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devlian [24]
3 years ago
12

If a missile is launched from the equator toward an object in the Northern Hemisphere, should it be aimed to the left, to the ri

ght, or directly at its target? Why?
Physics
1 answer:
nexus9112 [7]3 years ago
5 0

Answer:

Must be aimed to the left

Explanation:

Coriolis effect deflects wind or any moving item which has no contact with earth and has substantial distance to cover and moving at a substantial altitude. Coriolis effect deflectd the object to the right in the northern hemisphere. Therefore, missile must be aimed to the left, to nullify the coriollis effect.

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An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo
Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =<u>16321.0769 N/C</u>

The electric field strength at point (x,y) = (0cm, 20 mm) is =<u>35321.58999 N/C</u>

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=<u>16321.0769 N/C</u>

 Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷  0.000000152881

=<u>35321.58999 N/C</u>

8 0
3 years ago
How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to
finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is x=\sqrt{\frac{2\times EPE}{k}}

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

EPE=\frac{1}{2}kx^2

Rewriting the above expression in terms of 'x', we get:

x=\sqrt{\frac{2\times EPE}{k}}

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0
3 years ago
What kind of potential energy allows you to throw a ball
Aleks [24]
Force.................
3 0
3 years ago
Read 2 more answers
f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t
inysia [295]

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

h(t)=35t-0.83t^2

We know that the rate of change of displacement is equal to the velocity of an object.

v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t

Velocity of the arrow after 3 seconds will be :

v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

7 0
3 years ago
Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface tempe
stiks02 [169]

Answer:

6.0 × 10^{11} W/m^{2}

Explanation:

From Wien's displacement formula;

Q = e AT^{4}

Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = \frac{Q}{A} = eT^{4}

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × (1000)^{4}

                             = 0.6 × 1.0 × 10^{12}

                             = 6.0 × 10^{11} \frac{W}{m^{2} }

Therefore, the emissive intensity coming out of the surface is 6.0 × 10^{11} W/m^{2}.

3 0
3 years ago
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