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Artyom0805 [142]
2 years ago
7

Consider a compact car that is being driven

Physics
1 answer:
tigry1 [53]2 years ago
6 0
The answer is 2847.830m
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A 0.8 g object is placed in a 534 N/C uniform electric field. Upon being released from rest, it moves 12 m in 1.2 s. Determine t
Mice21 [21]

Explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

Electric field, E = 534 N/C

Distance, s = 12 m

Time, t = 1.2 s

We need to find the acceleration of the object. It can be solved as :

m a = q E.......(1)

m = mass of electron

a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

s=ut+\dfrac{1}{2}at^2

s=0+\dfrac{1}{2}at^2

a=\dfrac{2s}{t^2}

a=\dfrac{2\times 12\ m}{(1.2\ s)^2}

a = 16.67 m/s²

Now put the value of a in equation (1) as :

q=\dfrac{ma}{E}

q=\dfrac{0.0008\ kg\times 16.67\ m/s^2}{534\ N/C}

q = 0.0000249 C

or

q=2.49\times 10^{-5}\ C

Hence, this is the required solution.

5 0
3 years ago
An unbalanced force of 20N acts on a 4.0kg mass what is it's acceleration​
tankabanditka [31]

Hi there!

According to Newton's second law:

∑F = m · a, where:

∑F = net force (N = kgm/s²)

m = mass (kg)

a = acceleration (m/s²)

Rearrange to solve for acceleration:

F/m = a

20N / 4.0kg = 5 m/s²

4 0
3 years ago
How do you get 5 minutes to seconds
sergiy2304 [10]
There are 60 seconds in a minute.

This means that 5 minutes would be 60 times 5.
60×5 = 300

There are 300 seconds in 5 minutes.
5 0
3 years ago
Read 2 more answers
A transformer has a secondary voltage of 140 volts and a secondary current of 3.5 amps. if the primary current is 10 amps, what
Lynna [10]

For an ideal transformer power loss is assumed to be zero

i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage

this can be written in form of equation

V_1 i_1 = V_2 i_2

here we know that

V_2 = 140 volts

i_2 = 3.5 A

i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1

V_1 = 49 volts

So primary coil voltage is 49 Volts

7 0
3 years ago
If a tank filled with water contains a block and the height of the water above point A within the block is 0.6 meter, what's the
garri49 [273]
Given:\\\rho=1000 \frac{kg}{m^3}\\g=9.8 \frac{m}{s^2} \\h=0.6m\\\\Find:\\p=?\\\\Solution:\\\\p=\rho gh\\\\p=1000 \frac{kg}{m^3}\cdot 9.8 \frac{m}{s^2} \cdot0.6m=5880Pa=5.88kPa\\\\Correct\;is\;answer\;\;D
6 0
3 years ago
Read 2 more answers
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