In humans, which part of the brain contains 80 percent of its weight? cerebrum cerebellum spiral cord brain stem

A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s$

$\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}$

$\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}$

$\mu_{k} = 0.704$

b) The speed of the block is determined by using the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = x\cdot \sqrt{\frac{k}{m} }$

$v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }$

$v \approx 9.829\,\frac{m}{s}$

The radius of the circular loop is:

$\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}$

$\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}$

$R = 0.312\,m$

5 0

3 years ago

Airports at high elevations have longer runways for takeoffs and landings than do airports at sea level. One reason is that airc

Sati [7]

-- Airfoils (wings) develop less lift in less-dense air.
So the aircraft needs a longer take-off roll before it can lift off.
Also, its landing speed is greater, even with full flaps.

-- Air resistance (friction) is lower in less-dense air.
So the aircraft needs a longer landing roll before the spoilers
can reduce its speed sufficiently. (Brakes can't do it all.)

3 0

2 years ago

A 3 V battery is connected to a device of resistance 150mΩ. The current drawn by the device from this supply is------ A

Ket [755]

Answer:

20amperes

Explanation:

V= current × resistance

current= voltage ÷ resistance= 3v÷ (150×10^-3)= 20amps

4 0

2 years ago

A child is bouncing an a trampoline. The child has a mass of 30.5 kg . She leaves the surface of the trampoline with an initial

WINSTONCH [101]

Answer:

v = 60.7 m / s

Explanation:

This is an exercise that we can solve using the moment equations.

To begin we must create a system that is formed by the child and the block. For this system the forces during the crash are internal, for local moment it is conserved. Let's write the moment before and after the crash

Before

p₀ = m v₀ + 0

After

$p_{f}$ = (m + M) v

Where m and M are the masses of the child and the block, respectively. Notice that when they collide they are joined by which the mass is the sum of the two

p₀ = pf

m v₀ = (m + M) v

v = v₀ m / (m + M) (1)

We must find the speed of the child at 1.60 m, for this we use the law of conservation of energy, calculate the energy at two points: where it jumps from the trampoline and just before grabs the block

When he jumps off the trampoline

Em₁ = K = ½ m v₁²

Just before taking the block

Em₂ = K + U = ½ m v₂² + mg y

Em₁ = Em₂

½ m v₁² = ½ m v₂² + mg y

Let's calculate the speed with which it reaches the point of the crash

v₂² = v₁² - 2 g y

v₂ = √ (v₁² - 2 g y)

v₂ = √ (10.2² - 2 9.8 1.60)

v₂ = 72.7 m / s

This is the speed with which it reaches the point where the block is, vo = v₂ = 72.7 m/s, with this value we can substitute and calculate in equation (1) of conservation of the moment

v = vo m / (m + M)

v = 72.7 30.5 / (30.5 + 6.00)

v = 60.7 m / s

6 0

2 years ago

A set of charged plates is

cluponka [151]

Answer:

5.49×10¯⁴ m²

Explanation:

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

Distance (d) = 2.22×10¯⁴ m

Charge (Q) = 5.24×10¯⁹ C

Potential difference (V) = 240 V

Permittivity of free space (ε₀) = 8.85×10¯¹² F/m

Area (A) =?

Thus, the area of the plate can be obtained as follow:

Q = ε₀AV /d

5.24×10¯⁹ = 8.85×10¯¹² × A × 240 / 2.22×10¯⁴

5.24×10¯⁹ = 2.12×10¯⁹ × A / 2.22×10¯⁴

Cross multiply

2.12×10¯⁹ × A = 5.24×10¯⁹ × 2.22×10¯⁴

Divide both side by 2.12×10¯⁹

A = (5.24×10¯⁹ × 2.22×10¯⁴) / 2.12×10¯⁹

A = 5.49×10¯⁴ m²

Thus, the area of the plates is 5.49×10¯⁴ m².

8 0

2 years ago