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Blababa [14]
3 years ago
5

As a skydiver accelerates downward, what force increases? A. Gravity B. Thrust C. Air resistance D. Centripetal

Physics
1 answer:
dybincka [34]3 years ago
7 0

Answer:

(A) Gravity is you're answer.

Explanation:

When an object or human is falling at an increased rate, The force of gravity is taking place.

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The rules and expectations concerning correct or polite behavior is called _______________________.
aalyn [17]
The answer would be etiquette!
7 0
3 years ago
Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
4 years ago
A wave with a frequency of 14Hz has a wavelength of 3 meters. At what speed will this wave travel​
topjm [15]

Answer:

42 m/s

Explanation:

by definition of the velocity(speed) of a wave,

The velocity of light, v, is the product of its wavelength, λ , and its frequency, f.

V= fλ

frequency - number of occurances in a unit time

(check the graph)

velocity=frequency*wave.length\\=14Hz*3m\\=42 ms^{-1}

6 0
3 years ago
The resistance (R) of a copper wire varies directly as its length (L). Write this relation as a formula using k as the constant
Drupady [299]

Answer:

Explanation:

According to ohm's law current flowing in a conductor is directly proportional to the voltage applied across two end of conductor.

i.e. V\propto R

V=R I

where R=resistance

R\propto L

R\propto \frac{1}{d^2}

whee L and d are length and Diameter

thus R=k \frac{L}{d^2}

where k=constant of Variation

6 0
3 years ago
What charge appears on each plate of a 10.0 μF parallel plate capacitor, when it is charged to 110 V?
elena-14-01-66 [18.8K]

Answer:

charge, q = ± 1.1 mC

Given:

Capacitance, C = 10.0\micro F = 10.0\times 10^{- 6} F

Voltage, V = 110 V

Solution:

The charge on the capacitor plates can be calculated by using the definition of capacitance as :

q ∝ V

where

q = charge

V = potential difference or Voltage

Therefore,

q = CV

Now, charge, q :

q = 10.0\times 10^{- 6}\times 110 = 1100\micro C = 1.1 mC

Therefore, the charge on the positive plate is:

q = + 1.1 mC

the charge on the negative plate is:

q = - 1.1 mC

8 0
3 years ago
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