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3 years ago

An object is weighed at different locations on the

Physics

2 answers:

Step2247 [10]3 years ago

6 0

Answer:

Explanation:

The weight will always be different while mass is described as the stuff inside an object, and that stays the same.

Such as it weighs differently in space.

creativ13 [48]3 years ago

4 0

Answer:

C. The mass is constant while the weight varies

Explanation:

happy to help ya! :)

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A 62.2-kg person, running horizontally with a velocity of 3.80 m/s, jumps onto a 19.7-kg sled that is initially at rest. (a) Ign

jek_recluse [69]

Answer:

a) $v = 2.886\,\frac{m}{s}$ , b) $\mu_{k} = 0.014$

Explanation:

a) The final speed is determined by the Principle of Momentum Conservation:

$(62.2\,kg)\cdot (3.80\,\frac{m}{s} ) = (81.9\,kg)\cdot v$

$v = 2.886\,\frac{m}{s}$

b) The deceleration experimented by the system person-sled is:

$a = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(2.886\,\frac{m}{s} \right)^{2}}{2\cdot (30\,m)}$

$a = -0.139\,\frac{m}{s^{2}}$

By using the Newton's Laws, the only force acting on the motion of the system is the friction between snow and sled. The kinetic coefficient of friction is:

$-\mu_{k}\cdot m\cdot g = m\cdot a$

$\mu_{k} = -\frac{a}{g}$

$\mu_{k} = -\frac{\left(-0.139\,\frac{m}{s^{2}} \right)}{9.807\,\frac{m}{s^{2}} }$

$\mu_{k} = 0.014$

3 0

3 years ago

If you have 240.0 mL of water at 25.00 °C and add 100.0 mL of water at 95.00 °C, what is the final temperature of the mixture? U

Rus_ich [418]

Answer: The final temperature of the mixture is 45.6°C

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

When volume is 240.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{240.0mL}\\\\\text{Mass of water}=(1g/mL\times 240.0mL)=240g$

When volume is 100.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{100.0mL}\\\\\text{Mass of water}=(1g/mL\times 100.0mL)=100g$

When two water solutions at different temperature are mixed, the amount of heat released by water present at higher temperature will be equal to the amount of heat absorbed by water present at lower temperature..

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of water solution 1 = 240 g

$m_2$ = mass of water solution 2 = 100 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of water solution 1 = 25°C

$T_2$ = initial temperature of water solution 2 = 95°C

c = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

$240\times 4.186\times (T_{final}-25)=-[100\times 4.186\times (T_{final}-95)]\\\\T_{final}=45.6^oC$

Hence, the final temperature of the mixture is 45.6°C

7 0

3 years ago

Read 2 more answers

laser light hits two very narrow slits that are separated by 0.1mm adn is viewwed on a screen 2m downstream. Sketch on the axis

Irina-Kira [14]

Answer:

d y / x = m λ

Explanation:

When the laser beam, which is a coherent light, hits the slits, part of each beam passes through each slit.

When this is observed on a screen that is quite far from the slits, a series of intense linear separated by dark areas. The explanation for this distribution of the light pattern is that when adding the rays that come out of the slits they travel different distances, which introduces a difference in optical path and if this difference is an integer multiple of the wavelength, a bright line

d sin θ = m λ

Where d is the distance between the slits (0.1 mm)

Also, since the angle of the measurements is small, we can approximate the tangent

tan θ = y / x = sin θ /sin θ

sint θ = y / x

Substituting into the equation

d y / x = m λ

This expression gives the location of the bright lines on the screen

5 0

3 years ago

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During the deceleration of an ascending elevator, the normal force on the feet of a passenger is _____ her weight. During the de

gizmo_the_mogwai [7]

Answer: Smaller than ; larger than

Explanation:

When the elevator is moving in the upward direction, then the force acting on it is negative in nature because of

N= mg +ma, (g is gravity and a is acceleration)

here ma is negative so the N= mg-ma

Hence, it feels smaller than its original weight.

When the elevator is moving downward , then the force acting will be positive in nature

N= mg+ma,

here ma will be positive so it feels larger the original weight of passenger.

7 0

3 years ago

Which motion listed below matches the graph?

mote1985 [20]

Bro I really think it might be c

7 0

3 years ago

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