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3 years ago

15

24- Which of the fallowing devices make use a lens

2 answers:

mezya [45]3 years ago

6 0

Answer:

d.

Explanation:

all because the projector is a lens and the magnifying also the microscope

Luda [366]3 years ago

5 0

Answer would be D because they all use it

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Which volcanic hazard can block the sunlight and temporarily cool the Earth’s surface?

BaLLatris [955]

If a volcano epulses massive amounts of dust into the atmosphere, those two things will/can happen.
The events will last until the dust lays down on the earth.

8 0

3 years ago

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A car travels north for 20 km. The car then travels south for 35 km.

PSYCHO15rus [73]

Answer: -15 km

Explanation:

8 0

3 years ago

Charge q is 1 unit of distance away from the source charge S. Charge p is two times further away. The force exerted between S an

insens350 [35]

Answer : The correct option is, (d) 4 times

Solution :

According to the Coulomb's law, the electrostatic force of attraction or repulsion between two charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between the the charges.

Formula used :

$F=k_e\frac{q_1q_2}{r^2}$

where,

F = electrostatic force of attraction or repulsion

$k_e$ = Coulomb's constant

$q_1$ and $q_2$ are the charges

r = distance between two charges

First we have to calculate the force exerted between S and q when the distance between the charge is 1 unit and let us assumed that the charge be 'q'

$F_{sq}=k_e\frac{qq}{1^2}=k_e\times q^2$ ..........(1)

Now we have to calculate the force exerted between S and p when the distance between the charge is 2 unit at the same charge.

$F_{sp}=k_e\frac{qq}{2^2}=k_e\frac{q^2}{4}$ ...........(2)

Equation equation 1 and 2, we get

$\frac{F_{sq}}{F_{sp}}=\frac{1}{4}$

$F_{sq}=4\times F_{sp}$

Therefore, the force exerted between S and q is 4 times the force exerted between S and p.

5 0

3 years ago

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8) Find the X and Y component of 10degree vector that has 5N.

Deffense [45]

Answer:

Fx = 4.92 [N]

Fy = 0.868 [N]

Explanation:

Let's take the 10 degrees as a measure from the horizontal component to the vector.

Thus taking the components in the X & y axes respectively:

Fx = 5*cos(10) = 4.92 [N]

Fy = 5*sin(10) = 0.868 [N]

3 0

3 years ago

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