All categories

3 years ago

Which of these would make the best telescope?

Physics

1 answer:

Simora [160]3 years ago

5 0

Answer:A i think or D but its not c or b

Explanation:

You might be interested in

"Two identical positive charges exert a repulsive force of 6.7 × 10−9 N when separated by a distance 3.5 × 10−10 m. Calculate th

Tamiku [17]

Answer:

The charge of each charge is $3.02\times10^{-19} C$

Explanation:

When yo have two charged particles they interact exerting an electrostatic force in the other particles, the magnitude of the electrostatic force between two particles is:

$F_{e}=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}$ (1)

with q1 and q2 the charges, r the distance between them and k the Coulomb's constant ( $k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}$ )

Because the charges we're dealing are identical positive q1=q2, then (1) is:

$F_{e}=k\frac{\mid q^2 \mid}{r^{2}}$

Using the values the problem give us:

$6.7\times10^{-9}=8.98755\times10^{9}\frac{\mid q^2 \mid}{(3.5\times10^{-10})^{2}}$

solving for q:

$q=\sqrt{\frac{(3.5\times10^{-10})^{2}6.7\times10^{-9}}{8.98755\times10^{9}}}$

$q= 3.02\times10^{-19} C$

8 0

3 years ago

If an object has a mass of 1417g and it is moved 47 meters in 90 seconds, how much power was used?

astraxan [27]

Mass of the object is given as

$m = 1417 g = 1.417 kg$

now the speed of object is given as

$v = \frac{d}{t}$

here we know that

$d = 47 m$

$t = 90 s$

now we will have

$v = \frac{47}{90} = 0.52 m/s$

now we will have kinetic energy of the object as

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(1.417)(0.52)^2$

$KE = 0.19 J$

now the power is defined as rate of energy

so here we can find power as

$P = \frac{KE}{t}$

$P = \frac{0.19}{90} = 2.14\times 10^{-3} W$

so above is the power used for the object

3 0

4 years ago

A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0

4 years ago

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

5 0

4 years ago

Three fire hoses are connected to a fire hydrant. Each hose has a radius of 0.014 m. Water enters the hydrant through an undergr

Makovka662 [10]

Answer:

Explanation:

The total fluid mass can be obtained by multiplying the mass flow rate by the time flow rate.

Mass flow rate is given as

m = ρAv

Where

m is mass flow rate

ρ is density

A is area and it is given as πr²

v is velocity

Then,

M = mt

Where M is mass and t is time

Them,

M = ρAv × t

M = ρ× πr² × v × t

Given that, .

Radius of pipe is

r = 0.089m

velocity of pipe is

v = 3.3m/s

Time taken is

t = 1 hour = 3600 seconds

Density of water is

ρ = 1000kg/m³

M = ρ× πr² × v × t

M = 1000 × π × 0.089² × 3.3 × 3600

M = 295,628.52 kg

M = 2.96 × 10^5 kg

8 0

4 years ago