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Tema [17]
2 years ago
9

Convert 3 hours 21 minutes to decimal hours.

Physics
1 answer:
Oxana [17]2 years ago
7 0

Total hours is 3.35 hours

<u>Explanation:</u>

Given:

convert 3 hours 21 minutes to decimal hours

We know:

1 hour = 60 minutes

and

1 minute = 60 seconds

1 minute = 1 / 60 hours

So,

21 minutes = \frac{21}{60} hours

                  = 0.35 hours

Total hours would be = 3 hours + 0.35 hours

                                    = 3.35 hours

Therefore, total hours is 3.35 hours

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When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6
Stella [2.4K]

Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 = +8.4\mu C

q_2 = +5.6 \mu C

force between two charges is given as

F = 0.66 N

now we have

F = \frac{kq_1q_2}{r^2}

0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}

r = 0.8 m

so it is separated by 80 cm distance

3 0
3 years ago
Read 2 more answers
Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

6 0
2 years ago
A 62 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 12 m. She falls a total of 3
Andrew [12]

Answer:

k = 104.46 N/m

Explanation:

Here we can use energy conservation

so we will have

initial gravitational potential energy = final total spring potential energy

as we know that she falls a total distance of 31 m

while the unstretched length of the string is 12 m

so the extension in the string is given as

x = L - L_o

x = 31 - 12 = 19 m

so we have

mgH = \frac{1}{2}kx^2

62(9.81)(31) = \frac{1}{2}k (19^2)

k = 104.46 N/m

5 0
2 years ago
Gauges all very according the vehicle , model make and year and other factors. By turning the vehicle to the ___ position you ca
AleksandrR [38]

Answer:

By turning the vehicle "ON" position you can check to see if the gauges light works.        

When we switch ON or turn a key to ON the engine, we can find all the gauges working or not.

4 0
3 years ago
A parallel-plate capacitor has plates of area A. The plates are initially separated by a distance d, but this distance can be va
ohaa [14]

Answer:

d' = d /2

Explanation:

Given that

Distance = d

Voltage =V

We know that energy in capacitor given as

U=\dfrac{1}{2}CV^2

C=\dfrac{\varepsilon _oA}{d}

U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2

If energy become double U' = 2 U then d'

U'=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2\times \dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2 d ' = d

d' = d /2

So the distance between plates will be half on initial distance.

7 0
3 years ago
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