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3 years ago

State 1 pascal pressure

Physics

1 answer:

cricket20 [7]3 years ago

6 0

Answer:

1 pascal pressure is define as the one newton force is given which covers 1 square metre area

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Can anyone please help meee!!!

Neko [114]

Answer:

The centripetal acceleration is given by

a = v 2 r = 10 2 25 = 4 m s − 2

Using Newton's Second Law, the centripetal force acting is

F = m a = 900 ⋅ 4 =3600 N

Explanation:

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2 years ago

A circuit has a resistance of 8 Ohms. the voltage supplied to the circuit is 14 volts. what is the current flowing through it?

valina [46]

Answer:

e jiio7ruetw5258o0pomnmitwq2690okm ,,kkkiiht44456

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2 years ago

A car is traveling at an average speed of 70 m/s. How many km would the car travel in 6.5 hrs. ?

docker41 [41]

Answer:

<h2>38,769.23 miles</h2>

Explanation:

given:

A car is traveling at an average speed of 70 m/s.

find:

How many km would the car travel in 6.5 hrs. ?

solution:

distance = velocity over time

let velocity = 70 m/s

time = 6.5 hrs.

convert velocity 70 m/s into m/h for consistency of units.

<u>70 mi. </u> x <u>3600 sec.</u> = 252,000 mi/hour

sec. 1 hr.

now plugin values into the formula d = v/t

d = <u>252,000 miles/hour </u>

6.5 hours

d = 38,769.23 miles

therefore, the distance travelled in 6.5 hours with a speed of 70 m/s is 38,769.23 miles

5 0

3 years ago

The electric potential at the dot in the figure is 3160 V. What is charge q?

PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

$V = \frac{kQ}{r}$

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

$V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V$

Lower left charge's potential:

$V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V$

Add the two, and subtract from the total EP at the point:

$3160 + 1123.75 - 2247.5 = 2036.25$

The remaining charge must have a potential of 2036.25 V, so:

$2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}$

5 0

2 years ago

1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0

2 years ago

State 1 pascal pressure​

State 1 pascal pressure