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3 years ago

Which is an example of speed?

Physics

1 answer:

Sauron [17]3 years ago

3 0

Answer:

Definition: Speed is a way of measuring how quickly something is moving or being done, or something moving fast. An example of speed is a car being driven 45 miles per hour. An example of speed is someone cleaning a room in 10 minutes. An example of speed is how quickly a jaguar runs.

Explanation:

You didnt give examples so i just gave the definition :)

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Mester Exam 1 11 of 35

frez [133]

Please show picture of diagrams

8 0

3 years ago

A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca

svetoff [14.1K]

Answer:

$q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Explanation:

Given that $L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V$ , we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

$E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000 \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}$

#To find the particular solution:

$Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Hence the charge at any time, t is $q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

6 0

3 years ago

How to answer this question?

laila [671]

Answer:

measure the vector diagram first

5 0

2 years ago

For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

pickupchik [31]

The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.

In that particular situation, you can prove it like this: 

initial velocity is Vo 
launch angle is α 

initial vertical velocity is 
Vv = Vo×sin(α) 

horizontal velocity is 
Vh = Vo×cos(α) 

total time in the air is the the time it needs to fall back to a height of 0 m, so 
d = v×t + a×t²/2 
where 
d = distance = 0 m 
v = initial vertical velocity = Vv = Vo×sin(α) 
t = time = ? 
a = acceleration by gravity = g (= -9.8 m/s²) 
so 
0 = Vo×sin(α)×t + g×t²/2 
0 = (Vo×sin(α) + g×t/2)×t 
t = 0 (obviously, the projectile is at height 0 m at time = 0s) 
or 
Vo×sin(α) + g×t/2 = 0 
t = -2×Vo×sin(α)/g 

Now look at the horizontal range. 
r = v × t 
where 
r = horizontal range = ? 
v = horizontal velocity = Vh = Vo×cos(α) 
t = time = -2×Vo×sin(α)/g 
so 
r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) 
r = -(Vo)²×sin(2α)/g 

To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. 

dr/dα = d[-(Vo)²×sin(2α)/g] / dα 
dr/dα = -(Vo)²/g × d[sin(2α)] / dα 
dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα 
dr/dα = -2 × (Vo)² × cos(2α) / g 

Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when 
cos(2α) = 0 
2α = 90° 
α = 45°

4 0

3 years ago

A spring of force constant 1500 Nm-l is acted

olchik [2.2K]

Answer:

1.876 J

Explanation:

First, let’s calculate the compression of the spring from the Hooke’s law:

F=kx,

here, F=75 N is the force acted on the spring, k=1500 N⁄m is the force constant of the spring, x is the compression of the spring.

Then, we get:

x=F/k=(75 N)/(1500 N/m)=0.05 m.

Finally, we can find the potential energy stored in the spring:

PE=1/2 kx^2=1/2∙1500 N/m∙(0.05 m)^2=1.875 J.

correct my answer if it's wrong ^^

7 0

3 years ago

Which is an example of speed?​

Which is an example of speed?