A ray of sunlight is passing from diamond into crown glass; theangle of incidence is 31.00°.

A block of mass m1=3.7kg on a smooth inclined plane of angle 30 is connected by a cord over a small frictionless pulley to a sec

kari74 [83]

consider the forces on mass m₁ on the incline plane :

parallel to incline , force equation is given as

T - m₁ g Sin30 = m₁ a

T = m₁ g Sin30 + m₁ a eq-1

consider the force on mass m₂ on the incline plane :

m₂ g - T = m₂ a

T = m₂ g - m₂ a eq-2

Using eq-1 and eq-2

m₂ g - m₂ a = m₁ g Sin30 + m₁ a

inserting the values

(2.3 x 9.8) - 2.3 a = (3.7 x 9.8) Sin30 + 3.7 a

a = 0.74 m/s²

3 0

3 years ago

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-forme

Sindrei [870]

I need to know this answer

5 0

4 years ago

As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

MrMuchimi

Answer: $37^{\circ}$ North of west

Explanation:

Given

40,000-ton luxury line traveling 20 knots towards west and

60,000 ton freighter traveling towards North with 10 knots

suppose v is the common velocity after collision

conserving momentum in west direction

$20\times 40,000=(20,000+60,000)v\cos \theta$

suppose the final velocity makes \theta angle with x axis

$v\cos \theta =10----1$

Conserving Momentum in North direction

$10\times 60,000=80,000\times v\sin\theta$

$v\sin \theta =\frac{15}{2}----2$

divide 1 and 2

$\tan \theta =\frac{3}{4}$

$\theta =37^{\circ}$

so search in the area $37^{\circ}$ North of west to find the ship

3 0

3 years ago

A certain parallel-plate capacitor is filled with a dielectric for which Κ = 5.5 .The area of each plate is 0.034 m2 , and the p

Nesterboy [21]

Answer:

The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J

Explanation:

Given that,

dielectric constant k = 5.5

the area of each plate, A = 0.034 m²

separating distance, d = 2.0 mm = 2 x 10⁻³ m

magnitude of the electric field = 200 kN/C

Capacitance of the capacitor is calculated as follows;

$C = \frac{k \epsilon A}{d} = \frac{5.5*8.85*10^{-12}*0.034}{2*10^{-3}} = 8.275 *10^{-10} \ F$

Maximum potential difference:

V = E x d

V = 200000 x 2 x 10⁻³ = 400 V

Maximum energy that can be stored in the capacitor:

E = ¹/₂CV²

E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²

E = 6.62 x 10⁻⁵ J

Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J

4 0

3 years ago

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.

dalvyx [7]

Answer:

(a) $W_c=127.008 J$

(b) $W_g=148.176 J$

(c) K.E. = 21.168 J

(d) $v=3.4293m.s^{-1}$

Explanation:

Given:

mass of a block, M = 3.6 kg

initial velocity of the block, $u=0 m.s^{-1}$

constant downward acceleration, $a_d= \frac{g}{7}$

$\Rightarrow$ That a constant upward acceleration of $\frac{6g}{7}$ is applied in the presence of gravity.

∴ $a=- \frac{6g}{7}$

height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

$F_c= M\times a$

$F_c=3.6\times (-6)\times\frac{9.8}{7}$

$F_c=-30.24 N$

∴Work by the cord on the block,

$W_c= F_c\times d$

$W_c= -30.24\times 4.2$

We take -ve sign because the direction of force and the displacement are opposite to each other.

$W_c=-127.008 J$

(b)

Force on the block due to gravity:

$F_g= M.g$

∵the gravity is naturally a constant and we cannot change it

$F_g=3.6\times 9.8$

$F_g=35.28 N$

∴Work by the gravity on the block,

$W_g=F_g\times d$

$W_g=35.28\times 4.2$

$W_g=148.176 J$

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

$K.E.= W_g+W_c$

$K.E.=148.176-127.008$

K.E. = 21.168 J

(d)

From the equation of motion:

$v^2=u^2+2a_d\times d$

putting the respective values:

$v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }$

$v=3.4293m.s^{-1}$ is the speed when the block has fallen 4.2 meters.

6 0

3 years ago