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Dmitrij [34]
3 years ago
7

Can u plzz help meee

Physics
2 answers:
KATRIN_1 [288]3 years ago
4 0
The answer is
h: sB and Sb
hd : sb
hr : SB
Vlada [557]3 years ago
4 0
The answer is h:sb and sb
Hd:sb
Hr:sb
You might be interested in
Write the relation between horsepower and Watt?​
Vadim26 [7]

Explanation:

Horsepower is related to watt as: 1 horsepower = 746 W = 750 W approximately. 1 horsepower (HP) = 750 watt (approx.) The power is one watt when one joule of work is done in one second.

8 0
3 years ago
Read 2 more answers
An unmarked police car traveling a constant 95 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the polic
Montano1993 [528]

Answer:

Speed of the speeder will be 28 m/sec

Explanation:

In first case police car is traveling with a speed of 90 km/hr

We can change 90 km/hr in m/sec

So 90km/hr=\frac{90\times 1000m}{3600sec}=25m/sec

Car is traveling for 1 sec with a constant speed so distance traveled in 1 sec = 25×1 = 25 m

After that car is accelerating with a=2m/sec^2 for 7 sec

So distance traveled by car in these 7 sec

S=ut+\frac{1}{2}at^2=25\times 7+\frac{1}{2}\times 2\times 7^2=224m

So total distance traveled by police car = 224 m

This distance is also same for speeder

Now let speeder is moving with constant velocity v

so 224=\left ( 1+7 \right )v

v = 28 m/sec

4 0
4 years ago
Read 2 more answers
A,b, e are complete. Help on the others would be so appreciated!!
bixtya [17]

Answer:

serie Ceq=0.678 10⁻⁶ F  and the charge Q = 9.49 10⁻⁶ C

Explanation:

Let's calculate all capacity values

a) The equivalent capacitance of series capacitors

    1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5

    1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2

    1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2

    1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225

    1 / Ceq = 1,147

    Ceq = 0.678 10⁻⁶ F

b) Let's calculate the total system load

   Dv = Q / Ceq

   Q = DV Ceq

   Q = 14 0.678 10⁻⁶

   Q = 9.49 10⁻⁶ C

In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C

c) The potential difference

   ΔV = Q / C5

   ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶

   ΔV = 1,725 ​​V

d) The energy stores is

    U = ½ C V²

    U = ½ 0.678 10-6 14²

    U = 66.4 10⁻⁶ J

e) Parallel system

   Ceq = C1 + C2 + C3 + C4 + C5

   Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶

   Ceq = 22.7 10⁻⁶ F

f) In the parallel system the voltage is maintained

   Q5 = C5 V

   Q5 = 5.5 10⁻⁶ 14

   Q5 = 77 10⁻⁶ C

g) The voltage is constant V5 = 14 V

h) Energy stores

   U = ½ C V²

   U = ½ 22.7 10-6 14²

   U = 2.2 10⁻³ J

8 0
3 years ago
A 10 kilogram lump of rock weighs 16N on the Moon.
qwelly [4]

Answer:

c

Explanation:

it's the only one that makes sense

4 0
2 years ago
Read 2 more answers
How much tension is in a rope if it pulls a 5-kg bucket filled with water with an upward acceleration of 1m/s^2
liraira [26]
F=ma
Tension - weight = mass x acceleration
T - 5(9.81) = 5 x 1
T = 5 + 5(9.81)
T = 54.05 N
T ≈ 54 N
4 0
4 years ago
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