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Tomtit [17]
4 years ago
12

Object A weighs 750 N on earth. Object B weighs 750 N on Jupiter.

Physics
1 answer:
Katarina [22]4 years ago
5 0

The object A has the greater mass compared to object B.

<u>Explanation: </u>

The weight of any object on any planet is the measurement of gravity’s influence acting on the mass of the object. So for Earth, the acceleration will be acting on the object A’s mass (m) in Earth leading to the weight of the object A as 750 N.

While the acceleration of Jupiter will be acting on the object B’s mass kept in Jupiter to attain the weight of 750 N. So, the mass of both the objects at their respective planet will vary depending on the acceleration of each planet. We can check this as below:

            \text { object A's weight }=\text { m of object A } \times \text { Acceleration of Earth}

So,

           750 \mathrm{N}=\text { m of object } A \times 9.8 \mathrm{m} / \mathrm{s}^{2}

Thus,

          \text { m of object } A=\frac{750}{9.8}=76.5 \mathrm{kg}

Similarly for object B,

          \text { Weight of object } B=m \text { of object } B \times \text { Acceleration due to gravity of Jupiter }

          750 = m \text { of object } B \times 24.79 \mathrm{m} / \mathrm{s}^{2}

Thus,

          \text { m of the object } B=\frac{750}{24.79}=30.25 \mathrm{kg}

Thus, the mass of object A is greater than the mass of object B.

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A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface
givi [52]

Answer:

a = 0.009 J

b = 0.19 m/s

c = 0.005 J and 0.004 J

Explanation:

Given that

Mass of the object, m = 0.5 kg

Spring constant of the spring, k = 20 N/m

Amplitude of the motion, A = 3 cm = 0.03 m

Displacement of the system, x = 2 cm = 0.02 m

a

Total energy of the system, E =

E = 1/2 * k * A²

E = 1/2 * 20 * 0.03²

E = 10 * 0.0009

E = 0.009 J

b

E = 1/2 * k * A² = 1/2 * m * v(max)²

1/2 * m * v(max)² = 0.009

1/2 * 0.5 * v(max)² = 0.009

v(max)² = 0.009 * 2/0.5

v(max)² = 0.018 / 0.5

v(max)² = 0.036

v(max) = √0.036

v(max) = 0.19 m/s

c

V = ±√[(k/m) * (A² - x²)]

V = ±√[(20/0.5) * (0.03² - 0.02²)]

V = ±√(40 * 0.0005)

V = ±√0.02

V = ±0.141 m/s

Kinetic Energy, K = 1/2 * m * v²

K = 1/2 * 0.5 * 0.141²

K = 1/4 * 0.02

K = 0.005 J

Potential Energy, P = 1/2 * k * x²

P = 1/2 * 20 * 0.02²

P = 10 * 0.0004

P = 0.004 J

4 0
3 years ago
What property do liquids and gases share
vodka [1.7K]

Answer:

A is correct

Explanation:

6 0
3 years ago
Consider the following three statements: (i) For any electro-magnetic radiation, the product of the wavelength and the frequency
Scilla [17]

Answer:

A and B

Explanation:

The relation between frequency and wavelength is shown below as:

c=frequency\times Wavelength

c is the speed of light having value 3\times 10^8\ m/s

Thus, the product of the wavelength and the frequency is constant and equal to 3\times 10^8\ m/s

<u>Option A is correct.</u>

Given, Frequency = 1\times 10^{18}\ Hz

Thus, Wavelength is:

Wavelength=\frac{c}{Frequency}

Wavelength=\frac{3\times 10^8}{1\times 10^{18}}\ m

Wavelength=3\times 10^{-10}\ m

Also, 1 m = 3\times 10^{-10} Å

So,

<u>Wavelength = 3.0 Å</u>

<u>Option B is correct.</u>

As stated above, the speed of electromagnetic radiation is constant. Hence, each radiation of the spectrum travels with same speed.

<u>Option C is incorrect.</u>

3 0
3 years ago
Read 2 more answers
A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent
azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

  • <em>Length of the string, L = 100 cm</em>

<em />

The wavelengths of the constituent travelling waves is calculated as follows;

L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}

for first mode: n = 1

\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm

for second mode: n = 2

\lambda = \frac{2L}{2} = L = 100 \ cm

For the third mode: n = 3

\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm

For fourth mode: n = 4

\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50  \ cm

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

5 0
3 years ago
As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i
11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

Explanation:

Given that,

Height = 1.94  m

Bounced height = 1.48 m

Time interval t=14.86\times10^{-3}\ s

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

9.8\times1.94=\dfrac{1}{2}\times v^2

v=\sqrt{2\times9.8\times1.94}

v=6.166\ m/s

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2'

Put the value into the formula

9.8\times1.48=\dfrac{1}{2}\times v^2'

v'=\sqrt{2\times9.8\times1.48}

v'=5.38\ m/s

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

a=\dfrac{v-v'}{t}

Put the value into the formula

a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}

a=776.98\ m/s^2

a=0.77\times10^{3}\ m/s^2

Hence,The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

6 0
3 years ago
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