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torisob [31]
3 years ago
8

On June 23, which area will sunlight hit more directly?

Physics
1 answer:
Julli [10]3 years ago
8 0
<span>At any particular time of day on June 23, sunlight will fall on the
northern hemisphere more directly than the southern hemisphere,
and on the north pole more directly than the south pole. (In fact,
at that time, the south pole will be halfway through a 6-month period
of no sunlight at all.) 

On June 23, the sun will appear to be directly over the parallel
of 23.5</span>° north latitude<span>, or nearly so ... the farthest north that the
sun ever appears to be.</span>
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A 5.00 kilogram mass is traveling at 100. meters per second. Determine the speed of the mass after an impulse with a magnitude o
faltersainse [42]

m = mass = 5 kg

v_{i} = initial velocity = 100 m/s

v_{f} = final velocity = ?

I = impulse = 30 Ns

Using the impulse-change in momentum equation

I = m(v_{f} - v_{i})

30 = 5 (v_{f} - 100)

v_{f} = 106 m/s

5 0
3 years ago
Read 2 more answers
What are the smallest parts that make up matter?
Alecsey [184]
The smallest parts that make up different types of matter are called atoms.
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3 0
4 years ago
A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3
4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT

8 0
3 years ago
For a home sound system, two small speakers are located so that one is 52 cm closer to the listener than the other. What is the
galina1969 [7]

Answer:

The first frequency of audible sound in the speed sound is

f = 662 Hz

Explanation:

vs = 344 m/s

x =  52 cm * 1 / 100m = 0.52m

The wave length is the distance between the peak and peak so

d = 2x

d = 2*0.52 m

d = 1.04 m

So the frequency in the speed velocity is

f = 1 / T

f = vs / x = 344 m/s / 0.52m

f ≅ 662 Hz

7 0
4 years ago
3. What will happen to the black bass and blue gill as the floor of the ponds fills with organic
yKpoI14uk [10]

Answer: die

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