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Novosadov [1.4K]

2 years ago

13

You are riding in an elevator that is accelerating upward. Suppose you stand on a scale. The reading on the scale is __________.

2 answers:

enyata [817]2 years ago

6 0

Less than true weight

Darina [25.2K]2 years ago

5 0

Answer: A

Explanation: Your body gets pulled down, causing ur body to have more weight but not mass.

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A hot metal bolt of mass 0.06 kg and specific heat capacity 899 J/kg°C is dropped into a calorimeter containing 0.16 kg water at

skad [1K]

Answer:

Explanation:

Given

mass of hot metal $m_1=0.06\ kg$

specific heat $c=889\ J/kg-^{\circ}C$

mass of water $m_2=0.16\ kg$

Temperature pf water $T_2=20^{\circ}C$

Final Temperature $T=25^{\circ}C$

let $T_1$ be the temperature of hot metal ball

Heat lost by heat metal bolt is gained by water in calorimeter

Heat lost by hot metal bolt $Q_1=m\times c\times \Delta T$

$Q_1=0.06\times 889\times (T-25)$

Heat gained by water $Q_2=m_2\times c_w\times \Delta T$

$Q_2=0.16\times 4184\times (25-20)$

$Q_1=Q_2$

$0.06\times 889\times (T-25)=0.16\times 4184\times (25-20)$

$T_1=25+62.75$

$T_1=87.75^{\circ}C$

7 0

2 years ago

What is the magnetic flux linkage, in units of Weber, for a coil of 360 turns and cross sectional area of 0.133 m^2 when the mag

zalisa [80]

Answer:

83.3 Wb

Explanation:

The magnetic flux linkage through the coil is given by:

$N\phi = BAN sin \theta$

where

B is the magnetic field strength

A is the cross sectional area

N is the number of turns in the coil

$\theta$ is the angle between the direction of the field and the normal to the coil

In this problem:

B = 1.74 T

A = 0.133 m^2

N = 360

$\theta=90^{\circ}$

Therefore, the magnetic flux linkage is

$N\phi = (1.74 T)(0.133 m^2)(360) sin 90^{\circ}=83.3 Wb$

7 0

3 years ago

You are driving due north on i-81 to come to jmu with a speed of 10 m/s, suddenly you realize you forgot your book. You make a u

Readme [11.4K]

first we make a U turn and travel towards home in t = 20 s

so the distance of home from initial position is

$d_1 = v*t_1$

$d_1 = 10*20 = 200 m$

Now after picking up the book we travel back with speed 20 m/s

so again after t = 20 s the displacement is given as

$d_2 = v*t = 20*20 = 400 m$

so the net displacement is given as

$\vec d = \vec d_2 - \vec d_1$

$\vec d = 400 - 200 = 200 m$

so it will be displaced by total displacement 200 m

8 0

3 years ago

Rita conducts an experiment on how the amount of precipitation each fall affects

hodyreva [135]

Answer:

Explanation:

As spring season is a yearly phenomenon so, Rita should organize her data on yearly basis. Firstly, she should plan the procedure of her experiment and collect the data according to it. Secondly, identify the attribute of each object of her experiment. Thirdly, she can organize and segregate her data in tabular form, graphical form or diagrammatically.

5 0

3 years ago

The Sun has a mass of 1.99x10^30 kg and a radius of 6.96x10^8 m. Calculate the acceleration due to gravity, in meters per second

just olya [345]

Answer:

$g=274\ m/s^2$

Explanation:

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

The radius of the Sun, $r=6.96\times 10^8\ m$

We need to find the acceleration due to gravity on the surface of the Sun. It is given by the formula as follows :

$g=\dfrac{GM}{r^2}\\\\g=\dfrac{6.67\times 10^{-11}\times 1.99\times 10^{30}}{(6.96\times 10^8)^2}\\\\g=274\ m/s^2$

So, the value of acceleration due to gravity on the Sun is $274\ m/s^2$ .

8 0

2 years ago