All categories

2 years ago

6. How much voltage would be necessary to generate 100.5 amps of current in a circuit that

Physics

1 answer:

dlinn [17]2 years ago

7 0

Answer:

572.85 v

Explanation:

V = IR

V = 100.5 * 5.7 = 572.85 v

You might be interested in

Copernicus incorrectly assumed that the planets?

hammer [34]

Nicholas Copernicus correctly assumed that the planets revolved around the sun but he incorrectly assumed that the planets followed a perfect circle orbit around the sun. It was later on discovered by Johannes Kepler that the planets moved around the sun following an elliptical orbit.

6 0

3 years ago

¿A qué velocidad debe circular un auto de carreras para recorrer 87 km en 20 min? (pasar a metros por segundo m/s)

Alexxx [7]

Answer:

Velocidad en m / s = 72,25 m / s

Explanation:

Dado

Distancia a recorrer por el coche de carreras = 87 Km

1 km = 1000 m

Por lo tanto, 87 km = 87000 m

Tiempo necesario para viajar 87 km / 87000 metros = 20 minutos o 20 * 60 = 1200 segundos

Velocidad en m / s = 87000/1200

Velocidad en m / s = 72,25 m / s

7 0

3 years ago

I will gib brainlyest or whatever.

astraxan [27]

Answer:

Range of the projectile: approximately $1.06 \times 10^{3}\; {\rm m}$ .

Maximum height of the projectile: approximately $80\; {\rm m}$ (approximately $45.0\; {\rm m}$ above the top of the cliff.)

The projectile was in the air for approximately $7.07\; {\rm s}$ .

The speed of the projectile would be approximately $155\; {\rm m \cdot s^{-1}}$ right before landing.

(Assumptions: drag is negligible, and that $g = 9.81\; {\rm m\cdot s^{-1}}$ .)

Explanation:

If drag is negligible, the vertical acceleration of this projectile will be constantly $a_{y} = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}$ . The SUVAT equations will apply.

Let $\theta$ denote the initial angle of elevation of this projectile.

Initial velocity of the projectile:

vertical component: $u_{y} = u\, \sin(\theta) = 153\, \sin(11.2^{\circ}) \approx 29.71786\; {\rm m\cdot s^{-1}}$
horizontal component: $u_{x} = u\, \cos(\theta) = 153\, \cos(11.2^{\circ}) \approx 150.086\; {\rm m\cdot s^{-1}}$ .

Final vertical displacement of the projectile: $x_{y} = (-35)\; {\rm m}$ (the projectile landed $35\: {\rm m}$ below the top of the cliff.)

Apply the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ to find the final vertical velocity $v_{y}$ of this projectile:

${v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}$ .

$\begin{aligned} v_{y} &= -\sqrt{{u_{y}}^{2} + 2\, a_{y} \, x_{y}} \\ &= -\sqrt{(29.71786)^{2} + 2\, (-9.81)\, (-35)} \\ &\approx (-39.621)\; {\rm m\cdot s^{-1}}\end{aligned}$ .

(Negative since the projectile will be travelling downward towards the ground.)

Since drag is negligible, the horizontal velocity of this projectile will be a constant value. Thus, the final horizontal velocity of this projectile will be equal to the initial horizontal velocity: $v_{x} = u_{x}$ .

The overall final velocity of this projectile will be:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \sqrt{(150.086)^{2} + (-39.621)^{2}} \\ &\approx 155\; {\rm m\cdot s^{-1}} \end{aligned}$ .

Change in the vertical component of the velocity of this projectile:

$\begin{aligned} \Delta v_{y} &= v_{y} - u_{y} \\ &\approx (-39.621) - 29.71786 \\ &\approx 69.3386 \end{aligned}$ .

Divide the change in velocity by acceleration (rate of change in velocity) to find the time required to achieve such change:

$\begin{aligned}t &= \frac{\Delta v_{y}}{a_{y}} \\ &\approx \frac{69.3386}{(-9.81)} \\ &\approx 7.0682\; {\rm s}\end{aligned}$ .

Hence, the projectile would be in the air for approximately $7.07\; {\rm s}$ .

Also the horizontal velocity of this projectile is $u_{x} \approx 150.086\; {\rm m\cdot s^{-1}}$ throughout the flight, the range of this projectile will be:

$\begin{aligned}x_{x} &= u_{x}\, t \\ &\approx (150.086)\, (7.0682) \\ &\approx 1.06 \times 10^{3}\; {\rm m} \end{aligned}$ .

When this projectile is at maximum height, its vertical velocity will be $0$ . Apply the SUVAT equation $v^{2} - u^{2} = 2\, a\, x$ to find the maximum height of the projectile (relative to the top of the $35\; {\rm m}$ cliff.)

$\begin{aligned}x &= \frac{{v_{y}}^{2} - {u_{y}}^{2}}{2\, a} \\ &\approx \frac{0^{2} - 29.71786^{2}}{2\, (-9.81)} \\ &\approx 45.0\; {\rm m}\end{aligned}$ .

Thus, the maximum height of the projectile relative to the ground will be approximately $45.0\; {\rm m} + 35\; {\rm m} = 80\; {\rm m}$ .

5 0

1 year ago

okay so i already answered this question A. what is the density of an 84.7g sample of an unknown substance if the sample occupie

Lera25 [3.4K]

The density of the substance is obtained by dividing its mass by its volume. Density is an intensive property which means that its value does not depend on the amount of substance and will always stay the same for same conditions.
d = 84.7 g / 46.7 cm³ = 7.75 g / x cm³
The value of x is approximately 4.27 cm³.

7 0

3 years ago

(a) Without the wheels, a bicycle frame has a mass of 6.75 kg. Each of the wheels can be roughly modeled as a uniform solid disk

adelina 88 [10]

Answer:

a) Ktotal = 71.85 J

b) Ktotal = 71.85 J

Explanation:

a) The total kinetic energy is that of the total mass of the bicycle plus the rotational kinetic energy of the two wheels. The linear speed of the circumference of the wheels matches the forward speed of the bicycle, so their angular speed is

ω = v/r

The moment of inertia of one solid disk bicycle wheel is

I = 0.5*m₂*r²

And the rotational kinetic energy of one wheel is

Kr = 0.5*I*ω² = 0.5*(0.5*m₂*r²)*(v/r)² = 0.25*m₂*v²

The total kinetic energy is then that of the frame and wheels plus the rotational kinetic energy.

Ktotal = 0.5*(m₁ + 2*m₂)*v² + 2*(0.25*m₂*v²)

⇒ Ktotal = 0.5*v²*(m₁ + 3*m₂)

where

m₁ = 6.75 Kg

m₂ = 0.820 kg

v = 3.95 m/s

then

⇒ Ktotal = 0.5*(3.95 m/s)²*(6.75 Kg + 3*0.820 kg)

⇒ Ktotal = 71.85 J

b) We can apply the same equation obtained before

⇒ Ktotal = 0.5*v²*(m₁ + 3*m₂)

where

m₁ = 675 Kg

m₂ = 82.0 kg

v = 0.395 m/s

then

⇒ Ktotal = 0.5*(0.395 m/s)²*(675 Kg + 3*82 kg)

⇒ Ktotal = 71.85 J

3 0

3 years ago