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Debora [2.8K]
2 years ago
15

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.
Physics
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
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