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2 years ago

How much power is needed to lift a 20-kg object to a height of 4.0 m in 2.5 seconds?

Physics

1 answer:

4vir4ik [10]2 years ago

5 0

Answer:

313.92w

Explanation:

Formula for power:

P=W/∆t = Fv

Givens:

m=20kg

∆y=4.0m

∆t=2.5s

a=9.81m/s²

In order to find power, we first need to solve for work.

W=Fd (force*displacement), f=mg

W=mg∆y

W=(20kg)(9.81m/s²)(4.0m)

W=784.8J

P=W/∆t

P=784.8J/2.5s

P=313.92 watts

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Find the uniform acceleration that causes a car's velocity to change from 20.0 m/s to 105 m/s in and 12.0 s

Sauron [17]

Answer:

7.08 m/s²

Explanation:

Given:

v₀ = 20.0 m/s

v = 105 m/s

t = 12.0 s

Find: a

v = at + v₀

105 m/s = a (12.0 s) + 20.0 m/s

a = 7.08 m/s²

5 0

3 years ago

Maurice pulls on the end of a spring scale. He lets go of the end and observes the spring snap back into place. What force resto

andreev551 [17]

Elastic is the right answer.

3 0

3 years ago

Read 2 more answers

Which SI unit is the correctly abbreviated for describing the mass on an object.

djverab [1.8K]

S.I. Unit of mass is Kilogram which is denoted by Kg

In short, Your Answer would be Option C

Hope this helps!

4 0

3 years ago

There are four springs stretched by the same mass.

brilliants [131]

Spring C stretches 100 cm.

Explanation:

The spring constant is simply the stiffness of the spring. The higher the spring constant the more stiff the spring is.

Spring constant shows the force needed to stretch a spring from it's equilibrium position. If a material requires more force to cause it to stretch, it will have a high spring constant.

According to hooke's law "the force needed to extended an elastic material is directly proportional to its extension"

F = ke

k is the spring constant

e is the extension

We see that the spring that stretches by 100 is the less stiff compared to other springs. It has the smallest spring constant.

Learn more;

Force brainly.com/question/8882476

#learnwithBrainly

8 0

2 years ago

Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun

iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

$E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})$ (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

$E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV$

B. The energy of the emitted photon is given by the following formula:

$E=h\frac{c}{\lambda}$ (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

$2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J$

Next, you use the equation (2) and solve for λ:

$\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm$

C. The radius of the orbit is given by:

$r_n=n^2a_o$ (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

$r_5=5^2(2.380)=59.5$

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0

3 years ago