Look at the diagram shown below. The reading on ammeter 1 is 0.15 A. What will the reading on ammeter 2 be
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Answer:

A

Explanation:if the things have the same charge they will repel makeing the ball move away form its positively charged guy = that the coil gun works.

Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 50 × 9.8 × 600 = 294,000

We have the final answer as

Hope this helps you

Answer:

scientists often prefer to measure mass rather than weight because weight is affected by gravity while mass is not. Mass is always the same.

Explanation:

Weight is a force rather than a measure of the amount of matter something contains. The same amount of matter can have different weights depending on where it is. Mass directly measures the amount of matter present and remains the same no matter where it's measured.

Answer:

The energy of motion is known as kinetic energy.

Explanation:

Convert km to meters:

1km = 1000 m

1.5 km x 1000 = 1500 m

1500m / 4.8 m/s = 312.5 seconds

312.5 seconds / 60 seconds per minute =5.2 minutes = 5 minutes 12.5 seconds

Answer:

C. Mechanical Energy

Explanation:

The total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy.

Answer:

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

GPE = 25 × 10 × 4 = 1000

We have the final answer as

Hope this helps you

39°C=1.28L

8°C=? Less

=8°C/39°C×1.28L

=8/39×1.28L

=10.24/39

=512/195

=2.6L

Answer:

One of Canada's best-known poems, “In Flanders Fields” was written on May 2, 1915, when Canadian serviceman John McCrae was stationed at an army hospital in Flanders, Belgium, during World War I.

Explanation:

Answer:

Answer → Distance is 5 × 10⁴ km

Explanation:

Gravitational formular

[tex]{ \tt{F = \frac{GMm}{ {r}^{2} } }} \\ [/tex]

For the first case;

[tex]{ \tt{F _{1} = \frac{GMm}{2.5 \times {10}^{4} } }} \\ [/tex]

For the second case;

[tex]{ \tt{F _{2} = \frac{GMm}{r _{2}} }} \\ [/tex]

but F2 = ½F1

Therefore, F1 = 2F2

Hence:

[tex]{ \tt{ \frac{GMm}{2.5 \times {10}^{4} } =2 \times \frac{GMm}{r _{2} } }} \\ \\ { \tt{ \frac{1}{2.5 \times {10}^{4} } = \frac{2}{r _{2} } }} \\ \\ { \tt{r _{2} = 2 \times 2.5 \times {10}^{4} }} \\ \\ { \underline{ \tt{ \: \: r _{2} = 5 \times {10}^{4} \: km \: \: }}}[/tex]

Answer: Distance is 5 × 10⁴ km

Explanation:

Temperature inversion leads to an increase in temperature as altitude increases.

The term temperature inversion refers to a situation in which a layer of warm air lies over a layer of cool air. This is also referred to as thermal inversion. This occurs when the air below to loose heat rapidly.

One of the effects of temperature inversion is reduction in visibility. So, thermal inversion leads to an increase in temperature as altitude increases.

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Answer:

Oh its fine i am not in college but its still fine

Explanation:

From Rutherford  and Marsden scattering experiment, we can conclude that the particle shown in red will come straight back from the foil because it is deflected by the charge in the gold nucleus.

Rutherford proposed Planetary model atom, which visualized an atom to consists of a positively charged heavy core called the nucleus around which negatively charged electrons circle in orbits much as planets move round the sun

Thus, from Rutherford  and Marsden scattering experiment, we can conclude that the particle shown in red will come straight back from the foil because it is deflected by the charge in the gold nucleus.

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the answer is reflected or u can say repelled

Let the ball at the ground be A and that at the top be B.

Assume that you see the motion of A while sitting at B.

You(B) are obviously at rest w.r.t. yourselves. However you(B) and A have the same acceleration in the same direction. Thus acceleration of A w.r.t. you(B) is 0.

Thus, to you, it will appear as if A is travelling towards you with an uniform speed of 20m/s.

Dividing the distance 20m with that speed, we get that A reaches you(B) in 1s.

Now, you have considered these motions and time by assuming you were stationed at B. However, the time taken for A to meet B shouldn't be dependent on which reference frame you assume. Thus, time taken for A and B to collide is 1s.

Since B falls freely, he covers a distance of  1/2(g)(1²)=5  (assuming g = 10m/s²)

Thus, they meet at a height of (20–5)m = 15m from the ground.

this is my attachment answer hope it's helpful to you

Answer:

if your saying can it? then yes or if you are asking what type of magnetic force  then its fr its actual self magnetic force

Explanation:

Please find attached photograph for your answer.

Hope it helps.

Do comment if you have any query.

The initial velocity of the runner is 5.25 m/s.

You've probably seen that when there's a lot of traffic and there are more bikes blocking us, we tend to slow down the speed of our bikes.

Therefore, deceleration is defined as a reduction in speed as the body moves away from the beginning location. The opposite of acceleration is deceleration.

Given parameters:

Deceleration of the runner: a = 0.25 m/s².

Time interval: t = 5 second.

Final velocity: v = 4 m/s.

We have to find initial velocity of the runner: u = ?

From definition of deceleration, it can be written that:

deceleration = (initial velocity - final velocity) time interval

0.25 = (u -4)/5

u =  5.25 m/s.

Hence, initial velocity of the runner is 5.25 m/s.

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Answer:

the bird has 550J of potential energy

Explanation:

PEtotal=PE+KE

1000J=PE+450J

subtracting the kinetic energy from the total we get:

1000J-450J=PE

550J=PE

Answer:

"work against the force of the air in the tire"

The air in the tire provides a force opposing the force of the air provided by the plunger.

The extension of the spring when the ball is allowed to hang straight down, motionless is 0.0032 m.

The given parameters;

The radius of the circular path when spring is stretched is calculated as;

R = l₁ + x

R = 0.201 + 0.0176

R = 0.2186 m

The spring constant is calculated as follows;

[tex]F = ma\\\\Kx = \frac{mv^2}{R} \\\\K = \frac{mv^2}{Rx} \\\\K = \frac{(3.41)^2 m}{0.2186 \times 0.0176} \\\\K = 3,022.4 m \ N/m[/tex]

The extension of the spring when the ball is allowed to hang straight down, motionless;

[tex]F = mg\\\\Kx=mg\\\\x = \frac{mg}{K} \\\\x = \frac{9.8 m}{3022.4 m} \\\\x = 0.0032 \ m[/tex]

Thus, the extension of the spring when the ball is allowed to hang straight down, motionless is 0.0032 m.

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Answer:

Explanation:

Ax= 20*sin37

Ay =20*cos37

Bx=15*cos40

By =  -15*sin40

Cx = -6*cos60

Cy = -6*sin60

Answer:

Ax = 12 cos 53

Ay = 12 sin 53

Bx = 15 cos 320 = 15 cos 40

By = 15 sin 320 = -15 sin 40

Cx = 6 cos 240 = -6 cos 60

Cy = 6 sin 240 = -6 sin 60

Cuando el niño hace girar 0.7 vueltas por segundo una pelota de hule que se encuentra sujeta mediante un cordón de 0.75 m de longitud, tenemos que:

a) El periodo de la pelota es 1.43 segundos.

b) La velocidad lineal de la pelota es 3.3 m/s.

a) El periodo de la pelota está dado por:

[tex] T = \frac{2\pi}{\omega} [/tex]

En donde:

ω: es la velocidad angular

Dado que la pelota da 0.7 vueltas (revoluciones) cada segundo, la velocidad angular es:

[tex] \omega = \frac{0.7 \:rev}{s}*\frac{2\pi rad}{1 \:rev} = 4.40 rad/s [/tex]

Entonces, el periodo es:

[tex]T = \frac{2\pi}{\omega} = \frac{2\pi}{4.40 rad/s} = 1.43 s[/tex]

b) La velocidad lineal de la pelota se puede calcular usando la siguiente ecuación:

[tex] v = \omega r [/tex]

En donde:

r: es el radio de la circunferencia = longitud del cordón = 0.75 m

[tex]v = \omega r =4.40 rad/s*0.75 m = 3.3 m/s[/tex]

Por lo tanto, la velocidad lineal de la pelota es 3.3 m/s.

Puedes encontar mas aquí:

Espero que te sea de utilidad!

Explanation:

Uhhhh here is an example

a=vt. =27 m/s10 s. =2.7 m/s2.

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

Acceleration of the car is ~

[tex] \boxed{1.5 \: \: m/s}[/tex]

[tex] \large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}[/tex]

We know that,

Acceleration = rate of change in velocity ~

that is ~

where,

now, let's solve ~

안녕하세요

answer is 355 meters ...

Answer:

0.355m

Explanation:

To convert from cm to m, divide by 100

35.5÷100=0.355m

Answer:

Explanation:

The energy equation is E=hv

Hi there!

A)

The angle that will produce the same range is the compliment of 35°.

Thus, kicking the ball at 55° will result in the same range.

We can prove this by using the derived range equation:

[tex]R = \frac{v^2sin2\theta}{g}[/tex]

An angle of 35° yields:

[tex]R = \frac{v^2sin(2*35)}{g} = .939R[/tex]

An angle of 55° yields:

[tex]R = \frac{v^2sin(2*55)}{g} = .939R[/tex]

Both are the same, thus indicating that 55° produces the same range.

B)

The angle of 55° will produce the higher ball because the VERTICAL component of the ball's velocity is greater compared to kicking the ball at 35° degree.

sin(55) > sin(35)

C)

The angle of 55° will result in the ball being in the air the longest because when a ball is in the air (assuming no air resistance), the ball experiences an acceleration due to gravity of -9.8 m/s², causing the vertical velocity to decrease until it eventually reaches 0 m/s at the top of its path. A greater initial vertical velocity means that it will take longer for the ball to fall.

We can prove this using:

vf = vi + at

0 = vy - 9.8t

vy/9.8 = t

Greater vy (vertical component of velocity) ⇒ greater time taken.

D)

The angle that would result in the furthest range is 45°.

We can prove this using calculus. Recall the above range equation:

[tex]R = \frac{v^2sin2\theta}{g}[/tex]

We can take the derivative and use the first-derivative test to find its critical point:

[tex]\frac{dR}{d\theta} = \frac{v^22cos2\theta}{g} = 0[/tex]

Evaluate:

[tex]v^22cos2\theta = 0 \\\\cos2\theta = 0 \\\\2\theta = 90^o\\\\\boxed{\theta = 45^o}[/tex]

A. The angle that will produce the same range as a 35° kick is 53°.

B. The angle that produces the highest ball is 45°.

C. The angle that produces the ball that is in the air the longest is 45°.

D. There is no single angle that will produce the furthest range for all initial velocities and accelerations due to gravity.

A. The range of a projectile is the horizontal distance it travels before it hits the ground. The range of a projectile is determined by the initial velocity, the angle of projection, and the acceleration due to gravity.

For a given initial velocity, the range of a projectile is maximized when the angle of projection is 45°. However, if the ground is not level, the range of a projectile can be maximized at other angles.

In the case of a soccer ball kicked at an angle of 35°, the range will be maximized at an angle of 53°. This is because the range of a projectile is maximized when the vertical component of the initial velocity is equal to the horizontal component of the initial velocity.

The angle of 53° is the angle that produces a vertical component of the initial velocity that is equal to the horizontal component of the initial velocity when the ball is kicked at an angle of 35°.

Therefore, the angle that will produce the same range as a 35° kick is 53°.

B. The angle that produces the highest ball is 45°.

C. The angle that produces the ball that is in the air the longest is 45°.

D. The angle that produces the furthest range depends on the initial velocity and the acceleration due to gravity. There is no single angle that will produce the furthest range for all initial velocities and accelerations due to gravity.

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