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3 years ago

Question 1 2.5 cm= mm

Physics

2 answers:

Nookie1986 [14]3 years ago

8 0

There’s 10mm in a cm: 22mm

Elden [556K]3 years ago

4 0

2.5cm=25mm

i hope this helps!

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Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0

BaLLatris [955]

Answer:

Part a)

$r = 6.96 \times 10^6 m$

Part b)

$F_g = 3.004 \times 10^3 N$

Part c)

$a = 0.88 m/s^2$

Part d)

$v = \frac{2480}{2} = 1240 m/s$

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

$v = 2480 m/s$

now we will have

$v = \sqrt{\frac{GM_{mars}}{r}}$

now we have

$M_{mars} = 6.4191 \times 10^{23} kg$

$R_{mars} = 3.397 \times 10^6 m$

now we have

$2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}$

$r = 6.96 \times 10^6 m$

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

$F_g = \frac{GM_{mars} m}{r^2}$

$F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}$

$F_g = 3.004 \times 10^3 N$

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

$a = \frac{F_g}{m}$

$a = \frac{3004}{3400}$

$a = 0.88 m/s^2$

Part d)

As we know by III law of kepler

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

here we know that T2 = 8 T1

$(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}$

$\frac{r_1}{r_2} = (\frac{1}{2})^2$

so we have

$r_2 = 4r_1$

as we know that speed is given as

$v = \sqrt{\frac{GM}{r}}$

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

$v = \frac{2480}{2} = 1240 m/s$

7 0

4 years ago

Calculate the internal axial load at a point D if length L=7 ft. The part is subjected to loads P1=632 lbs, P2=888 lbs (applied

liraira [26]

Answer:

- 256 lbs

Explanation:

The internal axial load at point D can be calculated as the change in the subjected loads. if the magnitude of the horizontal direction = zero

$EF_x = 0$ ; Then:

internal axial load at point D = Δ P

= -(P₂ - P₁)

= - ( 888 lbs - 632 lbs)

= - 256 lbs

5 0

3 years ago

In an experiment, it took 300s to increase the temperature of 0.1kg of the liquid from 25°C to 50°C. During this period, the ene

klemol [59]

Answer:

<h2>4000 $\textbf{J}\text{ }\textbf{kg}^{\textbf{-1}}\text{ }\textbf{K}^{\textbf{-1}}$ </h2>

Explanation:

The temperature of 0.1 kg of liquid rises from 25°C to 50°C in 300 sec. Energy of 13,600 J was supplied during this time. Appartus was losing energy at the rate of 12 J/sec.

Let us assume the Specific heat capacity as $s$ .

As there is no state change from liquid to gas, only Specific heat capacity is involved. Also, work done is approximately zero because volume does not change much. So,

Energy gained = Energy required to rise the temperature

Energy gained by liquid = $13600\text{ }J-(300\text{ }sec)\times(12\text{ }\frac{J}{sec})=10000\text{ }J$

$10000\text{ }J=m\times s\times\Delta T=(0.1\text{ }kg)\times(s)\times(323K-298K)=2.5s\text{ }kgK\\s=4000\text{ }\frac{J}{kg.K}$

∴ Specific heat capacity of liquid = 4000 $\frac{J}{kg.K}$

4 0

3 years ago

1- Write Gauss’ Law.

Alecsey [184]

1)The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field.

4 0

4 years ago

1.) What is the process called when a liquid changes to a

nevsk [136]

Answer:

1.)Evaporation, 2.) solid

Explanation:

hope this helps can i get brainliest

8 0

3 years ago