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nirvana33 [79]
2 years ago
12

I'm completely lost in physics, the Kinematics and dynamics units.. My exam is coming up.. If a biker travelling at 6.4m/s sees

biker B 34m ahead on the road travelling 4.7m/s. How far will biker B get before biker A catches him?
Physics
1 answer:
Alexeev081 [22]2 years ago
8 0
D=rt
when biker A catches biker B, the time they've been riding is the same, so 
t=t, or d/r=d/r
the rates are 6.4 and 4.7, so
d/6.4=d/4.7
biker B is 34m ahead, so 
(d+34)/6.4=d/4.7
multiply both sides by 6.4*4.7:
4.7(d+34)=6.4d
4.7d+=6.4d+159.8
1.7d=159.8
d=94 meters

Another way to think of it is that biker A gains 1.7 meters on B every second (6.4-4.7=1.5), so the time it'll take for him to gain 34 meters is 34/1.7=20 seconds. In that time, biker B travels 4.7*20=94 meters
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Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field dE at the center of curvature of the arc is

dE = k\dfrac{dQ}{r^2}cos(\theta ) <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where r is the radius of curvature.

Now

dQ = \lambda rd\theta,

where \lambda is the charge per unit length, and it has the value

\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.

Thus, the electric field at the center of the curvature of the arc is:

E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta  }{r^2} cos(\theta)

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.

Now, we find \theta_1 and \theta_2. To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

fraction = \dfrac{3.0m}{2\pi (2.3m)}  = 0.2076

and this is  

0.2076*2\pi =1.304 radians.

Therefore,

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.

evaluating the integral, and putting in the numerical values  we get:

E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\

\boxed{ E = 31.329N/C.}

4 0
3 years ago
Twenty grams of a solid at 70°C is place in 100 grams of a fluid at 20°C. Thermal equilibrium is reached at 30°C.
zaharov [31]

Answer:

c. is more than that of the fluid.

Explanation:

This problem is based on the conservation of energy and the concept of thermal equilibrium

heat= m s \Delta T&#10;

m= mass

s= specific heat

\DeltaT=change in temperature

let s1= specific heat of solid and s2= specific heat of liquid

then

Heat lost by solid= 20(s_1)(70-30)=800s_1&#10;

Heat gained by fluid=100(s_2)(30-20)=1000s_2&#10;

Now heat gained = heat lost

therefore,

1000 S_2=800 S_1

S_1=1.25 S_2

so the specific heat of solid is more than that of the fluid.

8 0
3 years ago
a typical cmall flashlight contains two batteries each having na emf of 2.0 v connected in series with a bulb havin ga resistanc
Helen [10]

Answer:

P = 0.25 W

Explanation:

Given that,

The emf of the battry, E = 2 V

The resistance of a bulb, R = 16 ohms

We need to find the power delivered to the bulb. We know that, the formula for the power delivered is given by :

P=\dfrac{V^2}{R}\\\\P=\dfrac{2^2}{16}\\\\=0.25\ W

So, 0.25 W power is delivered to the bulb.

5 0
3 years ago
a child is swinging on swing, describe what happens to both the kinectic energy and potential enegry of the child as she swings
Igoryamba

Answer:

The K.E is maximum when the child is at the vertical position and the P.E is maximum at the extreme deviated position from the vertical.

Explanation:

  • A child is swinging on swing up and down has both kinetic and potential energy.
  • The total mechanical energy of the system is conserved throughout the system. At any instant the total mechanical energy is given by,

                                      E = K.E + P.E

  • The K.E is maximum when the child is at the vertical position.
  • The P.E is maximum at the extreme deviated position from the vertical.
  • And when K.E is maximum P.E becomes minimum and vice versa as per the law of conservation of energy.
5 0
3 years ago
you stop the stopwatch at 4.0 s, but you notice a short time later that the same ant is at 0.81 m on the meter stick. Assuming t
telo118 [61]

The time elapsed since you stopped the stopwatch is 0.41 s.

<em>Your question is not complete, it seems to be missing the following information;</em>

"The velocity of the ant is 2 m/s"

The given parameters;

  • velocity of the ant, v = 2 m/s
  • change in position of the ant, Δx = 0.81 m
  • initial time, t₁ = 4 s
  • time when the ant was noticed, = t₂

Velocity is defined as the change in displacement per change in time of motion of an object.

v = \frac{\Delta x}{\Delta t} = \frac{\Delta x}{t_2 - t_1} \\\\t_2 -t_1 = \frac{\Delta x}{v} \\\\t_2 - 4 = \frac{0.81}{2} \\\\t_2 - 4 = 0.405\\\\t_2 = 0.405 + 4\\\\t_2 = 4.405 \approx 4.41 \ s

The time elapsed since you stopped the stopwatch is calculated as;

t_{elapsed} = 4.41 \ s - 4\ s = 0.41 \ s

Thus, the time elapsed since you stopped the stopwatch is 0.41 s.

Learn more here: brainly.com/question/18153640

5 0
3 years ago
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