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Tomtit [17]
3 years ago
10

If a student is walking against the wind, would the student be traveling faster or slower than a person walking with the wind an

d why?
Physics
1 answer:
Ivahew [28]3 years ago
5 0

Answer:

Slower

Explanation:

Slower because the air would be pushing you back, and the other person would be walking faster because they are being pushed forward by the wind

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How does frequency relate to pitch?
tensa zangetsu [6.8K]
The sensation of a frequency is commonly referred to as the pitch of a sound. A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave. ... That is, two sound waves sound good when played together if one sound has twice the frequency of the other.
7 0
2 years ago
PLEASE PLEASE PLEASE A mad scientist places massive amounts of charge on basketball sized aluminum balls. The charge on the ball
bazaltina [42]

Answer:

4.2 x 10⁷N

Explanation:

Given parameters:

Charge on ball:

             q₁  = 3C

              q₂ = 14C

Distance between balls  = 9000m

Unknown:

Force acting on the two balls

Solution:

The force experienced by the two charges is given by coulombs law. It is mathematically expressed as;

                      F  = \frac{k q_{1} q_{2} }{r^{2} }

where k  = 9 x 10⁹Nm²/C²

           q is the charges

             r is the distance

Input the variables and solve;

                 

        F  = \frac{9 x 10^{9} x 3 x 14 }{9000}  = 4.2 x 10⁷N

8 0
3 years ago
How much force does an 88kg astronaut exert on his chair while accelerating straight up at 10 m/s^2
Fed [463]
Force=mass*acceleration.         So 88kg*10 m/s^2=880 newtons
6 0
3 years ago
•. Parking orbit is -
notsponge [240]

Answer:

Parking orbit is -

a. The path along which a plane travels

b. Orbit of a polar satellite

c. Orbit of geostationary satellite

d. Orbit of the earth.

Explanation:

7 0
3 years ago
Read 2 more answers
As a science fair project, you want to launch an 800 g model rocket st raight up and hit a horizontally moving target as it pass
fredd [130]

Sum the forces in the y (upward) direction

\sum F_y = ma

F_t +F_w = ma

15N - (0.800kg)(9.81m/s^2) = 0.800kg * a

a = 8.94 m/s^2

Applying the kinematic equations of linear motion we have that the displacement as a function of the initial speed, acceleration and time is

s = v_0t + \frac{1}{2} at^2

30 = 0 +\frac{1}{2} (8.94) t^2

t = 2.59 s

Again through the kinematic equation of linear motion that describes velocity as the change of displacement in a given time, we have to

v = \frac{d}{t} \rightarrow d = vt

d = (15m/s)(2.59s)

d = 38.85m

Therefore the horizontal distance between the target and the rocket should be 38.83m

4 0
3 years ago
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