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3 years ago

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• How much work isrequired to lift a 2kgobject 2m high?

1 answer:

pychu [463]3 years ago

7 0

Answer You need to consider that the gravity on earth is 9.8 m/s/s. This means any object you let go on the earths surface will gain 9.8 m/s of speed every second. You need to apply a force on the object in the opposite direction to avoid this acceleration. If you are pushing something up at a constant speed, you are just resisting earths acceleration. The more massive and object is, the greater force is needed to accelerate it. The equation is Force = mass*acceleration. So for a 2kg object in a 9.8 m/s/s gravity you need 2kg*9.8m/s/s = 19.6 Newtons to counteract gravity. Work or energy = force * distance. So to push with 19.6 N over a distance of 2 meters = 19.6 N*2 m = 39.2 Joules of energy. There is an equation that puts together those two equations I just used and it is E = mgh

The amount of Energy to lift an object is (mass) * (acceleration due to gravity) * (height)

:Hence, the Work done to life the mass of 2 kg to a height of 10 m is 196 J. Hope it helps❤️❤️❤️

Explanation:

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Light of a given wavelength is used to illuminate the surfacce of a metal, however, no photoelectrons are emitted. In order to c

Jlenok [28]

Answer:

B. use light of a shorter wavelength.

Explanation:

We know that

$E= \frac{hc}{\lambda}$

h= plank's constant

c= speed of light

λ= wavelength of the incident light

so, in order to have sufficient energy for for the emission of electron, the incident's radiation must have λ small enough.

B. use light of a shorter wavelength.

6 0

3 years ago

RC time constant circuit if R 50 KOC-21 a TOSS c. 1.05 s . what is the expected RC value b. 10.55 d. 0.105 s

Afina-wow [57]

Answer:

Time constant of RC circuit is 0.105 seconds.

Explanation:

It is given that,

Resistance, $R=50\ K\Omega=5\times 10^4\ \Omega$

Capacitance, $C=2.1\ \mu F=2.1\times 10^{-6}\ F$

We need to find the expected time constant for this RC circuit. It can be calculated as :

$\tau=R\times C$

$\tau=5\times 10^4\times 2.1\times 10^{-6}$

$\tau=0.105\ s$

So, the time constant for this RC circuit is 0.105 seconds. Hence, this is the required solution.

7 0

3 years ago

A 31.7 kg kid initially at rest slides down a frictionless water slide at 53.2 degrees, how fast is she moving in 3.45 s later?

Murrr4er [49]

Answer:

34.55 m/s

Explanation:

8 0

3 years ago

On the sonometer shown below, a horizontal cord of length 5 m has a mass of 1.45 g. When the cord was plucked the wave produced

Korolek [52]

Answer:

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

$v =f\lambda$

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

$\mu = \frac{m}{l}$

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

$\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m$

Now, for the tension we use the formula:

$v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T$

<u>T = 0.015 N</u>

<u></u>

(b)

The mass to be hung is:

$T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\$

<u>M = 1.53 x 10⁻³ kg = 1.53 g</u>

4 0

3 years ago

PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

horsena [70]

It will be cloudy and there will be rain.

Hope this helps

5 0

3 years ago

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