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3 years ago

7

How much of earths gravitational force acts on everyone and everything onboard of the international space station?

1 answer:

liraira [26]3 years ago

4 0

B) Approx. 90%

The actual value is 0.89g

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when an object is charged by contact, what kind of charge does the object have compared with the charge on the on the object giv

sveta [45]

Answer:

the charge that is given by the object is positive charge and the object which is taking the charge is negetively charged

Explanation:

5 0

3 years ago

A bird flies 2 km to the south then 0.2 km to the north . What is the displacement of the bird?

lidiya [134]

The displacement of the bird is 1.8 km...because displacement is the distanse of the object from mean position to the travelled position.....hope it helps

8 0

2 years ago

Read 2 more answers

In which of these situations would there be the least kinetic energy?

Alexandra [31]

C. a truck rolling down hill

7 0

2 years ago

Linear Thermal Expansion (in one dimension)

Leokris [45]

Answer:

1) $\Delta L= 0.612\ m$

2) a. $\Delta V_G=0.57\ L$

b. $\Delta V_S=0.021\ L$

c. $V_0=0.549\ L$

Explanation:

1)

given initial length, $L=1275\ m$

initial temperature, $T_i=-15^{\circ}C$

final temperature, $T_f=25^{\circ}C$

coefficient of linear expansion, $\alpha=12\times 10^{-6}\ ^{\circ}C^{-1}$

<u>∴Change in temperature:</u>

$\Delta T=T_f-T_i$

$\Delta T=25-(-15)$

$\Delta T=40^{\circ}C$

We have the equation for change in length as:

$\Delta L= L.\alpha. \Delta T$

$\Delta L= 1275\times 12\times 10^{-6}\times 40$

$\Delta L= 0.612\ m$

2)

Given relation:

$\Delta V=V.\beta.\Delta T$

where:

$\Delta V$ = change in volume

V= initial volume

$\Delta T$ =change in temperature

initial volume of tank, $V_{Si}=60\ L$

initial volume of gasoline, $V_{Gi}=60\ L$

initial temperature of steel tank, $T_{Si}=15^{\circ}C$

initial temperature of gasoline, $T_{Gi}=15^{\circ}C$

coefficients of volumetric expansion for gasoline, $\beta_G=950\times 10^{-6}\ ^{\circ}C$

coefficients of volumetric expansion for gasoline, $\beta_S=35\times 10^{-6}\ ^{\circ}C$

a)

final temperature of gasoline, $T_{Gf}=25^{\circ}C$

∴Change in temperature of gasoline,

$\Delta T_G=T_{Gf}-T_{Gi}$

$\Delta T_G=25-15$

$\Delta T_G=10^{\circ}C$

Now,

$\Delta V_G= V_G.\beta_G.\Delta T_G$

$\Delta V_G=60\times 950\times 10^{-6}\times 10$

$\Delta V_G=0.57\ L$

b)

final temperature of tank, $T_{Sf}=25^{\circ}C$

∴Change in temperature of tank,

$\Delta T_S=T_{Sf}-T_{Si}$

$\Delta T_S=25-15$

$\Delta T_S=10^{\circ}C$

Now,

$\Delta V_S= V_S.\beta_S.\Delta T_S$

$\Delta V_S=60\times 35\times 10^{-6}\times 10$

$\Delta V_S=0.021\ L$

c)

Quantity of gasoline spilled after the given temperature change:

$V_0=\Delta V_G-\Delta V_S$

$V_0=0.57-0.021$

$V_0=0.549\ L$

8 0

2 years ago

An aluminum sphere (specific gravity = 2.70) falling through water reaches a terminal speed of 4.58 cm/s. What is the terminal s

enyata [817]

Answer:2.69 cm/s

Explanation:

We know that drag force

Fd=6 πμ r v

μ=Dynamic viscosity

r=radius

v=terminal velocity

Bouncy force

Fb= ρ V g

$rho_w=density\ of\ water$

V=volume

At the condition of terminal velocity

when going down ( aluminum)

$Fb+ Fd= m g$

$Fd_1 = m g - Fb_1$ ---------1

when going up ( air)

$Fb_2= mg +Fd$

$Fd_2= Fb_2 - mg$ -----------2

m =mass of object

$m= density\times volume$

$\rho_a=density\ of\ Aluminium$

$\rho _b=density\ of\ bubble$

Divide 1 and 2

$\dfrac{v_1}{v_2}=\dfrac{mg-F_b_1}{F_b_2-mg}$

$\dfrac{4.58}{v_2}=\dfrac{\rho _a-\rho _w}{\rho _w-\rho _b}$

divide by $\rho _w$ in R.H.S

$\dfrac{4.58}{v_2}=\dfrac{2.7-1}{1.-0.0012}$

$v_2=\dfrac{4.58}{1.7}=2.69\ cm/s$

3 0

2 years ago