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soldier1979 [14.2K]
2 years ago
13

A meteoroid is traveling east through the atmosphere at 18. 3 km/s while descending at a rate of 11.5 km/s. What is its speed, i

n km/s?

Physics
1 answer:
Annette [7]2 years ago
6 0

Answer:

The speed of meteoroid is 21.61 km/s in south-east.

Explanation:

Given that,

A meteoroid is traveling through the atmosphere at 18.3 km/s. while descending at a rate of 11.5 km/s it means 11.5 km/s in south.

We need to draw a diagram

Using Pythagorean theorem

AC^2=AB^2+BC^2

AC^2=(18.3)^3+(11.5)^2

AC=\sqrt{(18.3)^2+(11.5)^2}

AC=21.61\ km/s

Hence, The speed of meteoroid is 21.61 km/s in south-east.

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A 1.2 kg basketball is thrown upwards. What is the potential energy of the basketball at the top of its path if it reaches a hei
Dahasolnce [82]

Answer:

Answer is 183.6 J

Explanation:

Using the Physics reference sheet the formula for Potential energy is

(mass) x (gravity) x (height)

Mass= 1.2

Gravity I used is 9.81 (use 10 to get the answer most schools use)

Height= 15.6

5 0
2 years ago
One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F
Alja [10]

Answer:

\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}

f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}

b) Wave speed

(i) Calculate the wavelength

In a  fundamental vibration, the length of the string is half the wavelength.

\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}

(b) Calculate the speed s

\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}

\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}

\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}

4 0
3 years ago
Human-Powered Flight Human-powered aircraft require a pilot to pedal, as in a bicycle, and produce a sustained power output of a
alukav5142 [94]

Answer:

# of Snickers bars 2

Explanation:

Power output= 0.30 HP

=0.3*746

= 0.30 HP (746 W=1.00 HP)

= 224 W

time required 2 h 49 m = 10140 seconds

Since power is work divided by time, then work is:

Work done by the jet = P*t

= 224 *(10140)

= 2.3 MJ (2.3 x 10^{6} J)

Converting MJ to Cal

2.3 MJ=549 Cal

# of Snickers bars = 549 Cal / 280 Cal

= 2.0 bars (rounded from 1.96)

8 0
3 years ago
A train pulls away from a station with a constant acceleration of 0.42 m/s2. A passenger arrives at a point next to the track 6.
Rina8888 [55]

Answer:

2.69 m/s

Explanation:

Hi!

First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:

x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m

So, the position as a function of time is:

xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m

Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:

xP(t)=V*t

In order for the passenger to catch the train

xP(t)=xT(t)

(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t

To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:

0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2

This equation give us the minimum velocity the passenger must have in order to catch the train:

V^2 - 7.22534(m/s)^2 = 0

V^2 = 7.22534(m/s)^2

V = 2.6879 m/s

4 0
3 years ago
A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.
Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

\Delta K = \Delta E_{p} = q\Delta V

<u>Where</u>:

K: is the kinetic energy

E_{p}: is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV                                    

\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J

Now, the kinetic energy of the particle as it moves through point B is:

\Delta K = K_{f} - K_{i}

K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!      

8 0
3 years ago
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