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2 years ago

14

Pls say this a doubt i have

1 answer:

hodyreva [135]2 years ago

4 0

Question four bulbs A,B,C and D are connected in a circuit shown in the figure below, the letters X, Y and Z represent three switches. Which switch is used to operate switch A separately?
Answer: x

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If Cell fragments are found in a rock sample it was most likely that The Rock formed

Ira Lisetskai [31]

5 is 4) shallow water, 6 is 2)sedimentary i think but I'm not 100% sure and 7 is 21%

7 0

3 years ago

Can you respond this two questions, please? :

Andrews [41]

Where's the diagram for question 1?

6 0

3 years ago

The speed of an arrow fired from a compound

san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

$X=v_{ox}*t$

$v_{0x} =v_0cos(\alpha)$

Y axis:

$Y= Y_0 +v_{y0} t - \frac{g}{2} t^2$

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, $Y=0$ , then

$0= Y_0 +v_{y0} t - \frac{g}{2} t^2$

$0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

Caculate the segcond degree equation to obtain the two possible values for t:

$t_1= 10.18 \\t_2= -0.04046$

But, in physics, time it could not be negative, so we take $t_1= 10.18$

Caculate now:

$X=79m/s*cos(\39)*10.18s= 624.996 m$

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

$v_0= 79m/s+13m/s= 92m/s$

Using the same procedure that item A, caculate X

First, we need to know the new time

$0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

And we obtain:

$t_1=11.845s\\t_2=-0.041s$

One more time, we take the positive time: $t_1=11.845s$

Finally:

$X=92m/s *cos(39)*11.845s=846.887 m$

6 0

3 years ago

High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of th

murzikaleks [220]

Answer:

Lifetime = 4.928 x 10^-32 s

Explanation:

(1 / v2 – 1 / c2) x2 = T2

T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225

T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6

T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s

4 0

3 years ago

You drive 6.00 km at 50.0 km/h and then another 6.00kmat 900 km/h Your average speed over

Cerrena [4.2K]

Explanation:

average speed = total distance travelled / total time travelled

time to travel the first 6km: 6 / 50 = 3/25 (h)

time to travel the next 6km: 6 / 90 = 1/15 (h)

[I think there's problem in the question 'cause 900km/h sounds impossible for normal person to travel in normal condition]

The total time: 3/25 + 1/15 = 14/75 (h)

Average speed over the 12 km drive will be:

$\frac{12}{ \frac{14}{75} } = \frac{450}{7} = 64.3 \: km{h}^{ - 1}$

8 0

3 years ago