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IrinaVladis [17]

2 years ago

9

A 10-A current is maintained in a simple circuit with atotal resistance of 209 ohm. What net charge passes through any point in

a circuit during a 1-minute interval?

1 answer:

DerKrebs [107]2 years ago

3 0

Answer:

I = Q / t definition of current I

Q = i t = 10 Coul/sec * 60 sec = 600 Coul

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A mosquito flies toward you with a velocity of 2.4 km/h [E]. If a distance of 35.0 m separates you and the mosquito initially, a

tangare [24]

First convert the speed of mosquito to m/s:

So the mosquito is flying at (2,400/3,600) m/s, or ⅔ m/s.

<span>

Since you are moving at 2m/s, so this makes the closing velocity between you and the mosquito to be 2⅔ m/s. </span>

Therefore the mosquito will hit your sunglasses at:<span>

35 m / (2⅔ m/s) = 13⅛ seconds.

2.0 m/s * 13⅛ s = 26¼ m from your initial position.

<span>⅔ m/s * 13⅛ s = 8¾ m from the mosquito's initial position. </span></span>

7 0

4 years ago

A wood block is sliding up a wood ramp. if the ramp is very steep, the block will reverse direction at its highest point and sli

DanielleElmas [232]

<span>Answer: So it gets to the top of the ramp and stops. The parallel force pushing it down the ramp is mg sin Î¸, but for it to move, the frictional force must be overcome. This frictional force is ÎĽmg cos Î¸, where ÎĽ is the coefficient of static friction. For movement, then, mg sin Î¸ > ÎĽmg cos Î¸ ==> tan Î¸ > ÎĽ ==> Î¸ > arctan 0.5 = 26.565Â° ==> Î¸ = 27Â°</span>

5 0

3 years ago

A 2kg mass is moving at 3m/s. What is its kinetic energy?

Illusion [34]

<h2><u>KINETIC ENERGY</u></h2>

<h3>Problem:</h3>

» A 2kg mass is moving at 3m/s. What is its kinetic energy?

<h3>Answer:</h3>

$\color{hotpink} \bold{9 \: J} \\$

— — — — — — — — — —

<h3>Formula:</h3>

To calculate the velocity of a kinetic energy, we can use formula

$\underline{ \boxed{ \tt KE = \frac{1}{2} m{v}^{2} } }$

where,

v is the velocity in m/s
KE is the kinetic energy in J (joules)
m is the mass in kg

— — —

Based on the problem, the givens are:

KE (Kinetic energy) = ? (unknown)
m (mass) = 2 kg
v (velocity) = 3 m/s

<h3>Solution:</h3>

To get the velocity, substitute the givens in the formula above then solve.

$\: \: \tt KE = \frac{1}{2} m{v}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \: KE = \frac{1}{2} \times 2 \times {(3)}^{2} \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: KE = \frac{1}{2} \times(2\times 9) \\ \tt KE = \underline{ \boxed{ \blue{ \tt9 \: J}}}$

Therefore, the kinetic energy is 9 Joules.

3 0

2 years ago

Two objects are thrown vertically upward, first one, and then, a bit later, the other. Is it (a) possible or (b) impossible that

Studentka2010 [4]

Answer:

No, it is impossible

Explanation:

Kinematics equation:

$Vf^{2} =Vo^{2} -2gy$

if height is maximum:

y=H and Vf=0

so:

<h3> $Vo^{2} =2gH$ </h3><h3> $H=Vo^{2} /2g$ </h3>

Analysis: From the last equation we see that the maximum height depends ONLY on the initial speed. This means that if both objects reach the same maximum height, then they necessarily need to have the SAME initial velocity. If they have the same initial velocity and in order to reach the maximum height at the SAME time the only way is that they are released at the SAME TIME.

5 0

3 years ago

Force due to gravity. While JWST is in orbit, the Earth will be at a distance of 1.494 x 109 m from the telescope, and the Sun w

Alona [7]

Answer:

the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

Explanation:

Given that;

distance of earth from the telescope r1 = 1.494 × 10⁹ m

and the Sun will be 1.49598 × 10¹¹ m further away

so r2 = r1 + 1.49598 × 10¹¹

r2 = 1.494 × 10⁹ m + 1.49598 × 10¹¹ m = 1.51092 × 10¹¹ m

mass of sun Ms = 1.9884 × 10³⁰ kg

mass of earth Me = 5.945× 10²⁴ kg

mass of JWST Mj = 6500 kg

What is the gravitational force JWST will feel from the Sun (strength and direction)?

the gravitational force of the sun will be attractive based on Newton law of gravitational force; so

Fjs = GMjMs / r2²

constant G = 6.674 × 10⁻¹¹ Nm²/kg²

Force on the JWST by the sun will be;

Fjs = GMjMs / r2² { leftward}

we substitute

Fjs = [(6.674 × 10⁻¹¹ Nm²/kg²)(6500 kg )(1.9884 × 10³⁰ kg)] / (1.51092 × 10¹¹ m)²

= (8.62587804 × 10²³) / ( 2.28287925 × 10²² )

= 37.785 Newton { leftward }

Therefore, the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

7 0

3 years ago