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melamori03 [73]
3 years ago
5

1. What was the electromagnetic spectrum ?

Physics
1 answer:
Maurinko [17]3 years ago
4 0
The answer is C. I hope this helps you.
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You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
1 year ago
Read 2 more answers
Cavity wall insulation costs £240 but will save you £32 each year. What is the payback time for cavity
likoan [24]

Answer:

Thus, the payback time for cavity wall insulation is 7.5 years

5 0
2 years ago
Since all objects are ‘weightless’ for an astronaut in orbit, is it possible for astronauts to tell whether an object is heavy o
ivanzaharov [21]
W=gm
where g - gravitation 
m - mass
w - weight
as gravitation equals to zero, multiplying by 0 gives W=0
It is not possible to tell whether and object is heavy or light 

4 0
3 years ago
Read 2 more answers
The AC voltage source is connected to an inductor and a resistor in series. If the frequency of the source is increased the curr
DaniilM [7]

Answer:

If the frequency of the source is increased the current in the circuit will decrease.

Explanation:

The current through the circuit is given as;

I = \frac{V}{Z}

Where;

V is the voltage in the AC circuit

Z is the impedance

Z = \sqrt{R^2 + X_L^2}

Where;

R is the resistance

X_L is the inductive reactance

X_L = ωL = 2πfL

where;

L is the inductance

f is the frequency of the source

Finally, the current in the circuit is given as;

I = \frac{V}{\sqrt{R^2 + (2\pi fL)^2} }

From the equation above, an increase in frequency (f) will cause a decrease in current (I).

Therefore, If the frequency of the source is increased the current in the circuit will decrease.

5 0
3 years ago
Use the Pythagorean theorem to answer the following question. A ping-pong ball is shot straight north from a popgun at 4.0 m/s.
MaRussiya [10]
Resultant is 5 m/s using the Pythagorean theorem<span />
4 0
3 years ago
Read 2 more answers
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