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Vesnalui [34]
3 years ago
14

How much time elapses before the two kittens collide

Physics
1 answer:
geniusboy [140]3 years ago
3 0

Answer:

6.8 s

Explanation:

Let's say the west end of the field is position 0 m and the east end of the field is position 250 m.

The position of kitten A is:

x = 0 + 25 t

And the position of kitten B is:

x = 250 - 12 t

The kittens collide when they have the same position:

25 t = 250 - 12 t

37 t = 250

t ≈ 6.8 s

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a magnet has a 20 cm magnetic field if a piece of metal is 18 cm from the magnet will it be attracted or not
mina [271]

Answer:

yes

Explanation:

The metal is closer than 20 cm to the magnet which is in the magnetic field.

4 0
3 years ago
The water flowing through a 2.0 cm (inside diameter) pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the thr
Verdich [7]

Answer:

a)54L/min

b)0.845

Explanation:

a) A x V=A_1V_1+ A_2V_2+A_3V_3

where suffix 1,2,3 refers to the three pipes.

            =27L/min+16L/min+11 L/min

            =54L/min

b) A x V=54L/min => \frac{\pi }{4} d^2 x v

   d= 2 cm

\frac{\pi }{4} d^2 x v = 54

v= \frac{4}{\pi } x \frac{54}{2^2}

-> A_1 x V_1=27L/min => \frac{\pi }{4} d_1^2 x v_1

d_1= 1.3cm

\frac{\pi }{4} d^2 x v_1 = 27

v_1= \frac{4}{\pi } x \frac{27}{1.3^2}

Next is to find the ratio of speed i.e \frac{v}{v_1}

\frac{4}{\pi } x \frac{54}{2^2} / \frac{4}{\pi } x \frac{27}{1.3^2} => \frac{54}{27} \frac{1.3^2}{2^2}

\frac{v}{v_1}= 0.845

8 0
3 years ago
Sound wave A has a lower frequency than sound wave B, but both waves
Keith_Richards [23]

Answer: C

Explanation: Amplitude controls loudness, and frequency controls pitch. The more frequent the higher pitch.

4 0
3 years ago
Read 2 more answers
On the way to school, the bus speeds up from 20 m/s to 36 m/s in 4 seconds. What distance
Tresset [83]
Answer B. 112 m



Step-by-Step Explanation

initial velocity u = 20 m /s
final velocity v = 36 m /s
time taken t = 4 s
acceleration = (v - U) / t
= (36 - 20) / 4
a=4m/s2
from the formula
7-u2=2as , sis distance covered
putting the values
362-202=2×4×s
1296 - 400 = 8 x S
S= 112 m
4 0
2 years ago
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
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