In what direction does your brain see light as traveling when it reflects off an object?

A projectile is fired from a very powerful cannon vertically upward from Earth's surface at an initial speed of 5.5 km/s . Assum

spayn [35]

Answer:

Maximum altitude above the ground surface is 2053 km.

Explanation:

See written workings in attached picture.

6 0

4 years ago

A rocket moves upward, starting from rest with an acceleration of +29.4 for 3.98 s. it runs out of fuel at the end of the 3.98 s

topjm [15]

U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s

Let h₁ = the height at time t, for t ≤ 3.98 s
Let h₂ = the height at time t > 3.98 s

Motion for t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s

Motion for t > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m

The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m

Answer: 931.4 m (nearest tenth)

6 0

4 years ago

Read 2 more answers

A block of mass 3 kg slides along a horizontal surface that has negligible friction except for one section, as shown above. The

ivanzaharov [21]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is $F_f = 3 * -4$

Explanation:

From the question we are told that

The mass of the block is $m = 3 \ kg$

The initial speed of the block before entering the rough section is $u = 5 \ m/s$

The time taken is t = 0.5 s

The speed with which the block leaves the rough section is $v = 3 \ m/s$

Generally from kinematic equation

$v = u + at$

=> $a = \frac{v - u }{t}$

=> $a = \frac{3 - 5 }{0.5}$

=> $a = - 4$

Generally the magnitude of the average frictional force is mathematically represented as

$F_f = m * a$

=> $F_f = 3 * -4$

=> $F_f = -12 \ N$

The negative sign show that the force is acting in the opposite direction to the direction of motion of the block

3 0

3 years ago

The average standard rectangular building brick has a mass of 3.10 kg and dimensions of 225 m x 112 m x 75 m. The gravitational

zzz [600]

Answer:

$P=3.61\times 10^{-3}\ Pa$

Explanation:

Given that,

Mass of a brick, m = 3.1 kg

The dimensions of the brick 225 m x 112 m x 75 m

We need to find the maximum pressure created by the brick. We know that, the force acting per unit area is called pressure exerted. It is given by the formula as follows :

$P=\dfrac{F}{A}$

F = mg

A = area with minimum dimensions i.e. 112 m x 75 m

Pressure is maximum when the area is least.

So,

$P=\dfrac{mg}{A}\\\\P=\dfrac{3.1\times 9.8}{112\times 75}\\\\P=3.61\times 10^{-3}\ Pa$

So, the maximum pressure created by brick is $3.61\times 10^{-3}\ Pa$ .

5 0

3 years ago

Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspe

ser-zykov [4K]

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

7 0

3 years ago