A car possesses 20,000 units of momentum. what would be the car's new momentum if ... its velocity was doubled?

Why are force fields used to describe magnetic force?

Illusion [34]

Answer:

If I'm not working I think the answer is C.

7 0

2 years ago

When water flows from a faucet, the water molecules tend to join together and form a stream. Which of the four fundamental force

Yuki888 [10]

The fundamental force responsible for the cohesion of the water molecules leaving the faucet is the electromagnetic force.
Electromagnetic forces act on particles that are electrically charged. Water molecules are polar, which means that they have a positively charged end and a negatively charged end. This polarity arises from the fact that oxygen pulls the electrons in the molecule towards itself and attains a negative charge, while the hydrogen atoms in the molecules are left with a positive charge.

3 0

2 years ago

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A 5-ft-tall person walks away from the wall at a rate of 2 ft/sec. A spotlight is located on the ground 40 ft from the wall. How

AveGali [126]

Answer:

The rate of change of the height is - 4 ft/s

Solution:

As per the question:

Height of the person, y = 5 ft

The rate at which the person walks away, $\frac{dx}{dt} = 4\ ft/s$

Distance of the spotlight from the wall, x = 40 ft

Now,

To calculate the rate of change in the height, $\frac{dy}{dt}$ of the person when, x = 10 m:

From fig 1.

$\Delta ABC[\tex] ≈ [tex]\Delta PQC[\tex]Thus[tex]\frac{BC}{AB} = \frac{PQ}{QC}$

$\frac{y}{40} = \frac{5}{x}$

xy = 200 (1)

Differentiating the above eqn w.r.t time t:

$x\frac{dy}{dt} + y\frac{dx}{dt} = 0$

Thus

$\frac{dy}{dt} = - \frac{y}{x}\frac{dx}{dt}$ (2)

From eqn (1):

When x = 10 ft

10y = 200

y = 20 ft

Using eqn (2):

$\frac{dy}{dt} = - \frac{20}{10}\times 2 = - 4\ ft$

8 0

1 year ago

On a warm summer day (31 ∘c), it takes 4.60 s for an echo to return from a cliff across a lake. on a winter day, it takes 5.00 s

xenn [34]

The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.

First of all, we need to calculate the speed of sound at temperature of $T=31^{\circ}C$ :
$v=(331+0.60 T) m/s = (331+0.6 \cdot 31) m/s = 349.6 m/s$

The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
$v= \frac{2L}{t}$
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
$L= \frac{vt}{2}= \frac{(349.6 m/s)/4.60 s)}{2}= 804.1 m$

6 0

2 years ago

A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the

Sedbober [7]

Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is

$f = \sqrt{ \frac{T}{4mL} }$ -- (A)

here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg 

Plug in the values in Equation (A)

so $f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }$ = 197.97Hz 

the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i

3 0

2 years ago

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