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julsineya [31]
4 years ago
7

in which activity is no work done? A. pushing a shopping cart B. climbing stairs C. holding a brick D. lifting a book

Physics
1 answer:
Vesnalui [34]4 years ago
3 0
Your answer is c holding a brick doesn't contain movement, but energy to grip on it.

hoped it helped!!!
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I think the answer is a
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4 years ago
A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and
irina [24]

The eccentricity of its orbit is $$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

<h3>What is mass?</h3>
  • Mass is a physical body's total amount of matter. It also serves as a gauge for the body's inertia or resistance to acceleration (change in velocity) in the presence of a net force. The strength of an object's gravitational pull to other bodies is also influenced by its mass.
  • The kilogram is the SI unit of mass (kg). In science and technology, a body's weight in a given reference frame is the force that causes it to accelerate at a rate equal to the local acceleration of free fall in that frame.
  • For instance, a kilogram mass weighs around 2.2 pounds at the surface of the planet. However, the same kilogram mass would weigh just about 0.8 pounds on Mars and about 5.5 pounds on Jupiter.
  • An object's mass is a crucial indicator of how much stuff it contains. Weight is a measurement of an object's gravitational pull. It is influenced by the object's location in addition to its mass. As a result, weight is a measurement of force.

The length of the semi-major axis is calculated as follows:

where, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$

$M=1.99 \times 10^{30} \mathrm{~kg}=$mass of sur

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

$$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

To learn more about mass, refer to:

brainly.com/question/3187640

#SPJ4

4 0
2 years ago
Which type of electromagnetic radiation is most likely to be ionizing
zlopas [31]

Answer:

Xrays. i hope that helps

Explanation:

4 0
3 years ago
Read 2 more answers
I need help with this pls
enyata [817]
Friction makes a moving object slower
8 0
3 years ago
A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im
vfiekz [6]

Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}

\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}

\dfrac{1}{v}=\dfrac{7}{690}

v=98.57\ cm

We need to calculate the height of the image

Using formula of height

\dfrac{h'}{h}=\dfrac{-v}{u}

Where, h' = height of image

h = height of object

\dfrac{h'}{4}=\dfrac{-98.57}{-30}

h'=\dfrac{98.57\times4}{30}

h'=13.14\ cm

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}

\dfrac{1}{v}=-\dfrac{10}{299}

v=-29.9\ cm

We need to calculate the height of the image

\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}

h=-\dfrac{29.9\times4}{13.0}

h=-9.2\ cm

The image is virtual and erect.

Hence, This is required solution.

6 0
3 years ago
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