A box is initially at rest on a frictionless inclined plane

Light travels through a vacuum at a speed of 3 x 10 m/s. What is the speed of

Andrew [12]

Answer:

v = c / n (n = 1 for air)

v = c / 1.33 = 3 * 10E8 m/s / 1.33 = 2.25 * 10E8 m/s

3 0

3 years ago

The highest barrier that a projectile can clear is 19.7 m, when the projectile is launched at an angle of 15.0o above the horizo

svlad2 [7]

Answer:

76 m /s

Explanation:

maximum height, H = 19.7 m

Angle of projection, θ = 15 degree

Let u be the projectile launch speed

Use the formula of maximum height

$H = u^{2}Sin\theta ^{2}/2g$

$19.7 = u^{2}Sin^{2}15/19.6$

u = 75.9199 = 76 m /s

6 0

3 years ago

For an object at rest, the normal force equals the gravitational force. What is the sum of these forces called?

Thepotemich [5.8K]

The sum of all forces on an object is called the "net" force on it.

For an object at rest, we call that sum of forces "zero".

8 0

3 years ago

Read 2 more answers

Una masa de 0,5 kg está sobre una pendiente inclinada 20º sujeta mediante una cuerda paralela a la pendiente que impide que desl

jeka94

Answer:

4.61 N

Explanation:

masa = 0.5 kg

ángulo de inclinación = 20°

Peso normal de la masa = mg

donde m = masa

g = aceleración debido a la gravedad = 9.81 m/s^2

Peso normal = 0.5 x 9.81 = 4.905 N

Si la masa se mantiene en su lugar mediante una cuerda paralela al plano, y no hay fricción en la masa, entonces

La fuerza sobre la cuerda = peso normal x cos ∅

donde ∅ = 20 °

La fuerza sobre la cuerda = 4.905 x cos 20°

==> 4.61 N

4 0

3 years ago

Points A and B lie within a region of space where there is a uniform electric field that has no x- or z-component; only the y-co

liraira [26]

Answer:

(a) Ey is negative

(b) The magnitude of the electric field is E = 171.429 V/m

(c) The potential difference between points B and C is 17.1429 V

Explanation:

(a) Here, we have the potentials given by;

$V_A - V_B = +12.0V$ with point A at y = 8.00 cm and point B at point y = 15.0 cm

where point B is at a higher potential than point A, that is the electric potential is from;

B with y = 15.0 cm to A with y = 8.0 cm which means

$E_y$ decreases as y increases or $E_y$ is negative.

(b) The magnitude of the electric field is given by

The work done to move a charge from B to A is

$W_{BA} = - \Delta U$ where

$\Delta U = U_a -U_b = q_0E(y_b-y_a)$

$V_{BA} = \frac{\Delta U}{q_0} = \frac{q_0E(y_b-y_a)}{q_0} = E(y_b-y_a)$

∴ $E = \frac{V_{BA}}{(y_b-y_a)}$

$E = \frac{12 \hspace{0.09cm}V}{(0.015\hspace{0.09cm} m -0.008\hspace{0.09cm} m)}$

E = 171.429 V/m

(c) Here we have point C x = 5.00 cm and y = 5.00 cm

Therefore we have the distance from B to C given by

$y_b-y_c = 15.00 \hspace{0.09cm}cm - 5.00 \hspace{0.09cm}cm = 10.00 \hspace{0.09cm} cm$

Where 10.00 cm = 0.01 m

E = V/Δy

Therefore, V = Δy·E

For $V_{BC}$ , Δy = $y_b-y_c = 0.01 \hspace{0.09cm} m$ and we have,

$V_{BC} = E\times (y_b-y_c)$

$V_{BC} = 171.429\times (0.015-0.005) = 17.1429\hspace{0.09cm}V$

7 0

3 years ago