\lambda}{d}\times m ....1

The location of the bright fringe is determined by :

y=L\times tan\theta

Where, L is the distance between the slit and the screen.

For small angle , sin\theta=tan\theta

So,  

Formula becomes:

y=L\times sin\theta

Using 1, we get:

y=L\times \frac {\lambda}{d}\times m

Thus, the distance between the central maximum is 3.00 cm

First bright fringe , m = 1 occur at 3.00 / 2 = 1.50 cm

Since,

1 cm = 0.01 m

y = 0.0150 m

Given L = 2.00 m

λ = 518 nm

Since, 1 nm = 10⁻⁹ m

So,

λ = 518 × 10⁻⁹ m

Applying the formula as:

0.0150\ m=2.00\ m\times \frac {518\times 10^{-9}\ m}{d}\times 1

<u>⇒ d, distance between the slits = 6.9066 × 10⁻⁵ m</u>

7 0
3 years ago
A sailboat picks up a gust of wind and accelerates to a speed of 6m/s in 16 seconds.If the initial velocity 1 m/s,what is the ac
noname [10]

Answer:

0.3125m/s

Explanation:

3 0
3 years ago
The gravitational potential energy of a particle of mass m moving under the influence of a fixed mass M is given by - , where G
djverab [1.8K]

-GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.

Given

A particle of mass m moving under the influence of a fixed mass's M, gravitational potential energy of formula  -GMm/r, where r is the separation between the masses and G is the gravitational constant of the universe.

As the Gravity Potential energy of particle = -GMm/r

Total energy of particle = Kinetic energy + Potential Energy

As we know that

Kinetic energy = 1/2mv²

Also, v is equals to square root of GM/r

v = √GM/r

Put the value of v in the formula of kinetic energy

We get,

Kinetic Energy = GMm/2r

Total Energy = GMm/2r + (-GMm/r)

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                     = -GMm/2r

Hence, -GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.

Learn more about Gravitational Potential Energy here brainly.com/question/15896499

#SPJ4

3 0
1 year ago
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