Answer:
[tex]E_{net} = (3.765\hat i - 1.207\hat j)kN/C[/tex]
Explanation:
Electric field due to long line charge on position of charge at x = 4 m is given as
[tex]E = \frac{2k\lambda}{r} \hat i[/tex]
so we have
[tex]\lambda = 0.30 \mu C/m[/tex]
now we have
[tex]E = \frac{2(9\times 10^9)(0.30 \mu C/m)}{4}[/tex]
[tex]E = 1350 N/C[/tex]
Now electric field due to the charge present at y = 2.0 m
[tex]E = \frac{kq}{r^2} \hat r[/tex]
[tex]E = \frac{(9\times 10^9)(6 \times 10^{-6})}{2^2 + 4^2}\times \frac{4\hat i - 2\hat j}{\sqrt{4^2 + 2^2}}[/tex]
[tex]E = 603.7 ( 4\hat i - 2\hat j)[/tex]
[tex]E = 2415\hat i - 1207.5 \hat j[/tex]
Now total electric field is given as
[tex]E_{net} = (1350\hat i) + (2415\hat i -1207.5\hat j)[/tex]
[tex]E_{net} = 3765\hat i - 1207.5 \hat j[/tex]
[tex]E_{net} = (3.765\hat i - 1.207\hat j)kN/C[/tex]
The sun is 1.5 × 108 km from Earth. The index of refraction for water is 1.349. How much longer would it take light from the sun to reach Earth if the space between them were filled with water rather than a vacuum?
Answer:
175s
Explanation:
time it takes sunlight to reach the earth in vacuum
C=light speed=299792458m/s
X=1.5x10^8km=1.5x10^11m
c=X/t
T1=X/c
T1=1.5X10^11/299792458=500.34s
time it takes sunlight to reach the earth in water:
First we calculate the speed of light in water taking into account the refractive index
Cw=299792458m/s/1.349=222233104.5m/s
T2=1.5x10^11/222233104.5m/s=675s
additional time it would take for the light to reach the earth
ΔT=T2-T1=675-500=175s
A water rocket can reach a speed of 95 m/s in 0.050 seconds from launch.
What is its average acceleration?
A loaf of bread (volume 3100 cm3) with a density of 0.90 g/cm3 is crushed in the bottom of the grocery bag into a volume of 1240 cm3. What is the density of the mashed bread? . g/cm3
Answer:
2,25 g/cm3
Explanation:
Hi, you have to know one thing for this.. Density = mass/Volume,
When you have the loaf of bread with 3100 cm3 and a density of 0.90 g/cm3, the mass of that bread is 2790 g because of if you isolate the variable mass from the equation you get.. mass= density x volume
Later, have on account the mass never changes, so you crush the bread and the mass is the same.. so when you have the mashed bread.. you know that the mass is 2790 g and the volume of the bag is 1240 cm3, so you apply the main equation.... density=2790 g / 1240 cm3 , so density = 2,25 g/cm3
Final answer:
The density of the mashed bread is 2.25 g/cm³, calculated by dividing the mass of the original loaf (found by multiplying its original density and volume) by the new volume after being crushed.
Explanation:
The density of the original loaf of bread is given as 0.90 g/cm³, and its initial volume is 3100 cm³. When the bread is crushed, its volume is reduced to 1240 cm³. Density is the ratio of the mass of a substance to its volume. Because the mass of the bread remains the same even when it is crushed, we can find the new density by using the mass from the original loaf. The mass can be calculated by multiplying the original density by the original volume. The calculated mass is then divided by the new volume.
Mass = Original density × Original volume = 0.90 g/cm³ × 3100 cm³ = 2790 grams
Now, we divide the mass by the new volume to get the density of the mashed bread:
Density of mashed bread = Mass / New volume = 2790 g / 1240 cm³ = 2.25 g/cm³.
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. When each of the spheres has lost half its initial charge, the magnitude of the electrostatic force will be
a) 1/16 Fb) 1/8 Fc) 1/2 Fd) 1/4 F
Answer: c) 1/4F
Explanation: In order to explain this result we must use the Coulomb force expression, that is:
F=(k*q1*q2)/d^2 where d is the distance between spheres, k a constant equal to 9*10^9N^2.m^2/C
after time each sphere lost haft of original charge and taking into account that F is directly proportional to charge of each sphere,
we have Fafter=(k*q1/2*q2/2)/d^2
then Fafter=(1/4)(k*q1*q2)/d^2=(1/4)Finitial
What is the boiling point of an benzene solution that freezes at 0.3 degrees C? Normal freezing point is 5.5 and the normal boiling point is 80.1 degrees C. Kfp = 5.12 K/m and Kbp = 2.53 K/m. Enter your answer to 2 decimal places.
Answer: The boiling point of an benzene solution is [tex]82.57^0C[/tex]
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=T_f^0-T_f=(5.5-0.3)^0C=5.2^0C=5.2K[/tex] = Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte)
[tex]K_f[/tex] = freezing point constant = [tex]5.12K/m[/tex]
m= molality
[tex]5.2=1\times 5.12\times m[/tex]
[tex]m=1.015[/tex]
Elevation in boiling point is given by:
[tex]\Delta T_b=i\times K_b\times m[/tex]
[tex]\Delta T_b=Tb-Tb^0=(Tb-80.1)^0C[/tex] = elevation in boiling point
i= vant hoff factor = 1 (for non electrolyte)
[tex]K_b[/tex] = boiling point constant = [tex]2.43K/m[/tex]
m= molality = 1.015
[tex](T_b-80.1)^0C=1\times 2.43\times 1.015[/tex]
[tex]T_b=82.57^0C[/tex]
Thus the boiling point of an benzene solution is [tex]82.57^0C[/tex]
Over a span of 8.0 seconds, an object moves 16 m to the left relative to where it began. If we treat left as the direction towards increasingly negative displacement, what is the average velocity of the object during this motion?
Answer:
-2 m/sec
Explanation:
We have given time t = 8 sec
As the object move in left relative to where it began and it is increasingly negative direction
So displacement = -16 m
We have to find the velocity
So velocity [tex]v=\frac{displacement}{time}=\frac{-16}{8}=-2m/sec[/tex]
As the velocity is a vector quantity so negative sign has significance its the direction of the velocity
A car traveling initially at +7.9 m/s accelerates uniformly at the rate of +0.72 m/s2 for a distance of 265 m. What is its velocity at the end of the acceleration? Answer in units of m/s. 10.0 points What is its velocity after it accelerates for 114 m? Answer in units of m/s
Answer:
in first case velocity =21.07 m/sec in second case velocity =15.05 m/sec
Explanation:
We have given initial velocity u = 7.9 m/sec
Acceleration [tex]a=0.72m/sec^2[/tex]
Distance S = 265 m
Now according to third law of motion [tex]v^2=u^2+2as[/tex] here v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance
So [tex]v^2=7.9^2+2\times 0.72\times 265=444.01[/tex]
v = 21.07 m/sec
In second case s =114 m
So [tex]v^2=7.9^2+2\times 0.72\times 114=226.57[/tex]
v =15.05 m/sec
Answer:
1. 21.07 m/s
2. 15.05 m/s
Given:
initial speed of the car, u = 7.9 m/s
distance covered by the car, d = 265 m
acceleration, a = 0.72[tex]m/s^{2}[/tex]
Solution:
To calculate the velocity at the end of acceleration, we use the third eqn of motion:
[tex]v^{2} = u^{2} + 2ad[/tex]
[tex]v^{2} = 7.9^{2} + 2\times 0.72\times 265[/tex]
[tex]v = \sqrt{7.9^{2} + 2\times 0.72\times 265} = 21.07 m/s[/tex]
Now,
Velocity after it accelerates for a distance for 114 m:
Here d = 114 m
Again, from third eqn of motion:
[tex]v^{2} = u^{2} + 2ad[/tex]
[tex]v = \sqrt{7.9^{2} + 2\times 0.72\times 114} = 15.05 m/s[/tex]
A person cannot see clearly objects closer than 2.00 m from her/him. What is the refractive power (in diopters) of the correction lenses that will allow her/him to clearly see objects located 30.0 cm, but not closer, from her eyes? Ignore the distance between her/his eyes and the lenses.
Answer:24
Explanation:
The refractive power of the correction lenses is - 2.833 diopters.
What is refractive power of lens?Refractive power in optics refers to how much light is converged or diverged by a lens, mirror, or other optical system. It is equivalent to the device's focal length's reciprocal.
Too much power in a myopic eye causes light to focus in front of the retina. A negative power is shown for this. In contrast, a hyperopic eye lacks insufficient strength, causing light to focus behind the retina when the eye is relaxed.
According to the question:
Object distance: u = 30.0m.
Image distance: v = 2.0 m = 200.0 cm.
Let the focal length of the lens be f, then:
1/f = 1/v - 1/u
= 1/200 - 1/30
f = - 35.29 cm
Hence, the refractive power of the correction lenses is = 100/f = 100/(-35.29) = - 2.833 diopters.
Learn more about refractive power here:
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An electric field is constant at every point on a square surface that is 0.80 m on a side. This field has a magnitude of 3.5 N/C and is oriented at an angle of 35° with respect to the surface, as the drawing shows. Calculate the electric flux ΦE passing through the surface.
Answer:
[tex]\phi = 1.28 Nm^2/C[/tex]
Explanation:
As we know that electric flux is defined as
[tex]\phi = \vec E . \vec A[/tex]
now we have
[tex]A = L^2[/tex]
[tex]A = 0.80^2 = 0.64 m^2[/tex]
[tex]E = 3.5 N/C[/tex]
also we know that electric field makes and angle of 35 degree with surface
so angle made by electric field with Area vector is given as
[tex]\theta = 90 - 35 = 55[/tex]
[tex]\phi = EAcos\theta[/tex]
[tex]\phi = (3.5)(0.64)cos55[/tex]
[tex]\phi = 1.28 Nm^2/C[/tex]
Final answer:
The electric flux through the surface is calculated using the formula ΦE = E · A · cos(θ), resulting in an electric flux of 1.84 N·m²/C for a uniform electric field of 3.5 N/C at an angle of 35° to a 0.80 m square surface.
Explanation:
The area of the surface (A) can be found by squaring the length of one side of the square. For a square that is 0.80 m on a side, A = (0.80 m)² = 0.64 m².
The electric flux is given by the equation ΦE = E · A · cos(θ), where E is the magnitude of the electric field, A is the area of the surface, and θ is the angle between the field and the normal to the surface.
Given that E = 3.5 N/C and θ = 35°, the flux through the surface is ΦE = 3.5 N/C · 0.64 m² · cos(35°).
The cosine of 35° is approximately 0.819. So, the electric flux is ΦE = 3.5 · 0.64 · 0.819 = 1.84 N·m²/C.
Jane is riding in a hot air balloon that is rising vertically at a constant speed of 3 m/s over a lake. She reaches out and drops a rock from the balloon when the distance from the rock to the water is 50 m. Use g = 10 m/s2, and let the up direction be positive. About how long after Jane drops the rock will it splash into the water?
Answer:
The rock will splash into the water [tex]3.47s[/tex] after Jane drops it.
Explanation:
Hi
Known data[tex]v_{i}=3m/s, y_{i}=50, y_{f}=0m[/tex] and [tex]g=10m/s^{2}[/tex].
We are going to use the formula below[tex]y_{f}=y_{i}+v_{i}t-\frac{1}{2} gt_^{2}[/tex]
Letting up direction be positive and computing with the known data, we have[tex]0=50+3t-\frac{1}{2} 10t_^{2}[/tex]
[tex]0=50+3t-5t_^{2}[/tex] a second-grade polynomial, we obtain two roots, so [tex]t_{1}=3.47s[/tex] and [tex]t_{2}=-2.87s[/tex], due negative root has no sense, we take the positive one, so the rock will splash into the water [tex]3.47s[/tex] after Jane drops it.