A certain x-ray tube requires a current of 7 mA at a voltage of 80 kV. The rate of energy dissipation is:

Write the velocity vectors in component form.

• initial velocity:

v₁ = 4 m/s at 45º N of E

v₁ = (4 m/s) (cos(45º) i + sin(45º) j)

v₁ ≈ (2.83 m/s) i + (2.83 m/s) j

• final velocity:

v₂ = 4 m/s at 10º N of E

v₂ = (4 m/s) (cos(10º) i + sin(10º) j)

v₂ ≈ (3.94 m/s) i + (0.695 m/s) j

The average acceleration over this 3-second interval is then

a = (v₂ - v₁) / (3 s)

a ≈ (0.370 m/s²) + (-0.711 m/s²)

with magnitude

||a|| = √[(0.370 m/s²)² + (-0.711 m/s²)²] ≈ 0.802 m/s²

and direction θ such that

tan(θ) = (-0.711 m/s²) / (0.370 m/s²) ≈ -1.92

→   θ ≈ -62.5º

which corresponds to an angle of about 62.5º S of E, or 27.5º E of S. To use the notation in the question, you could say it's E 62.5º S or S 27.5º E.

The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is .

Given :

• Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope.
• The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle.
• The horizontal surface on which block A (mass 2.10 kg) moves is frictionless.
• The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s.

a) First, determine the acceleration of the B block.

Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.

b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.

c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.

Now, substitute the values of the known terms in the above expression.

For more information, refer to the link given below:

brainly.com/question/2287912

Answer:

(a) 62.3 N

(b) 1.89 N

(c) 0.430 kg m²

Explanation:

(a) Find the acceleration of block B.

Δy = v₀ t + ½ at²

1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²

a = 0.90 m/s²

Draw a free body diagram of block B.  There are two forces:

Weight force mg pulling down,

and tension force Tb pulling up.

Sum of forces in the -y direction:

∑F = ma

mg − Tb = ma

Tb = m (g − a)

Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)

Tb = 62.3 N

(b) Draw a free body diagram of block A.  There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and tension force Ta pulling right.

Sum of forces in the +x direction:

∑F = ma

Ta = ma

Ta = (2.10 kg) (0.90 m/s²)

Ta = 1.89 N

(c) Draw a free body diagram of the pulley.  There are two forces:

Tension force Tb pulling down,

and tension force Ta pulling left.

Sum of torques in the clockwise direction:

∑τ = Iα

Tb r − Ta r = Iα

(Tb − Ta) r = I (a/r)

I = (Tb − Ta) r² / a

I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)

I = 0.430 kg m²

Answer:

67.5 cm

Explanation:

u = - 90 cm, v = 3 x u = 3 x 90 = 270 cm

let f be the focal length

Use lens equation

1 / f = 1 / v - 1 / u

1 / f = 1 / 270 + 1 / 90

1 / f = 4 / 270

f = 67.5 cm

Final answer:

To determine the focal length of the lens, we use the lens formula and set up an equation based on the given information. Solving for the image distance, we find that it is zero, indicating the image is formed at infinity. Therefore, the focal length of the lens is 90 cm.

Explanation:

To determine the focal length of the lens, we can use the lens formula:

1/f = 1/v - 1/u

Where f is the focal length, v is the image distance, and u is the object distance.

Given that the image distance triples when the object is moved from infinity to 90 cm in front of the lens, we can set up the following equation:

1/f = 1/(3v) - 1/(90)

Multiplying through by 90*3v, we get:

90*3v/f = 270v - 90*3v

90*3v/f = 270v - 270v

90*3v/f = 0

Simplifying further, we find that: v = 0

When the image distance is zero, it means the image is formed at infinity, so the lens is focused at the focal point. Therefore, the focal length of the lens is 90 cm.

Answer:

A. 29.7 m/s

B. 6.06 s

Explanation:

From the question given above, the following data were obtained:

Maximum height (h) = 45 m

A. Determination of the initial velocity (u)

Maximum height (h) = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 m/s (at maximum height)

Initial velocity (u) =.?

v² = u² – 2gh (since the ball is going against gravity)

0² = u² – (2 × 9.8 × 45)

0 = u² – 882

Collect like terms

0 + 882 = u²

882 = u²

Take the square root of both side

u = √882

u = 29.7 m/s

Therefore, the ball must be thrown with a speed of 29.7 m/s.

B. Determination of the time spent by the ball in the air.

We'll begin by calculating the time take to reach the maximum height. This can be obtained as follow:

Maximum height (h) = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) to reach the maximum height =?

h = ½gt²

45 = ½ × 9. 8 × t²

45 = 4.9 × t²

Divide both side by 4.9

t² = 45/4.9

Take the square root of both side

t = √(45/4.9)

t = 3.03 s

Finally, we shall determine the time spent by the ball in the air. This can be obtained as follow:

Time (t) to reach the maximum height = 3.03 s

Time (T) spent by the ball in the air =?

T = 2t

T = 2 × 3.03

T = 6.06 s

Therefore, the ball spent 6.06 s in the air.

Answer: positive

Explanation:

Gravity can be defined as the force with which the body is attracted towards the center of the earth, or towards any other body. If the force acting on the body is in the direction of displacement then the word done by the applicable force is positive. This causes the free fall of the ball under the influence of gravity is also positive.