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2 years ago

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A man who weighs 75kg on the surface of the earth whose mass is 6*10^24kg. If the radius of earth is 6480km ,calculate the force

of attraction between them.

1 answer:

Fiesta28 [93]2 years ago

5 0

Answer:

718.02N is the answer to the question

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A block of density pb = 9.50 times 10^2 kg/m^3 floats face down in a fluid of density pt = 1.30 times 10^3 kg/m^3. The block has

Nookie1986 [14]

Answer:

The depth and acceleration are 0.1919291 ft and 3.61 m/s².

Explanation:

Given that,

Density of block $\rho_{b} =9.50\times10^2\ kg/m^3$

Density of fluid $\rho_{t} =1.30\times10^3\ kg/m^3$

We need to calculate the depth

Using balance equation

$mg=\rho g V$ ....(I)

We know that,

The density is

$\rho=\dfrac{m}{V}$

$m=\rh0\times V$

Put the value of m in equation (I)

$\rho_{b}\times V\times g=\rho_{t}\times g\times V$

$\rho_{b}\times A\times H\times g=\rho_{t}\times g\times A\times h$

$h=\dfrac{\rho_{b}H}{\rho_{t}}$

Put the value into the formula

$h=\dfrac{9.50\times10^2\times8.00\times10^{-2}}{1.30\times10^3}$

$h= 5.85\ cm$

$h=0.1919291\ ft$

We need to calculate the acceleration

Using formula of net force

$F_{net}=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$ma=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$\rho_{b}\times V\times a=\rho_{t}\times g\times V- \rho_{b}\times g\times V$

$a=(\dfrac{\rho_{t}}{\rho_{b}}-1)g$ ....(II)

Put the value in the equation (II)

$a=(\dfrac{1.30\times10^3}{9.50\times10^2}-1)\times9.8$

$a=3.61\ m/s^2$

Hence, The depth and acceleration are 0.1919291 ft and 3.61 m/s².

4 0

2 years ago

Cellus<br> Find the x-component of this<br> vector:<br> 92.5 m<br> 32.0

Mazyrski [523]

Explanation:

x-component:

Vx = Vcos(theta)

= (92.5 m)cos(32.0)

= 78.4 m

5 0

2 years ago

The table below compares the cost of living in Philadelphia

Feliz [49]

It should be A sorry if I’m wrong

6 0

2 years ago

Read 2 more answers

When a battery, a resistor, a switch, and an inductor form a circuit and the switch is closed, the inductor acts to oppose the c

vladimir1956 [14]

Answer:Time constant gets doubled

Explanation:

Given

L-R circuit is given and suppose R and L is the resistance and inductance of the circuit then current is given by

$i=i_0\left [ 1-e^{-\frac{t}{\tau }}\right ]$

where $i_0$ is maximum current

i=current at any time

$\tau =\frac{L}{R}=time\ constant$

$\tau '=\frac{2L}{R}=2\tau$

thus if inductance is doubled then time constant also gets doubled or twice to its original value.

5 0

3 years ago

Racing cars driven by chris and kelly are side by side at the start of a race. the table shows the velocities of each car (in mi

Mamont248 [21]

Solution

distance travelled by Chris

\Delta t=\frac{1}{3600}hr.

X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}

=\frac{579.5}{3600}=0.161miles

Kelly,

\Delta t=\frac{1}{3600}hr.

X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}

=\frac{657.5}{3600}

\Delta X=X_{k}-X_{C}=0.021miles

4 0

3 years ago